Integrate sqrt(cos x)sin^3x

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Integrate sqrt(cos x)sin^3x

$\begingroup$ I tried to solve $\int \sin^3(x)\sqrt{\cos(x)}\,dx$ by setting it equal to $$\int \cos^{1/2}(x)\left(1-\cos^2x\right)\sin(x)\,dx $$ and then making $u=\cos(x)$ and $du=-\sin(x)$. I ended up with $$\frac{-2\cos^{3/2}(x)}{3} + \frac{\cos^2(x)}{2} + C$$ but the book's answer is $$\frac{2\cos^{7/2}(x)}{7} + \frac{2\cos^{3/2}(x)}{3} + C$$ Could you give a hint as to what I'm doing wrong? Here's my full work. $\endgroup$ 3 by M. Bourne We can use the trigonometric identities that we learned earlier to simplify the integration process. The main identities are shown here for reference: `cos^2x+sin^2x=1` `1+tan^2x=sec^2x` `1+cot^2x=csc^2x` `2\ cos^2x=1+cos 2x` `2\ sin^2x=1-cos 2x` The process that we use involves using the trigonometric ratios to simplify the expression, or to get the expression into a form that can be integrated. Integrating a Product of Powers of Sine and Cosine - one power odd To integrate a product of powers of sine and cosine, we use `cos^2x+sin^2x=1` if at least one of the powers is odd. Example 1 Integrate: `int3\ cos^3x\ dx`. Answer `int3\ cos^3x\ dx` `=3int(cos^2x)cos x\ dx` `=3int(1-sin^2x)cos x\ dx` `=3int(cos x\ -sin^2x\ cos x\) dx` Letting `u=sin x`, and thus `du=cos x\ dx` gives: `=3[sin x+K_1-intu^2du]` `=3[sin x-(u^3)/3+K]` `=3[sin x-(sin^3x)/3+K]` Integrating a Product of Powers of Sine and Cosine powers even We use `2\ cos^2x=1+cos 2x` or `2\ sin^2x=1-cos 2x` if the power of `sin x` or `cos x` is even. Example 2 Integrate: `intcos^2\ 2x\ dx`. Answer `intcos^2 2x\ dx` Let `u=2x`, then `du=2\ dx` This gives: `intcos^2u(du)/2 =1/2intcos^2 u\ du` `=1/2int(1+cos 2u)/2du` `=1/4int1+cos 2u\ du` `=1/4[u+(sin 2u)/2]+K` `=1/4[2x+(sin 2(2x))/2]+K` `=x/2+(sin 4x)/8+K` Example 3 Integrate: `6intcot^3x\ dx`. Answer We have: `6intcot^3x\ dx =6int(cot^2 x)cot x\ dx` `=6int(csc^2x-1)cot x\ dx` `=6(intcsc^2 x\ cot x\ dx` `{:-intcot x\ dx)` Take the `intcsc^2x\ cot x\ dx` integral first: Let `u = cot x`, then `du = -csc^2x\ dx` So `intcsc^2x\ cot x\ dx =-intu\ du` `=-u^2/2+C` `=(cot^2x)/2+C` For the second integral, from our table, we have `intcot x\ dx=ln\ |sin x|+C` Returning to the main question: `6intcot^3x\ dx =6(intcsc^2x\ cot x\ dx-` `{:intcot x\ dx)` `=6(-(cot^2x)/2-ln\ |sin x|)+K` `=-3\ cot^2x-6\ ln\ |sin x|+K` Application - Root Mean Square Value The root mean square value of the function y with respect to x is given by: `y_("rms")=sqrt(1/T int_0^T y^2dx` where T is the period of y. (See Period of Sine and Cosine if you are not sure about this.) Effective Current A common use of this concept is effective current. This is the value of the direct current that would produce the same quantity of heat energy in the same time as a certain alternating current. It is used in the design of heaters. Example 4 Find the root mean square (rms) value of i = 3 + 2 cos t. Answer In this case, T = 2. So `i_("rms") =sqrt(1/Tint_0^Ti^2dt)` `=sqrt(1/(2pi)int_0^(2pi)(3+2\ cos t)^2dt)` Now `(3+2\ cos t)^2=9+12\ cos t+4\ cos^2 t` Since `cos 2t = 2 cos^2 t - 1`, it follows that `cos^2t=(cos 2t+1)/2`. So `(3+2\ cos t)^2 =9+12 cos t+4((cos 2t+1)/2)` `=9+12 cos t+2(cos 2t+1)` `=11+12\ cos t+2\ cos 2t` So `int_0^Ti^2dt =int_0^(2pi)(11+12\ cos t+2\ cos 2t)dt` `=[11t+12\ sin t+sin 2t]_0^(2pi)` `=22pi` And, finally: `i_("rms") =sqrt(1/Tint_0^Ti^2dt) ` `=sqrt((22pi)/(2pi))` `=sqrt11` `=3.317` This is a graph of our cosine current with the RMS effective current shown. Graph of `i(t)=3+2cos(t)`, with the RMS current indicated by the dashed magenta line. Example 5 For a current i given by i = i0 sin t, show that the root-mean-square of the current for one period is `(i_0)/sqrt2.` Answer In this case, `T=(2pi)/omega` So `i_("rms")=sqrt(1/T int_0^Ti^2dt)` `=sqrt(omega/(2pi)int_0^((2pi)/omega)(i_0\ sin omegat)^2dt)` `=sqrt((omega(i _0)^2)/(2pi)int_0^((2pi)/omega)sin^2omegat\ dt)` Let's just take the integral part first, and put `sin^2omegat=(1-cos 2 omegat)/2` `int_0^(2pi//omega)sin^2omegat\ dt =1/2int_0^(2pi//omega)(1-cos 2 omegat)dt` `=1/2[t-(sin 2omega t)/(2omega)]_0^(2pi//omega)` `=1/2[((2pi)/omega-0)-(0-0)]` `=pi/omega` So `i_("rms")=sqrt((omega(i_0)^2)/(2pi)int_0^((2pi)/omega)sin^2omegat\ dt)` `=sqrt{(omega (i_0)^2)/(2pi)pi/omega}` `=sqrt(((i_0)^2)/2)` `=(i_0)/sqrt2` This is what the question required us to show. Exercises Integrate each of the given functions: 1. `int_(pi//3)^(pi//2)sqrt(cos x)\ sin^3x\ dx` Answer `int_(pi//3)^(pi//2)sqrt(cos x)\ sin^3x\ dx` Write `sin^3x ` `= sin^2x\ sin x` ` = (1 - cos^2x)(sin x)` So, taking the indefinite case first: `sqrt(cos x) sin^3x` `=cos^(1//2)x(1-cos^2x)(sin x)` `=(cos^(1//2)x-cos^(5//2)x)(sin x)` Put `u = cos x` then `du = -sin x\ dx` So `intsqrt(cos x)\ sin^3x\ dx =int(cos^(1//2)x-cos^(5//2)x)(sin x)dx` `=-int(u^(1//2)-u^(5//2))du` `=-(2u^(3//2))/3+(2u^(7//2))/7+K` `=-(2\ cos^(3//2) x)/3+(2\ cos^(7//2)x)/7+K` So we have for the definite case: `int_(pi//3)^(pi//2)sqrt(cos x)\ sin^3x\ dx =[-(2 cos^(3//2)x)/3+(2 cos^(7//2)x)/7]_(pi//3)^(pi//2)` `=[(0+0)-(-0.23570+0.02525)]` `=0.2104` The solution for Exercise 1 represents the area under the curve `y(x)=sqrt(cos x)\ sin^3x\ dx` between `pi/3 Graph of `y(x)=sqrt(cos x)\ sin^3x`, indicating the area under the curve from `x=pi/3` to `x=pi/2`. Zooming out that graph shows it's periodic (it repeats itself) with period `2pi`. There are holes in the graph because of the `sqrt(cos x)` part (we can't have the square root of a negative number). Graph of `y(x)=sqrt(cos x)\ sin^3x dx`, zoomed out to see its periodic nature. 2. `int_0^1sin^2 4x\ dx` Answer `int_0^1sin^2 4x\ dx` We recall that `2\ sin^2theta=1-cos 2theta` In this case, if ` = 4x`, `2 sin^2 4x` `=1-cos [2(4x)]` `=1-cos 8x` So `sin^2 4x=(1-cos 8x)/2` `int_0^1sin^2 4x dx =1/2int_0^1(1-cos 8x) dx` `=1/2[x-(sin 8x)/8]_0^1` `=1/2[(1-0.9894/8)-(0-0)]` `=0.4382` Here is the graph of the integration we just found, indicating the area under the curve `y=sin^2 4x`. Graph of `y(x)=sin^2 4x`, indicating the area under the curve from `x=0` to `x=1`. 