(ii) int cos x cos2x cos3xdx

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(ii) int cos x cos2x cos3xdx

I took a shot in the dark and assumed that this is similar to solving $\int e^{x}\sin{x}\ dx$, but wolfram is giving me a different answer than what I got, and on top of that, I tried to differentiate my result and am not getting back what I started with. It's putting into question whether I was doing previous questions right or not.. First step of my

attempt: let $u=\cos(2x),\ du=-2\sin(2x)\ dx$ let $dv=\cos(3x)\ dx,\ v=\frac{\sin(3x)}{3}$ $$\int\cos(2x)\cos(3x)\ dx=\frac{\cos(2x)\sin(3x)}{3}+\frac{2}{3}\int\sin(2x)\sin(3x)\ dx $$ Then I did IBP again: let $u=\sin(2x),\ du=2\cos(2x)\ dx$ let $dv=\sin(3x)\ dx, v=-\frac{cos(3x)}{3}$ $$=\frac{\cos(2x)\sin(3x)}{3}+\frac{2}{3}\left[-

\frac{\cos(3x)\sin(2x)}{3}+\frac{2}{3}\int\cos(2x)\cos(3x)\ dx\right]$$ From there, I simplify and re-arrange to get $$\frac{1}{3}\int\cos(2x)\cos(3x)\ dx=\frac{3\cos(2x)\sin(3x)-2\cos(3x)\sin(2x)}{9}$$ $$\int\cos(2x)\cos(3x)\ dx=\frac{3\cos(2x)\sin(3x)-2\cos(3x)\sin(2x)}{3}+C$$ So where did I go wrong? Wolfram says the answer should be

$$\int\cos(2x)\cos(3x)\ dx=\frac{1}{10}5\sin(x)+\sin(5x)+C$$ Gerd Altmann/Pixabay If you're trying to figure out what x squared plus x squared equals, you may wonder why there are letters in a math problem. That's because, in the case of an equation like this, x can be whatever you want it to be. To find out what x squared plus x squared equals,

you have to multiply x times itself. Then you add that number to itself to get your final answer.Examples of X Squared Plus X Squared Here are some examples of that equation to make it easier to understand. If x equals 2, then x squared, or x times itself, equals 4. Add four to itself, and you get 8. Therefore, 2 squared plus 2 squared equals 8. To use

another example, let's see what happens when x equals 3. In that case, x squared equals 9. Then, 9 plus 9 equals 18. The beauty of this equation is that x can equal anything, and you can solve it using whatever value you want for x. Math that Uses Letters We call mathematics that uses letters to take the place of different values algebra. Algebra uses

symbols in most cases, letters to represent quantities that don't necessarily have the same value all the time. These quantities are called variables, and you can figure out what those variables mean when you use algebra. Equations are like sentences that explain the relationships between numbers and variables. You figure out what the variables

in an equation are by solving it. When you solve an equation in algebra, you break it down to its simplest form and discover what the variables mean. A Brief History of Algebra Since ancient times, mathematicians have worked with unknown variables in different ways. Islamic scholars began to give the science of working with variables a name. They

called this type of math the "science of restoration and balancing," and the Arabic word for "restoration," or "al-jabru," became the root word for the word "algebra." As mathematicians in the Middle Ages experimented with the principles of algebra, they realized they could solve equations for two- and three-dimensional items, which led to even more

discoveries of what algebra could do. Modern scholars have found even more complex equations that algebra can solve. Algebra in Everyday Life You may have heard people say that you'll never use algebra in your everyday life, but you'd be surprised at how often you use algebra. Algebra comes in handy when you're trying to figure out how much a

group of items costs per item. When you're trying to figure out how to split a restaurant bill or how much gas you can buy for a certain amount. You can use algebra to figure out the dimensions of a room or even as you make up your shopping list. Algebra is a versatile form of math that you use more often than you might think and, sometimes, you

don't even realize that you're solving math problems. Why It's Important to Learn Algebra Learning algebra is important for more than just solving equations. Educators consider algebra the gateway to higher forms of math, so if you or your child wants to explore a career in science or technology, algebra can unlock so many more new ideas. Algebra

can also help students with critical thinking and logic skills. Using algebra is like exercise that helps make your brain stronger. Putting algebra to use in your everyday life can help you in so many ways. MORE FROM Updated: 05/02/2021 by Computer Hope X may refer to any of the following: 1. In general, an x is used to represent

a generic variable. For example, you may see Computer Hope and other companies write Microsoft Windows 3.x. In this case, the "x" can be replaced with any valid digit. For instance, it may represent the "1" in Windows 3.1 or the "11" in Windows 3.11. 2. When referring to a computer CD drive, an X refers to the transfer speed. For example, the

original 1X CD-ROM had a speed of 153,600 BPS, and a 24X CD-ROM has a BPS of 3,686,400 or 153,600 x 24. The higher the number, the faster that data is going to be read from the CD drive. See the CD-ROM definition for further information about transfer speeds. What does the x stand for on 32x? CD-ROM help and support. 3. Formerly Google X,

X is a subsidiary of the Alphabet conglomerate. See the X company page for further information. 4. In mathematics, an x may be used to represent a times sign. For example, 4 x 4 = 16. The times sign may also be represented as an interpunct (?) or an asterisk (*). See our multiply page for further information. 5. When shown in the top-right corner of

a window, the X is a button used to close a window. 6. X is a service found on some IRC networks that allows users to manage the chat rooms, called channels, that they have access to view. 7. Abbreviation for X Window System. See the x command page for further information about this command. 8. X is a key used with the keyboard shortcuts Alt+X,

