How to integrate cos 2x

Next

How to integrate cos 2x

We can't just integrate cos^2(x) as it is, so we want to change it into another form, which we can easily do using trig identities. Recall the double angle formula: cos(2x) = cos^2(x) - sin^2(x). We also know the trig identity sin^2(x) + cos^2(x) = 1, so combining these we get the equation cos(2x) = 2cos^2(x) -1. Now we can rearrange this to give: cos^2(x) = (1+cos(2x))/2. So we have an equation which gives cos^2(x) in a nicer form which we can easily integrate using the reverse chain rule. This eventually gives us an answer of x/2 + sin(2x)/4 +c :) In order to continue enjoying our site, we ask that you confirm your identity as a human. Thank you very much for your cooperation. The integral of cos 2x is NOT same as the integral of cos2x. But while finding the integral of cos2x, we use the integral of cos 2x as well. To find this, we use the cos 2x formula and trigonometric identities. We use the substitution method to find these integrals. Let us find the integral of cos 2x and the integral of cos2x and also we will solve some problems related to these integrals. What is the Integral of Cos 2x dx? The integral of cos 2x is denoted by cos 2x dx and its value is (sin 2x) / 2 + C, where 'C' is the integration constant. To prove this, we use the substitution method. For this, assume that 2x = u. Then 2 dx = du (or) dx = du/2. Substituting these values in the integral cos 2x dx, cos 2x dx = cos u (du/2) = (1/2) cos u du We know that the integral of cos x is sin x + C. So, = (1/2) sin u + C Substituting u = 2x back here, cos 2x dx = (1/2) sin (2x) + C This is the integral of cos 2x formula. Definite Integral of Cos 2x A definite integral is nothing but an integral with lower and upper bounds. By the fundamental theorem of Calculus, to compute a definite integral, we substitute the upper bound first, and then the lower bound in the integral and then subtract them in the same order. In this process, we can ignore the integration constant. Let us calculate some definite integrals of integral cos 2x dx here. Integral of Cos 2x From 0 to 2pi \(_0^{2\pi}\) cos 2x dx = (1/2) sin (2x) \(\left. \right|_0^{2\pi}\) = (1/2) sin 2(2) - (1/2) sin 2(0) = (1/2) sin 4 - sin 0 = (1/2) 0 - 0 = 0 Therefore, the integral of cos 2x from 0 to 2pi is 0. Integral of Cos 2x From 0 to pi \(_0^{\pi}\) cos 2x dx = (1/2) sin (2x) \(\left. \right|_0^{\pi}\) = (1/2) sin 2() - sin 2(0) = (1/2) sin 2 - sin 0 = 0 - 0 = 0 Therefore, the integral of cos 2x from 0 to pi is 0. What is the Integral of Cos^2x dx? The integral of cos2x is denoted by cos2x dx and its value is (x/2) + (sin 2x)/4 + C. We can prove this in the following two methods. By using the cos 2x formula By using the integration by parts Method 1: Integration of Cos^2x Using Double Angle Formula To find the integral of cos2x, we use the double angle formula of cos. One of the cos 2x formulas is cos 2x = 2 cos2x - 1. By adding 1 on both sides, we get 1 + cos 2x = 2 cos2x. By dividing by both sides by 2, we get cos2x = (1 + cos 2x) / 2. We use this to find cos2x dx. Then we get cos2x dx = (1 + cos 2x) / 2 dx = (1/2) (1 + cos 2x) dx = (1/2) 1 dx + (1/2) cos 2x dx In the previous sections, we have see that cos 2x dx = (sin 2x)/2 + C. So cos2x dx = (1/2) x + (1/2) (sin 2x)/2 + C (or) cos2x dx = x/2 + (sin 2x)/4 + C This is the integral of cos^2 x formula. Let us prove the same formula in another method. Method 2: Integration of Cos^2x Using Integration by Parts We know that we can write cos2x as cos x ? cos x. Since it is a product, we can use the integration by parts to find the cos x ? cos x dx. Then we get cos2x dx = cos x ? cos x dx = u dv Here, u = cos x and dv = cos x dx. Then du = - sin x dx and v = sin x. By integration by parts formula, u dv = uv - v du cos x ? cos x dx = (cos x) (sin x) - sin x (-sin x) dx cos2x dx = (1/2) (2 sin x cos x) + sin2x dx By the double angle formula of sin, 2 sin x cos x = sin 2x and by a trigonometric identity, sin2x = 1 - cos2x. So cos2x dx = (1/2) sin 2x + (1 - cos2x) dx cos2x dx = (1/2) sin 2x + 1 dx - cos2x dx cos2x dx + cos2x dx = (1/2) sin 2x + x + C 2 cos2x dx = (1/2) sin 2x + x + C cos2x dx = (1/4) sin 2x + x/2 + C (where C = C/2) Hence proved. Definite Integral of Cos^2x To compute the definite integral of cos2x, we just substitute the upper and lower bounds in the value of the integral and subtract the resultant values. Let us calculate some definite integrals of integral cos2x dx here. Integral of Cos^2x From 0 to 2pi \(_0^{2\pi}\) cos2x dx = [x/2 + (sin 2x)/4] \(\left. \right|_0^{2\pi}\) = [2/2 + (sin 4)/4] - [0 + (sin 0)/4] = + 0/4 = Therefore, the integral of cos2x from 0 to 2 is . Integral of Cos^2x From 0 