3. `intcot 4x\ csc^4 4x\ dx` Answer `int cot 4x\ csc^4 4x\ dx` We write the expression under the integral sign as follows: `cot 4x\ csc^4 4x\ dx` `=(csc^3 4x)\ cot 4x\ csc\ 4x` Then, let `u = csc\ 4x` and so we have `du = -4\ csc\ 4x\ cot 4x\ dx` That is, `-(du)/4 = csc\ 4x\ cot 4x\ dx` Now we can perform the integral: `int cot 4x\ csc^4 4x\ dx =int(csc^3 4x)\ cot 4x\ csc\ 4x\ dx` `=-1/4intu^3du` `=-u^4/16+K` `=-(csc^4 4x)/16+K` 4. `intsqrt(tan x)\ sec^4x\ dx` Answer Recall that `sec^2x=1+tan^2x` So we can write the part under the integral as: `sqrt(tan x)\ sec^4 x =(tan^(1//2) x)\ sec^2 x\ sec^2 x` `=(tan^(1//2) x )(1+tan^2 x )sec^2 x` `=(tan^(1//2) x+tan^(5//2)x)sec^2 x` Next, we put `u = tan x`, giving `du = sec^2x\ dx` So our integral becomes: `int (tan^(1//2) x+tan^(5//2)x)sec^2 x\ dx =int(u^(1//2)+u^(5//2))du` `=2/3u^(3//2)+2/7u^(7//2)+K` `=2/3tan^(3//2)x+2/7tan^(7//2)x+K` 5. `int_(pi//6)^(pi//3)(2dx)/(1+sin x` Answer `int_(pi//6)^(pi//3)(2\ dx)/(1+sin x)` We need to re-express this and we make use of a fairly common technique. We multiply the numerator (top) and denominator (bottom) of the fraction by the conjugate of the denominator. (The conjugate has the opposite sign in the middle. In this case, it's a minus sign.) We will make use of this important result: `sin^2 x+ cos^2 x = 1` `1/(1+sin x) =1/(1+sin x)xx(1-sin x)/(1-sin x)` `=(1-sin x)/(1-sin^2x)` `=(1-sin x)/(cos^2 x)` `=sec^2x-(sin x)/(cos^2 x)` Now we can integrate. Putting `u = cos x` in the right hand part, we have: `du = -sin x\ dx` So `int_(pi//6)^(pi//3)(2\ dx)/(1+sin x)` `=2int_(pi//6)^(pi//3) (sec^2x-(sin x)/(cos^2x))dx` `=2[tan x-1/(cos x)]_(pi//6)^(pi//3)` `=2[(1.73215-2)-` `{:(0.57735-1.15470)]` `=0.6190` Graph of `y(x)=2/(1+sin(x))`, indicating the area under the curve from `x=pi/6` to `x=pi/3`. Application - Length of a Curve The length s of the arc of a curve y = f(x) from x = a to x = b is given by: `s=int_a^bsqrt(1+((dy)/dx)^2dx` Find the length of the curve y = ln (cos x) from `x=0` to `x=pi/3`. Answer The curve in this problem is `y = ln cos x`. We need its derivative: `(dy)/(dx)=-(sin x)/(cos x)=-tan x` For this problem, we'll make use of an earlier result, `1+tan^2 x=sec^2 x` Applying the formula gives: `s=int_a^bsqrt(1+((dy)/dx)^2)dx` `=int_0^(pi//3)sqrt(1+(-tan x)^2)dx` `=int_0^(pi//3)sqrt(1+tan^2x) dx` `=int_0^(pi//3)sqrt(sec^2x) dx` `=int_0^(pi//3)sec x\ dx` `=[ln\ |sec x+tan x|]_0^(pi//3)` `=[ln\ |sec (pi/3)+tan (pi/3)|` `{:-|ln(1)+ln(0)|]` `=1.317` Here's the graph of the arc length we just found (in pink). I needed to take the absolute value of the `cos(x)` values, otherwise there would be gaps in the graph (whenever `cos(x)` was negative). Graph of `y(x)=ln|cos x|`, with the curve length we just found indicated in magenta (pink). Page 2 by M. Bourne Sometimes we meet an integration that is the product of 2 functions. We may be able to integrate such products by using Integration by Parts. If u and v are functions of x, the product rule for differentiation that we met earlier gives us: `d/(dx)(uv)=u(dv)/(dx)+v(du)/(dx)` Rearranging, we have: `u(dv)/(dx)=d/(dx)(uv)-v(du)/(dx)` Integrating throughout with respect to x, we obtain the formula for integration by parts: This formula allows us to turn a complicated integral into more simple ones. We must make sure we choose u and dv carefully. NOTE: The function u is chosen so that `(du)/(dx)` is simpler than u. Priorities for Choosing u When you have a mix of functions in the expression to be integrated, use the following for your choice of `u`, in order. 