Command+X, and Ctrl+X. 9. In chat, 'x' or 'X' is a way of representing a kiss. For example, "xoxo" is kiss, hug, kiss, and hug. 10. With regular expressions, "x" is a regular expression flag that allows spaces and comments. 11. With Microsoft Excel and other spreadsheet programs, "X" is the twenty-fourth column of a spreadsheet. To reference the first

cell in the column, you'd use "X1." 12. In the phonetic alphabet, "X" is often pronounced as "x-ray." 13. X is the twenty-fourth letter of the English alphabet. The letter "X" comes after "W" and is followed by the letter "Y." To create a capitalized "X" press the Shift key and X at the same time. On a U.S. QWERTY keyboard, the "X" key is on the bottom

row, to the right of "Z" and left of the "C" key. See our keyboard page for a visual example of all keyboard keys. Tip In ASCII the uppercase "X" is "088" in decimal (01011000 in binary). The lowercase "x" is "120" in decimal (01111000 in binary). Related information CD terms, Chat terms, Letter, x86 \[\int\text{ cos x cos 2x cos 3x dx}\] \[\int\text{

cos x . cos 2x . cos 3x dx}\]\[ \Rightarrow \frac{1}{2}\int\left[ 2 \cos 2x \cdot \cos x \right] \text{ cos 3x dx}\]\[ \Rightarrow \frac{1}{2}\int\left[ \text{ cos } \left( 2x + x \right) + \text{ cos } \left( 2x - x \right) \right] \text{ cos 3x dx} ..............\left[ \because 2\text{ cos }A\text{ cos B }= \cos \left( A + B \right) + \text{ cos }\left( A - B \right)

\right]\]\[ \Rightarrow \frac{1}{2}\int\left( \cos3x + \cos x \right) \text{ cos 3x dx }\]\[ \Rightarrow \frac{1}{2}\int \text{ cos }^2 \text{ 3x dx} + \frac{1}{2}\int\text{ cos } 3x \cdot \text{ cos x dx}\]\[ \Rightarrow \frac{1}{2}\int\left( \frac{1 + \text{ cos }6x}{2} \right)dx + \frac{1}{4}\int2 \text{ cos 3x} \cdot \text{ cos x dx} ...................\left[

\because \cos 2x = \cos^2 x - 1 \right]\]\[ \Rightarrow \frac{1}{4}\left[ x + \frac{\sin 6x}{6} \right] + \frac{1}{4}\int\left( \cos 4x + \cos 2x \right)dx\]\[ \Rightarrow \frac{1}{4}\left[ x + \frac{\sin 6x}{6} \right] + \frac{1}{4}\left[ \frac{\sin 4x}{4} + \frac{\sin 2x}{2} \right] + C\]\[ \Rightarrow \frac{x}{4} + \frac{\sin 6x}{24} + \frac{\sin 4x}{16}

+ \frac{\sin 2x}{8} + C\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 2 \[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\] \[\text{ Let I }= \int\left( \frac{\sin x + \cos x}{\sqrt{\sin 2 x}} \right)dx\]\[\text{ Putting sin x - cos x = t }\]\[ \Rightarrow \left( \cos x + \sin x \right)dx = dt\]\[\text{ Also} \left(

\text{ sin x} - \cos x \right)^2 = t^2 \]\[ \Rightarrow \sin^2 x + \cos^2 x - 2 \sin x \cos x = t^2 \]\[ \Rightarrow 1 - t^2 = \text{ sin }\left( 2x \right)\]\[ \therefore I = \int\frac{dt}{\sqrt{1 - t^2}}\]\[ = \sin^{- 1} t + C \left( \int\frac{dt}{\sqrt{a^2 - x^2}} = \sin^{- 1} \frac{x}{a} + C \right)\]` = \text{ sin}^{- 1} \text{ ( sin x - cos x }) + C

( t =

sin x - cos x ) ` Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 3 \[\int\frac{\sin x - \cos x}{\sqrt{\sin 2x}} \text{ dx }\] \[\text{ Let I } = \int\left( \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \right) dx\]\[\text{ Putting sin x + cos x = t}\]\[ \Rightarrow \left( \cos x - \sin x \right) dx = dt\]\[ \Rightarrow \left( \sin x - \cos x

\right) dx = - dt\]\[\text{ Also sin x + cos x = t}\]\[\text{ Squaring both sides,} \]\[ \left( \sin x + \cos x \right)^2 = t^2 \]\[ \Rightarrow \sin^2 x + \cos^2 x + 2 \sin x \cos x = t^2 \]\[ \Rightarrow 1 + \text{ sin 2x }= t^2 \]\[ \Rightarrow \text{ sin 2x} = t^2 - 1\]\[ \therefore I = \int\frac{- dt}{\sqrt{t^2 - 1}}\]\[ = - \text{ ln} \left| t + \sqrt{t^2 - 1}

\right| + C ..........\left( \because \int\frac{dt}{\sqrt{x^2 - a^2}} = \text{ ln}\left| x + \sqrt{x^2 - a^2} \right| + C \right)\]\[ = - \text{ ln} \left| \left( \sin x + \cos x \right) + \sqrt{\left( \sin x + \cos x \right)^2 - 1} \right| + C ..........\left( \because t = \sin x + \cos x \right)\]\[ = - \text{ ln }\left| \left( \sin x + \cos x \right) + \sqrt{\sin^2 x + \cos^2 x + 2