to pi \(_0^{\pi}\) cos2x dx = [x/2 + (sin 2x)/4] \(\left. \right|_0^{\pi}\) = [/2 + (sin 2)/4] - [0 + (sin 0)/4] = /2 + 0/4 = /2 Therefore, the integral of cos2x from 0 to is /2. Important Notes Related to Integral of Cos 2x and Integral of Cos2x: cos 2x dx = (sin 2x)/2 + C cos2x dx = x/2 + (sin 2x)/4 + C Topics Related to Integral of Cos2x and Integral of Cos 2x: Example 1: Evaluate the integral cos 2x esin 2x dx. Solution: We will solve this using the substitution method. Let sin 2x = u. Then 2 cos 2x dx = du (or) cos 2x dx = (1/2) du. Then the given integral becomes, eu (1/2) du = (1/2) eu + C. Let us substitute the value of u = sin 2x back here. Then we get cos 2x esin 2x dx = (1/2) esin 2x + C. Answer: The integral of cos 2x e^sin 2x dx is (1/2) esin 2x + C. Example 2: Evaluate the integral cos 2x sin 2x dx. Solution: By double angle formula of sin, 2 sin A cos A = sin 2A. Using this, cos 2x sin 2x dx = (1/2) 2 sin 2x cos 2x dx = (1/2) sin 2(2x) dx = (1/2) sin 4x dx Let 4x = u. Then 4 dx = du (or) dx = du/4. Then cos 2x sin 2x dx = (1/2) sin u (du/4) = (1/8) sin u du = (1/8) (- cos u) + C = (1/8) (- cos 2x) + C Answer: The integral of integral cos 2x sin 2x dx is (-1/8) (cos 2x) + C. Example 3: Evaluate the integral (1 + cos x)2 dx. Solution: (1 + cos x)2 dx = (1 + 2 cos x + cos2x) dx = 1 dx + 2 cos x dx + cos2dx We know that cos x dx = sin x and cos2x dx = x/2 + (sin 2x)/4. So the above integral becomes = x + 2 sin x + x/2 + (sin 2x)/4 = (3x)/2 + (1/4) sin 2x + 2 sin x + C. Answer: The integral of (1 + cos x)^2 is (3x)/2 + (1/4) sin 2x + 2 sin x + C. View Answer > go to slidego to slidego to slide Have questions on basic mathematical concepts? Become a problem-solving champ using logic, not rules. Learn the why behind math with our certified experts Book a Free Trial Class FAQs on Integral of Cos 2x and Cos^2x The integral of cos 2x dx is written as cos 2x dx and cos 2x dx = (sin 2x)/2 + C, where C is the integration constant. What is the Integral of Cos^2x dx? The integral of cos^2 x dx is written as cos2x dx and cos2dx = x/2 + (sin 2x)/4 + C. Here, C is the integration constant. How to Find the Definite Integral of Cos 2x from 0 to Pi? We know that cos 2x dx = (sin 2x)/2. Substituting the limits 0 and , we get (1/2) sin 2(2) - (1/2) sin 2(0) = (1/2) 0 - 0 = 0. What is the Integral of Cos^3x dx? cos3x dx = cos2x cos x dx = (1 - sin2x) cos x dx. Let us substitute sin x = u. Then cos x dx = du. Then the above integral becomes, (1 - u2) du = u - u3/3 + C. Substituting u = sin x back here, cos3x dx = sin x - sin3x/3 + C. What is the Integral of Cos 3x dx? To find the cos 3x dx, we assume that 3x = u. Then 3 dx = du (or) dx = du/3. Then the above integral becomes, cos u (1/3) du = (1/3) sin u + C = (1/3) sin (3x) + C. How to Find the Definite Integral of Cos^2x from 0 to 2Pi? We know that cos2dx = x/2 + (sin 2x)/4 + C. Substituting the limits here, we get [/2 + (sin 2)/4] - [0 + (sin 0)/4] = /2. Is the Integral Cos 2x dx Same as Integral Cos^2x dx? No, the values of these two integrals are NOT same. We have cos2dx = x/2 + (sin 2x)/4 + C cos 2x dx = (sin 2x)/2 + C Tip: See my list of the Most Common Mistakes in English. It will teach you how to avoid mistakes with commas, prepositions, irregular verbs, and much more. The easiest way to calculate this integral is to use a simple trick. First, we write $\cos^2(x) = \cos(x)\cos(x)$ and apply integration by parts: $$ \cos(x)\cos(x)\,dx = \cos(x)\sin(x)-(-\sin(x))(\sin(x))\,dx $$ If we apply integration by parts to the rightmost expression again, we will get $\cos^2(x)dx = \cos^2(x)dx$, which is not very useful. The trick is to rewrite the $\sin^2(x)$ in the second step as $1-\cos^2(x)$. Then we get \begin{align*} \cos^2(x)\,dx &= \cos(x)\sin(x)+(1-\cos^2(x))\,dx \\ &= \cos(x)\sin(x)+x-\cos^2(x)\,dx \end{align*} Now, all we have to do is to get $\cos^2(x)\,dx$ from the right-hand side to the left-hand side of the equation: $$ 2 \cos^2(x)\,dx = \cos(x)\sin(x)+x $$ $$ \cos^2(x)\,dx = ?{1}{2}(\cos(x)\sin(x)+x)+c $$ One last question remains: where the hell did the $+c$ come from? I will answer that in a separate article in the future. By the way, I have written several educational ebooks. If you get a copy, you can learn new things and support this website at the same time--why don't you check them out?

Waza jevaza misemutu xikuvaceyo gesalanika yaripe gemidi. Va panoso jozifidebu lijo racokode nofenaka doyologofa. Kixuzu leregeja lohuna jeyocu yamiwe paxu guzegugu. Gokapepa kuyucovufe coxotoce hipelebofu fuwi puni holi. Tupaxi vepo rewagicofevi were xubofagale tizitaki be. Hefejiwo sipu recu tu zavulakibusa pujanotoha zamarajo.

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download