1. Let `u = ln x` 2. Let `u = x^n` 3. Let `u = e^(nx)` Example 1 `intx\ sin 2x\ dx` Solution We need to choose `u`. In this question we don't have any of the functions suggested in the "priorities" list above. We could let `u = x` or `u = sin 2x`, but usually only one of them will work. In general, we choose the one that allows `(du)/(dx)` to be of a simpler form than u. So for this example, we choose u = x and so `dv` will be the "rest" of the integral, dv = sin 2x dx. We have `u = x` so `du = dx`. Also `dv = sin 2x\ dx` and integrating gives: `v=intsin 2x\ dx` `=(-cos 2x)/2` Substituting these 4 expressions into the integration by parts formula, we get (using color-coding so it's easier to see where things come from): `int \color{green}{\underbrace{u}}\ \ \ \color{red}{\underbrace{dv}}\ \ ` ` =\ \ \color{green} {\underbrace{u}}\ \ \ \color{blue}{\underbrace{v}} \ \ -\ \ int \color{blue}{\underbrace{v}}\ \ \color{magenta}{\underbrace{du}}` `int \color{green}{\fbox{:x:}}\ \color{red}{\fbox{:sin 2x dx:}} = \color{green}{\fbox{:x:}}\ \color{blue}{\fbox{:{-cos2x}/2:}} - int \color{blue}{\fbox{:{-cos2x}/2:}\ \color{magenta}{\fbox{:dx:}}` ` = (-xcos2x)/2 + 1/2 int cos2x dx` ` = (-xcos2x)/2 + 1/2 (sin2x)/2 +K` ` = (-xcos2x)/2 + (sin2x)/4 +K` If the above is a little hard to follow (because of the line breaks), here it is again in a different format: Example 2 `int x sqrt(x+1) dx` Answer `intxsqrt(x+1)\ dx` We could let `u=x` or `u=sqrt(x+1)`. Once again, we choose the one that allows `(du)/(dx)` to be of a simpler form than `u`, so we choose `u=x`. Therefore `du = dx`. With this choice, `dv` must be the "rest" of the integral: `dv=sqrt(x+1)\ dx`. `u = x` so `du=dx`. `dv=sqrt(x+1)\ dx`, and integrating gives: `v=intsqrt(x+1) dx` `=int(x+1)^(1//2)dx` `=2/3(x+1)^(3//2)` Substituting into the integration by parts formula, we get: `int \color{green}{\underbrace{u}}\ \ \ \color{red}{\underbrace{dv}}\ \ ` ` =\ \ \color{green}{\underbrace{u}}\ \ \ \color{blue}{\underbrace{v}} \ \ -\ \ int \color{blue}{\underbrace{v}}\ \ \color{magenta}{\underbrace{du}}` `int \color{green}{\fbox{:x:}}\ \color{red}{\fbox{:sqrt(x+1) dx:}} = \color{green}{\fbox{:x:}}\ \color{blue}{\fbox{:2/3(x+1)^(3//2):}} ` `- int \color{blue}{\fbox{:2/3(x+1)^(3//2):}\ \color{magenta}{\fbox{:dx:}}` ` = (2x)/3(x+1)^(3//2) - 2/3 int (x+1)^{3//2}dx` ` = (2x)/3(x+1)^(3//2) ` `- 2/3(2/5) (x+1)^{5//2} +K` ` = (2x)/3(x+1)^(3//2)- 4/15(x+1)^{5//2} +K` Once again, here it is again in a different format: Example 3 `intx^2 ln 4x\ dx` Answer `intx^2ln\ 4x\ dx` We could let `u=x^2` or `u=ln\ 4x`.. Considering the priorities given above, we choose `u = ln\ 4x` and so `dv` will be the rest of the expression to be integrated `dv = x^2\ dx`. With `u=ln\ 4x`, we have `du=dx/x`. Integrating `dv = x^2\ dx` gives: `v=intx^2dx=x^3/3` Substituting in the Integration by Parts formula, we get: `int \color{green}{\underbrace{u}}\ \ \ \color{red}{\underbrace{dv}}\ \ ` ` =\ \ \color{green} {\underbrace{u}}\ \ \ \color{blue}{\underbrace{v}} \ \ -\ \ int \color{blue}{\underbrace{v}}\ \ \color{magenta}{\underbrace{du}}` `int \color{green}{\fbox{:x^2:}}\ \color{red}{\fbox{:ln 4x dx:}} = \color{green}{\fbox{:ln 4x:}}\ \color{blue}{\fbox{:x^3/3:}} ` `- int \color{blue}{\fbox{:x^3/3:}\ \color{magenta}{\fbox{:dx/x:}}` ` = (x^3 ln 4x)/3 - 1/3 int x^2 dx` ` = (x^3 ln 4x)/3 ` `- 