\sin \cos x - 1} \right| + C\]\[ = - \text{ ln }\left| \sin x + \cos x + \sqrt{\sin 2 x} \right| + C\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 4 \[\int\frac{1}{\text{ sin} \left( x - a \right) \text{ sin } \left( x - b \right)} \text{ dx }\] \[\int\frac{1}{\text{ sin} \left( x - a \right) \cdot \text{ sin}\left( x - b \right)}dx\]\[

= \frac{1}{\text{ sin}\left( b - a \right)}\int\frac{\text{ sin}\left( b - a \right)}{\text{ sin}\left( x - a \right) \cdot \text{ sin }\left( x - b \right)} \text{ dx }\]\[ = \frac{1}{\text{ sin}\left( b - a \right)}\int\frac{\text{ sin}\left[ \left( x - a \right) - \left( x - b \right) \right]}{\text{ sin}\left( x - a \right) \cdot \text{ sin}\left( x - b \right)} \text{ dx }\]\[ = \frac{1}

{\text{ sin }\left( b - a \right)}\int\frac{\text{ sin }\left( x - a \right) \cdot \cos \left( x - b \right) - \text{ cos} \left( x - a \right) \text{ sin}\left( x - b \right)}{\text{ sin}\left( x - a \right) \cdot \text{ sin}\left( x - b \right)} \text{ dx }\]\[ = \frac{1}{\text{ sin}\left( b - a \right)}\int\left[ \frac{\text{ sin}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)}

{\text{ sin}\left( x - a \right) \cdot \text{ sin}\left( x - b \right)} - \frac{\text{ cos}\left( x - a \right) \text{ sin}\left( x - b \right)}{\sin \left( x - a \right) \text{ sin}\left( x - b \right)} \right] \text{ dx }\]\[ = \frac{1}{\text{ sin}\left( b - a \right)}\int\left[ \text{ cot}\left( x - b \right) - \text{ cot}\left( x - a \right) \right] \text{ dx }\]\[ = \frac{1}{\text{

sin}\left( b - a \right)}\int\text{ cot}\left( x - b \right) dx - \int\text{ cot}\left( x - a \right) \text{ dx }\]\[ = \frac{1}{\text{ sin}\left( b - a \right)}\left[ \text{ ln }\left| \text{ sin}\left( x - b \right) \right| - \text{ ln} \left| \text{ sin}\left( x - a \right) \right| \right] + C\]\[ = \frac{1}{\text{ sin}\left( b - a \right)}\left[ \text{ ln }\left| \frac{\text{ sin}\left( x - b

\right)}{\text{ sin}\left( x - a \right)} \right| \right] + C\]\[ = \frac{- 1}{\text{ sin}\left( a - b \right)}\left[ \text{ ln}\left| \frac{\text{ sin}\left( x - b \right)}{\text{ sin}\left( x - a \right)} \right| \right] + C\]\[ = \frac{1}{\text{ sin}\left( a - b \right)} \text{ ln }\left| \frac{\text{ sin}\left( x - a \right)}{\sin \left( x - b \right)} \right| + C\] Concept: Indefinite

Integral Problems Is there an error in this question or solution? Page 5 \[\int\frac{1}{\text{ cos }\left( x - a \right) \text{ cos }\left( x - b \right)} \text{ dx }\] \[\int\frac{1}{\text{ cos } \left( x - a \right) \cdot \text{ cos} \left( x - b \right)}dx\]\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\frac{\text{ sin }\left( a - b \right)}{\text{ cos}\left( x - a \right)

\cdot \text{ cos}\left( x - b \right)} dx\]\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\frac{\text{ sin }\left[ \left( x - b \right) - \left( x - a \right) \right]}{\text{ cos}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)} dx\]\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\frac{\text{ sin }\left( x - b \right) \cdot \text{ cos}\left( x - a \right) - \text{ cos}\left( x -

b \right) \cdot \text{ sin }\left( x - a \right)}{\text{ cos}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)}\]\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\left[ \frac{\text{ sin }\left( x - b \right) \cdot \text{ cos}\left( x - a \right)}{\text{ cos}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)} - \frac{\text{ cos}\left( x - b \right) \cdot \text{ sin }\left( x -

a \right)}{\text{ cos}\left( x - a \right) \cdot \text{ cos}\left( x - b \right)} \right] dx\]\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\left[ \text{ tan }\left( x - b \right) - \text{ tan }\left( x - a \right) \right] dx\]\[ = \frac{1}{\text{ sin }\left( a - b \right)}\int\text{ tan }\left( x - b \right) dx - \int\text{ tan } \left( x - a \right) dx\]\[ = \frac{1}{\text{ sin }\left( a

- b \right)}\left[ \text{ ln }\left| \text{ sec }\left( x - b \right) \right| - \text{ ln } \left| \text{ sec }\left( x - a \right) \right| \right] + C\]\[ = \frac{1}{\text{ sin }\left( a - b \right)}\left[ \text{ ln }\left| \text{ cos }\left( x - a \right) \right| - \text{ ln }\left| \text{ cos}\left( x - b \right) \right| \right] + C\]\[ = \frac{1}{\text{ sin }\left( a - b \right)}\left[ \text{ ln

}\left| \frac{\text{ cos}\left( x - a \right)}{\text{ cos}\left( x - b \right)} \right| \right] + C\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 6 \[\int\frac{\sin x}{\sqrt{1 + \sin x}} dx\] \[\text{ We have ,} \]\[I = \int\frac{\sin x}{\sqrt{1 + \sin x}} \text{ dx }\]\[I = \int\frac{2 \sin\frac{x}{2}\cos\frac{x}{2}}

