# How to know if integral is correct

• Mar 12th 2013, 02:40 PM
MichaelEngstler
How to know if integral is correct
Hello,
I have the following function I need to find an integral for:
sin(x)^3*cos(x)^5

I have done my way and got to this integral:
cos(x)^6/6 - cos(x)^8/8 +C

I checked the answer in my book and saw the answer was:
sin(x)^4/4 - sin(x)^6/3 + sin(x)^8/8 +C

I checked the answer in wolfram and got this:
Wolfram Mathematica Online Integrator

I understand each function has a infinite number of integrals, and they can be shown is many different ways.
But still, I cant find a reliable way to check if my answer is correct.
I would like to ask how do you deal with such a problem ?

Thanks a lot,
Michael.
• Mar 12th 2013, 05:03 PM
BobP
Re: How to know if integral is correct
The book answer has been achieved by means of the substitution u = sin x.
Mathematica has arrived at its result by making use of various trig identities to reduce powers to unity.
You seem to have used the substitution u = cos x, though I think that you've made a mistake, (a lost negative sign).
It should be possible to get from any one result to either of the other two by means of the various trig identities.
As to checking whether or not an integration is correct, the usual method is to differentiate the result to see whether it produces the thing that you started with.
• Mar 12th 2013, 05:12 PM
Prove It
Re: How to know if integral is correct
Quote:

Originally Posted by MichaelEngstler
Hello,
I have the following function I need to find an integral for:
sin(x)^3*cos(x)^5

I have done my way and got to this integral:
cos(x)^6/6 - cos(x)^8/8 +C

I checked the answer in my book and saw the answer was:
sin(x)^4/4 - sin(x)^6/3 + sin(x)^8/8 +C

I checked the answer in wolfram and got this:
Wolfram Mathematica Online Integrator

I understand each function has a infinite number of integrals, and they can be shown is many different ways.
But still, I cant find a reliable way to check if my answer is correct.
I would like to ask how do you deal with such a problem ?

Thanks a lot,
Michael.

\displaystyle \displaystyle \begin{align*} \int{\sin^3{(x)}\cos^5{(x)}\,dx} &= \int{\sin{(x)}\sin^2{(x)}\cos^5{(x)}\,dx} \\ &= \int{\sin{(x)}\left[ 1 - \cos^2{(x)} \right] \cos^5{(x)}\,dx} \\ &= \int{\sin{(x)}\left[ \cos^5{(x)} - \cos^7{(x)}\right] dx} \\ &= -\int{-\sin{(x)}\left[ \cos^5{(x)} - \cos^7{(x)}\right] dx} \end{align*}

Now let \displaystyle \displaystyle \begin{align*} u = \cos{(x)} \implies du = -\sin{(x)}\,dx \end{align*} and the integral becomes

\displaystyle \displaystyle \begin{align*} -\int{-\sin{(x)}\left[ \cos^5{(x)} - \cos^7{(x)} \right] dx} &= -\int{ u^5 - u^7\,du} \\ &= -\left( \frac{u^6}{6} - \frac{u^8}{8} \right) + C \\ &= -\left[ \frac{\cos^6{(x)}}{6} - \frac{\cos^8{(x)}}{8} \right] + C \end{align*}

You can enter (your answer)-(the book's answer) into WolframAlpha. If it turns up a constant, you're answer is correct. I tried it on Prove It's answer, and it came up with $\displaystyle \frac{1}{24}$.