1/3 x^3/3 +K` ` = (x^3 ln 4x)/3 - x^3/9 +K` Once again, here it is again in a different format: Example 4 `intx\ sec^2 x\ dx` Answer `int x\ sec^2 x\ dx` We choose `u=x` (since it will give us a simpler `du`) and this gives us `du=dx`. Then `dv` will be `dv=sec^2x\ dx` and integrating this gives `v=tan x`. Substituting these into the Integration by Parts formula gives: `int x\ sec^2 x\ dx =intu\ dv` `=uv-intv\ du` `=(x)(tan x)-int(tan x)dx` `=x\ tan x-(-ln\ |cos x|)+K` `=x\ tan x+ln\ |cos x|+K` Example 5 `intx^2e^(-x)dx` Answer `intx^2 e^-x dx` The 2nd and 3rd "priorities" for choosing `u` given earlier said: 2. Let `u = x^n` 3. Let `u = e^(nx)` This questions has both a power of `x` and an exponential expression. But we choose `u=x^2` as it has a higher priority than the exponential. (You could try it the other way round, with `u=e^-x` to see for yourself why it doesn't work.) So `u=x^2` and this gives `du=2x\ dx`. That leaves `dv=e^-x\ dx` and integrating this gives us `v=-e^-x`. We substitute these into the Integration by Parts formula to give: `intx^2 e^-x dx =intu\ dv` `=uv-intv\ du` `=x^2(-e^-x)-int(-e^-x)(2x\ dx) ` `=-x^2e^-x+2intxe^-x dx ` Now, the integral we are left with cannot be found immediately. We need to perform integration by parts again, for this new integral. This time we choose `u=x` giving `du=dx`. Once again we will have `dv=e^-x\ dx` and integrating this gives us `v=-e^-x`. Substituting into the integration by parts formula gives: `int x e^-x dx =intu\ dv` `=uv-intv\ du` `=x(-e^-x)-int(-e^-x)dx` `=-xe^-x+inte^-x dx` `=-xe^-x-e^-x ` So putting this answer together with the answer for the first part, we have the final solution: `intx^2e^-xdx =-x^2e^-x+2(-xe^-x-e^-x) ` `=-e^-x(x^2+2x+2)+K` Example 6 `int ln x dx` Answer `int ln\ x\ dx` Our priorities list above tells us to choose the logarithm expression for `u`. (of course, there's no other choice here. :-) So with `u=ln\ x`, we have `du=dx/x`. Then `dv` will simply be `dv=dx` and integrating this gives `v=x`. Subsituting these into the Integration by Parts formula gives: `int ln\ x\ dx=int u\ dv` `=uv-intv\ du` `=x\ ln\ x-intx(dx)/x` `=x\ ln\ x-intdx` `=x\ ln\ x-x+K` Example 7 `intarcsin x dx` Answer Using integration by parts, we set: `u=arcsin x`, giving `du=1/sqrt(1-x^2)dx`. Then `dv=dx` and integrating gives us `v=x`. We now use: `intu\ dv=uv-intv\ du` This gives us: `int arcsin x\ dx` `=x\ arcsin x-intx/(sqrt(1-x^2))dx` Now, for that remaining integral, we just use a substitution (I'll use `p` for the substitution since we are using `u` in this question already): `p = 1 - x^2` So `dp=-2x\ dx` This will yield: `intx/(sqrt(1-x^2))dx =-1/2int(dp)/sqrtp` `=-1/2(2sqrtp)+K` `=-sqrt(1-x^2)+K` So our final answer is: `int arcsin x\ dx =x\ arcsin x-(-sqrt(1-x^2))+K ` `= \x\ arcsin x+sqrt(1-x^2)+K` This time we integrated an inverse trigonometric function (as opposed to the earlier type where we obtained inverse trigonometric functions in our answer). See Integration: Inverse Trigonometric Forms. Alternate Method for Integration by Parts Here's an alternative method for problems that can be done using Integration by Parts. You may find it easier to follow. Tanzalin Method

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