{\sqrt{\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + \text{ 2 }\sin\frac{x}{2}\cos\frac{x}{2}}} \text{ dx }\]\[I = \int\frac{2 \sin\frac{x}{2}\cos\frac{x}{2}}{\sqrt{\left( \sin\frac{x}{2} + \cos\frac{x}{2} \right)^2}} \text{ dx }\]\[I = \int\frac{2 \sin\frac{x}{2}\cos\frac{x}{2}}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{ dx }\]\[I = \int\frac{1 +

2\sin\frac{x}{2} \cos\frac{x}{2} - 1}{\sin x + \cos x} \text{ dx }\]\[I = \int\frac{\sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} + 2\sin\frac{x}{2} \cos\frac{x}{2} - 1}{\sin x + \cos x} \text{ dx }\]\[I = \int\frac{\left( \sin\frac{x}{2} + \cos\frac{x}{2} \right)^2 - 1}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{ dx }\]\[I = \int\frac{\left( \sin\frac{x}{2} +

\cos\frac{x}{2} \right)^2}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{ dx }- \int\frac{1}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{ dx }\]\[I = \int\left( \sin\frac{x}{2} + \cos\frac{x}{2} \right) dx - \int\frac{1}{\sin\frac{x}{2} + \cos\frac{x}{2}} \text{ dx }\]\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) + C_1 - \frac{1}

{\sqrt{2}}\int\frac{1}{\frac{1}{\sqrt{2}}\left( \sin\frac{x}{2} + \cos\frac{x}{2} \right)} \text{ dx }\]\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) + C_1 - \frac{1}{\sqrt{2}}\int\frac{1}{\sin\frac{x}{2} cos\frac{\pi}{4} + \cos\frac{x}{2} sin\frac{\pi}{4}} \text{ dx }\]\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) + C_1 -

\frac{1}{\sqrt{2}}\int\frac{1}{\sin\left( \frac{x}{2} + \frac{\pi}{4} \right)} \text{ dx }\]\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) + C_1 - \frac{1}{\sqrt{2}}\int \text{ cosec} \left( \frac{x}{2} + \frac{\pi}{4} \right) \text{ dx }\]\[I = 2\left( - \cos\frac{x}{2} + \sin\frac{x}{2} \right) - \sqrt{2}\text{ log}\left| \text{ tan}\left( \frac{x}

{4} + \frac{\pi}{8} \right) \right| + C\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 7 \[\int\frac{\sin x}{\cos 2x} \text{ dx }\] \[\text{ Let I} = \int\frac{\sin x}{\text{ cos 2 x}} dx\]\[ = \int\left( \frac{\sin x}{2 \cos^2 x - 1} \right) dx ...................\left[ \because \cos 2x = 2 \cos^2 x - 1 \right] \]\[\text{

Putting cos x = t}\]\[ \Rightarrow - \text{ sin x dx = dt}\]\[ \Rightarrow \text{ sin x dx = - dt} \]\[ \therefore I = \int\frac{- dt}{2 t^2 - 1}\]\[ = \frac{1}{2}\int\frac{- dt}{t^2 - \frac{1}{2}}\]\[ = \frac{- 1}{2}\int\frac{dt}{t^2 - \left( \frac{1}{\sqrt{2}} \right)^2}\]` = -1 / 2 ? 1/ 2 ? 1/1\sqrt2 In |{t -1/\sqrt2}/{t+1/\sqrt2 }| + C ..... [

{1}/{x^2 - a^2} = {1}/{2a}\text{ ln } | {x - a}/{x + a} | + C ] `\[ = - \frac{1}{2\sqrt{2}} \text{ ln} \left| \frac{\sqrt{2}t - 1}{\sqrt{2}t + 1} \right| + C\]\[ = - \frac{1}{2\sqrt{2}} \text{ ln }\left| \frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1} \right| + C ............\left[ \because t = \cos x \right]\]\[ = \frac{1}{2\sqrt{2}} \text{ ln} \left| \frac{\sqrt{2}

\cos x + 1}{\sqrt{2} \cos x - 1} \right| + C\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 8\[\text{ Let I } = \int \tan^3 x \text{ dx }\]\[ = \int\tan x \cdot \tan^2 x\text{ dx }\]\[ = \int\tan x \left( \sec^2 x - 1 \right)dx\]\[ = \int\tan x \cdot \sec^2 x \text{ dx} - \int\text{ tan x dx }\]\[\text{ Putting tan x }= t\

in\ the\ Ist\ integral\]\[ \Rightarrow \text{ sec}^2 \text{ x dx }= dt\]\[ \therefore I = \int t \cdot dt - \int\text{ tan x dx }\]\[ = \frac{t^2}{2} - \text{ ln }\left| \sec x \right| + C\]\[ = \frac{\tan^2 x}{2} - \text{ ln }\left| \sec x \right| + C .............\left[ \because t = \tan x \right]\] Concept: Indefinite Integral Problems Is there an error in this question or

solution? Page 9\[\text{ Let I } = \int \text{ tan}^4 \text{ x dx }\]\[ = \int \tan^2 x \cdot \tan^2 \text{ x dx}\]\[ = \int\left( \sec^2 x - 1 \right) \tan^2 \text{ x dx}\]\[ = \int \sec^2 x \cdot \tan^2\text{ x dx }- \int \tan^2 \text{ x dx}\]\[ = \int \tan^2 x \cdot \sec^2 x - \int\left( \sec^2 x - 1 \right) dx\]\[\text{ Putting tan x } = \text{ t in the Ist

integral}\]\[ \Rightarrow \sec^2 \text{ x dx } = dt\]\[ \therefore I = \int t^2 \cdot dt - \int\left( \sec^2 x - 1 \right) dx\]\[ = \frac{t^3}{3} - \tan x + x + C\]\[ = \frac{\tan^3 x}{3} - \text{ tan x + x + C }..............\left( \because t = \tan x \right)\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 10\[\text{ Let I }=

\int \tan^5 \text{ x dx }\]\[ = \int \tan^3 x \cdot \tan^2\text{ x dx }\]\[ = \int \tan^3 x \left( \sec^2 x - 1 \right) dx\]\[ = \int \tan^3 x \cdot \sec^2 \text{ x dx} - \int \tan^3 \text{ x dx}\]\[ = \int \tan^3 x \cdot \sec^2 \text{ x dx} - \int\tan x \cdot \tan^2 \text{ x dx} \]\[ = \int \tan^3 x \cdot \sec^2 \text{ x dx} - \int\tan x \cdot \left( \sec^2 x - 1 \right)

dx\]\[ = \int \tan^3 x \cdot \sec^2 x dx - \int\tan x \cdot \sec^2\text{ x dx} + \int\tan x dx\]\[\text{ Putting tan x = t in the Ist and IInd integral} . \]\[ \Rightarrow \sec^2\text{ x dx} = dt\]\[ \therefore I = \int t^3 \cdot dt - \int t \cdot dt + \int\text{ tan x dx }\]\[ = \frac{t^4}{4} - \frac{t^2}{2} + \text{ ln} \left| \text{ sec x} \right| + C\]\[ =

\frac{\tan^4 x}{4} - \frac{\tan^2 x}{2} + \text{ ln} \left| \text{ sec x }\right| + C \left[ \because t = \text{ tan x} \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 11\[\text{ Let I }= \int \cot^4 \text{ x dx}\]\[ = \int \cot^2 x \cdot \cot^2 \text{ x dx}\]\[ = \int \cot^2 x \cdot \left( \text{ cosec}^2 x - 1

\right) \text{ dx}\]\[ = \int \cot^2 x \cdot \text{ cosec }^2 \text{ x dx} - \int \cot^2 \text{ x dx}\]\[ = \int \cot^2 x \cdot\text {cosec}^2 \text{ x dx}- \int\left( \text{cosec}^2 x - 1 \right) \text{ dx}\]\[ \text{ Putting cot x = t in the Ist integral}\]\[ \Rightarrow - \text{ cosec}^2 \text{ x dx} = dt\]\[ \therefore I = - \int t^2 dt - \int\left(

\text{cosec}^2 x - 1 \right) \text{ dx}\]\[ = \frac{- t^3}{3} + \text{ cot x + x + C}\]\[ = \frac{- \cot^3 x}{3} + \text{ cot x + x + C}................\left[ \because t = \text{ cot x} \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 12\[\text{ Let I } = \int \cot^5 \text{ x dx }\]\[ = \int \cot^2 x \cdot

\cot^3\text{ x dx }\]\[ = \int\left(\text{cosec}^2 x - 1 \right) \cot^3 \text{ x dx }\]\[ = \int \cot^3 x \cdot \text{cosec}^2\text{ x dx } - \int \cot^3 \text{ x dx }\]\[ = \int \cot^3 x \cdot \text{cosec}^2 \text{ x dx }- \int\cot x \cdot \cot^2 \text{ x dx }\]\[ = \int \cot^3 x \cdot \text{cosec}^2\text{ x dx }- \int\cot x \left( {cosec}^2 x - 1 \right)\text{ dx

}\]\[ = \int \cot^3 x \cdot \text{cosec}^2 \text{ x dx } - \int\cot x \cdot \text{ cosec}^2\text{ x dx }+ \int\cot\text{ x dx }\]\[\text{ Putting cot x = t in the Ist and IInd integral}\]\[ \Rightarrow - \text{cosec}^2\text{ x dx }= dt\]\[ \Rightarrow \text{cosec}^2 \text{ x dx }= - dt\]\[ \therefore I = - \int t^3 dt + \int t \cdot dt + \int\cot\text{ x dx

}\]\[ = - \frac{t^4}{4} + \frac{t^2}{2} + \text{ ln }\left| \sin x \right| + C\]\[ = - \frac{\cot^4 x}{4} + \frac{\cot^2 x}{2} + \text{ ln }\left| \sin x \right| + C .........\left( \because t = \cot x \right)\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 13 \[\int\frac{x^2}{\left( x - 1 \right)^3} dx\] \[\int\frac{x^2}

{\left( x - 1 \right)^3}\text{ dx }\]\[ = \int\left[ \frac{x^2 - 1 + 1}{\left( x - 1 \right)^3} \right]\text{ dx }\]\[ = \int\left[ \frac{\left( x - 1 \right) \left( x + 1 \right)}{\left( x - 1 \right)^3} + \frac{1}{\left( x - 1 \right)^3} \right]\text{ dx }\]\[ = \int\left[ \frac{x + 1}{\left( x - 1 \right)^2} + \frac{1}{\left( x - 1 \right)^3} \right]\text{ dx }\]\[ = \int\left[

\frac{x - 1 + 2}{\left( x - 1 \right)^2} + \frac{1}{\left( x - 1 \right)^3} \right]\text{ dx }\]\[ = \int\left[ \frac{1}{\left( x - 1 \right)} + \frac{2}{\left( x - 1 \right)^2} + \frac{1}{\left( x - 1 \right)^3} \right]\text{ dx }\]\[ = \int\frac{1}{x - 1}\text{ dx }+ 2\int \left( x - 1 \right)^{- 2} \text{ dx }+ \int \left( x - 1 \right)^{- 3} \text{ dx }\]\[ = \text{ ln} \left|

x - 1 \right| + 2 \left[ \frac{\left( x - 1 \right)^{- 2 + 1}}{- 2 + 1} \right] + \left[ \frac{\left( x - 1 \right)^{- 3 + 1}}{- 3 + 1} \right] + C\]\[ = \text{ ln} \left| x - 1 \right| - \frac{2}{\left( x - 1 \right)} - \frac{\left( x - 1 \right)^{- 2}}{2} + C\]\[ = \text{ ln }\left| x - 1 \right| - \frac{2}{x - 1} - \frac{1}{2 \left( x - 1 \right)^2} + C\] Concept: Indefinite

Integral Problems Is there an error in this question or solution? Page 14 \[\int x\sqrt{2x + 3} \text{ dx }\] \[ \text{ Let I }= \int \text{ x}\sqrt{2x + 3} \text{ dx }\]\[ \text{ Putting 2x + 3 = t }\]\[ \Rightarrow x = \frac{t - 3}{2}\]\[ \Rightarrow 2dx = dt\]\[ \Rightarrow dx = \frac{dt}{2}\]\[ \therefore I = \frac{1}{2}\int\left( \frac{t - 3}{2} \right)

\sqrt{t} \text{ dt }\]\[ = \frac{1}{4}\int\left( t - 3 \right) \sqrt{t} \text{ dt}\]\[ = \frac{1}{4}\int\left( t^\frac{3}{2} - 3 t^\frac{1}{2} \right) \text{ dt }\]\[ = \frac{1}{4}\left[ \frac{t^\frac{3}{2} + 1}{\frac{3}{2} + 1} - 3 \frac{t^\frac{1}{2} + 1}{\frac{1}{2} + 1} \right] + C\]\[ = \frac{1}{4} \times \frac{2}{5} t^\frac{5}{2} - \frac{3}{4}

\times \frac{2}{3}\text t^\frac{3}{2} + C\]\[ = \frac{1}{10} \text{ t}^\frac{5}{2} - 2 t^\frac{3}{2} + C\]\[ = \frac{1}{10} \left( 2x + 3 \right)^\frac{5}{2} - \frac{1}{2} \left( 2x + 3 \right)^\frac{3}{2} + C .........\left[ \because t = 2x + 3 \right]\]\[ = \frac{1}{10} \left( 2x + 3 \right)^\frac{5}{2} - \frac{1}{2} \left( 2x + 3 \right)^\frac{3}{2} +

C\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 15 \[\int\frac{x^3}{\left( 1 + x^2 \right)^2} \text{ dx }\] \[\text{ Let I } = \int\frac{x^3}{\left( 1 + x^2 \right)^2}\text{ dx }\]\[ = \int\frac{x^2 \times x}{\left( 1 + x^2 \right)^2}\text{ dx }\]\[\text{ Putting 1 + x}^2 = t \]\[ \Rightarrow x^2 = t - 1\]\[

\Rightarrow 2x\text{ dx } = dt\]\[ \Rightarrow \text{ x dx }= \frac{dt}{2}\]\[ \therefore I = \frac{1}{2}\int\frac{\left( t - 1 \right)}{t^2}dt\]\[ = \frac{1}{2}\int\left( \frac{1}{t} - \frac{1}{t^2} \right)\text{ dt }\]\[ = \frac{1}{2}\int\frac{dt}{t} - \frac{1}{2}\int t^{- 2} \text{ dt }\]\[ = \frac{1}{2} \text{ ln} \left| t \right| - \frac{1}{2}\left[ \frac{t^{-

2 + 1}}{- 2 + 1} \right] + C\]\[ = \frac{1}{2} \text{ ln } \left| t \right| + \frac{1}{2t} + C\]\[ = \frac{1}{2} \text{ ln }\left| 1 + x^2 \right| + \frac{1}{2 \left( 1 + x^2 \right)} + C...... \left( \because t = 1 + x^2 \right)\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 16 \[\int x \sin^5 x^2 \cos x^2 dx\] \

[\text{ Let I} = \int x \cdot \sin^5 x^2 \cdot \cos x^2 \text{ dx }\]\[\text{ Putting sin x}^2 = t\]\[ \Rightarrow \text{ cos} \left( x^2 \right) \times 2x \text{ dx } = dt\]\[ \Rightarrow \text{ cos} \left( x^2 \right) \cdot x \text{ dx }= \frac{dt}{2}\]\[ \therefore I = \frac{1}{2}\int t^5 \cdot dt\]\[ = \frac{1}{2} \left[ \frac{t^6}{6} \right] + C\]\[ =

\frac{t^6}{12} + C\]\[ = \frac{\sin^6 x^2}{12} + C .....\left[ \because t = \sin x^2 \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 17\[\text{ Let I }= \int \sin^5 x \text{ dx }\]\[ = \int \sin^4 x \cdot \text{ sin x dx}\]\[ = \int \left( \sin^2 x \right)^2 \text{ sin x dx}\]\[ = \int \left( 1 - \cos^2 x \right)^2

\text{ sin x dx}\]\[ = \int\left( \cos^4 x - 2 \cos^2 x + 1 \right) \text{ sin x dx}\]\[\text{ Putting cos x = t}\]\[ \Rightarrow - \text{ sin x dx} = dt\]\[ \Rightarrow \text{ sin x dx} = - dt\]\[ \therefore I = - \int\left( t^4 - 2 t^2 + 1 \right) dt\]\[ = - \int t^4 dt + 2\int t^2 dt - \int dt\]\[ = \frac{- t^5}{5} + \frac{2 t^3}{3} - t + C\]\[ = \frac{- \cos^5 x}{5} +

\frac{2}{3} \text{ cos}^3 x - \cos x + C .......\left[ \because t = \cos x \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 18\[\text{ Let I }= \int \cos^5 x \text{ dx }\]\[ = \int \cos^4 x \cdot \text{ cos x dx}\] \[ = \int \left( \cos^2 x \right)^2 \text{ cos x dx} \]\[ = \int \left( 1 - \sin^2 x \right)^2 \text{ cos x

dx}\]\[\text{ Putting sin x = t}\]\[ \Rightarrow \text{ cos x dx} = dt\]\[ \therefore I = \int \left( 1 - t^2 \right)^2 \cdot dt\]\[ = \int\left( t^4 - 2 t^2 + 1 \right) dt\]\[ = \int t^4 \cdot dt - 2\int t^2 dt + \int dt\]\[ = \frac{t^5}{5} - 2 \times \frac{t^{2 + 1}}{2 + 1} + t + C\]\[ = \frac{t^5}{5} - \frac{2}{3} t^3 + t + C\]\[ = \frac{\sin^5 x}{5} - \frac{2}

{3} \sin^3 x + \sin x + C ........\left[ \because t = \sin x \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 19 \[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\] \[ \text{ Let I} = \int\sqrt{\sin x} \cdot \cos^3 \text{ x dx }\]\[ = \int\sqrt{\sin x} \cdot \left( \cos^2 x \right) \cdot \text{ cos x dx }\]\[ = \int\sqrt{\sin x}

\left( 1 - \sin^2 x \right) \cdot \text{ cos x dx}\]\[\text{ Putting sin x} = t\]\[ \Rightarrow \text{ cos x dx }= dt\]\[ \therefore I = \int\sqrt{t} \left( 1 - t^2 \right) \cdot dt\]\[ = \int t^\frac{1}{2} dt - \int t^\frac{1}{2} \cdot t^2 dt\]\[ = \int t^\frac{1}{2} dt - \int t^\frac{5}{2} dt\]\[ = \frac{t^\frac{3}{2}}{\frac{3}{2}} - \frac{t^\frac{7}{2}}

{\frac{7}{2}} + C\]\[ = \frac{2}{3} t^\frac{3}{2} - \frac{2}{7} t^\frac{7}{2} + C\]\[ = \frac{2}{3} \text{ sin }^\frac{3}{2} \text{ x }- \frac{2}{7} \text{ sin }^\frac{7}{2} \text{ x }+ C ..........\left[ \because t = \text{ sin x }\right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 20 \[\int\frac{\sin 2x}

{\sin^4 x + \cos^4 x} \text{ dx }\] \[\text{ Let I } = \int\frac{\sin 2x}{\sin^4 x + \cos^4 x}dx\]\[ = \int\frac{2 \text{ sin x }\cdot \text{ cos x dx}}{\sin^4 x + \cos^4 x}\]\[\text{Dividing numerator and denominator by} \cos^4 x\]\[ \Rightarrow \int\frac{2 \frac{\text{ sin x }\cdot \text{ cos x}}{\cos^4 x}dx}{1 + \tan^4 x}\]\[ \Rightarrow

\int\frac{2 \tan x \cdot \text{ sec}^2 x dx}{1 + \left( \tan^2 x \right)^2}\]\[\text{ Putting tan}^2 x = t\]\[ \Rightarrow 2 \tan x \cdot \text{ sec}^2 \text{ x dx}\]\[ \therefore I = \int\frac{dt}{1 + t^2}\]\[ = \tan^{- 1} t + C\]\[ = \tan^{- 1} \left( \text{ tan}^2 x \right) + C......... \left[ \because t = \tan {}^2 x \right]\] Concept: Indefinite Integral

Problems Is there an error in this question or solution? Page 21\[\text{ Let I }= \int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx}\] \[\text{ Putting x = a sec } \] \[ \Rightarrow \text{ dx = a sec tan \text{ d}} \] \[ \therefore I = \int\frac{a \sec\theta \tan \text{ d} }{\sqrt{a^2 \sec^2 \theta - a^2}}\] \[ = \int\frac{{a \sec\theta\tan \text{

d} }}{a \cdot \tan\theta}\] \[ = \int\sec\tan \text{ d} \] \[ = \text{ ln }\left| \sec\theta + \tan\theta \right| + C\] \[ = \text{ ln} \left| \sec\theta + \sqrt{\sec^2 \theta - 1} \right| + C\] \[ = \text{ ln }\left| \frac{x}{a} + \sqrt{\left( \frac{x}{a} \right)^2 - 1} \right| + C\] \[ = \text{ ln} \left| \frac{x + \sqrt{x^2 - a^2}}{a} \right| + C\] \[ = \text{ ln}

\left| x + \sqrt{x^2 - a^2} \right| - \text{ ln a} + C\] \[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| + C'\] \[\text{ where C' = C }- \text{ ln a }\]Page 22\[\text{ Let I } = \int\frac{dx}{\sqrt{x^2 - a^2}}\] \[\text{ Putting x} = a \tan \theta\] \[ \Rightarrow dx = a \sec^2 \text{ d }\] \[ \therefore I = \int\frac{a \cdot se c^2\text{ d }}

{\sqrt{a^2 \tan^2 \theta + a^2}}\] \[ = \int\frac{a \sec^2 \theta \cdot d\theta}{a\sqrt{1 + \tan^2 \theta}}\] \[ = \int\frac{\sec^2 \theta \cdot \text{ d }}{\sec\theta}\] \[ = \int\sec\theta \cdot d\theta\] \[ = \int\sec\theta \cdot d\theta\] \[ = \text{ ln } \left| \sec\theta + \tan\theta \right| + C\] \[ = \text{ ln }\left| \sec\theta + \sqrt{\sec^2 \theta - 1}

\right| + C\] \[ = \text{ ln }\left| \frac{x}{a} + \sqrt{\frac{x^2}{a^2} - 1} \right| + C\] \[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| - \ln a + C\] \[ = \text{ ln }\left| x + \sqrt{x^2 - a^2} \right| + C'\] \[\text{ where C' = C - ln a }\]Page 23 \[\int\frac{1}{4 x^2 + 4x + 5} dx\] \[\text{ Let I }= \int\frac{dx}{4 x^2 + 4x + 1 + 4}\]\[ = \int\frac{dx}

{\left( 2x \right)^2 + 2 \times 2x + 1 + 22}\]\[ = \int\frac{dx}{\left( 2x + 1 \right)^2 + 2^2}\]\[\text{ Putting }\left( 2x + 1 \right) = t\]\[ \Rightarrow 2 \text{ dx = dt }\]\[ \Rightarrow dx = \frac{dt}{2}\]\[ \therefore I = \frac{1}{2}\int\frac{dt}{t^2 + 2^2}\]\[ = \frac{1}{2} \times \frac{1}{2} \text{ tan}^{- 1} \left( \frac{t}{2} \right) + C\]\[ =

\frac{1}{4} \text{ tan}^{- 1} \left( \frac{2x + 1}{2} \right) + C ....................\left[ \because t = \left( 2x + 1 \right) \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 24 \[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\] \[\int\frac{1}{x^2 + 4x - 5}dx\]\[ = \int\frac{1}{x^2 + 4x + 4 - 4 - 5}dx\]\[ = \int\frac{1}

{x^2 + 4x + 4 - 3^2}dx\]\[ = \int\frac{1}{\left( x + 2 \right)^2 - 3^2}dx\]\[ = \frac{1}{2 \times 3} \text{ ln} \left| \frac{x + 2 - 3}{x + 2 + 3} \right| + C ................. \left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]\[ = \frac{1}{6} \text{ ln } \left| \frac{x - 1}{x + 5} \right| + C\] Concept:

Indefinite Integral Problems Is there an error in this question or solution? Page 25\[\text{ We have,} \]\[I = \int\frac{1}{1 - x - 4 x^2}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - - x^2 \frac{x}{4}}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - \left( x^2 + \frac{x}{4} \right)}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - \left\{ x^2 + + \left(

\frac{1}{8} \right)^2 - \left( \frac{1}{8} \right)^2 \frac{x}{4} \right\}}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - \left( x + \frac{1}{8} \right)^2 + \frac{1}{64}}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} + - \left( x + \frac{1}{8} \right)^2 \frac{1}{64}}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{16 + 1}{64} - \left( x + \frac{1}{8}

\right)^2}dx\] \[ = \frac{1}{4}\int\frac{1}{\left( \frac{\sqrt{17}}{8} \right)^2 - \left( x + \frac{1}{8} \right)^2}dx\]\[ = \frac{1}{4} \times \frac{1}{2 \times \frac{\sqrt{17}}{8}} \text{ ln }\left| \frac{\frac{\sqrt{17}}{8} + x + \frac{1}{8}}{\frac{\sqrt{17}}{8} - x - \frac{1}{8}} \right| + C .................\left[ \because \int\frac{1}{a^2 - x^2}dx =

\frac{1}{2a}\text{ ln }\left| \frac{a + x}{a - x} \right| + C \right]\]\[ = \frac{1}{\sqrt{17}} \text{ ln }\left| \frac{\frac{\sqrt{17} + 1}{8} + x}{\frac{\sqrt{17} - 1}{8} - x} \right| + C\]Page 26 \[\int\frac{1}{3 x^2 + 13x - 10} \text{ dx }\] \[\int\frac{1}{3 x^2 + 13x - 10}dx\]\[ = \frac{1}{3}\int\frac{1}{x^2 + \frac{13}{3}x - \frac{10}{3}}dx\]\[ =

\frac{1}{3}\int\frac{1}{x^2 + \frac{13 x}{3} + \left( \frac{13}{6} \right)^2 - \left( \frac{13}{6} \right)^2 - \frac{10}{3}}dx\]\[ = \frac{1}{3}\int\frac{1}{\left( x + \frac{13}{6} \right)^2 - \frac{169}{36} - \frac{10}{3}}dx\]\[ = \frac{1}{3}\int\frac{1}{\left( x + \frac{13}{6} \right)^2 - \frac{169 - 120}{36}}dx\]\[ = \frac{1}{3}\int\frac{1}

{\left( x + \frac{13}{6} \right)^2 - \left( \frac{17}{6} \right)^2}dx\]\[ = \frac{1}{3} \times \frac{1}{2 \times \frac{17}{6}} \text{ ln } \left| \frac{x + \frac{13}{6} - \frac{17}{6}}{x + \frac{13}{6} + \frac{17}{6}} \right| .............\left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]\[ =

\frac{1}{17} \text{ ln}\left| \frac{x - \frac{2}{3}}{x + 5} \right| + C\]\[ = \frac{1}{17} \text{ ln }\left| \frac{3x - 2}{3x + 15} \right| + C\] Concept: Indefinite Integral Problems Is there an error in this question or solution?

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