# Math Help - Integrals Involving Powers of Trig Functions.

1. ## Integrals Involving Powers of Trig Functions.

Well new subject in my Calculus 2 class, and I do not understand why I was wrong on my homework. First I'll show you my work and then I'll show the correct answer from wolfram which uses a formula (reduction Formula) I have yet to learn. I only used wolfram to check the answer.

The Start,

$\int sin^5(x)cos^8(x)dx$

Break them up:
$\int sin^4(x)cos^8(x)sin(x)dx$

Break sin up even further so I can use Trig Identities:
$\int (sin^2(x))^2cos^8(x)sin(x)dx$

Replace sin squared with a trig identity:
$\int (1-cos^2(x))^2cos^8(x)sin(x)dx$

Substitution:
$u=cos(x)$

$du=-sin(x)dx$

Replace cos(x) with u:
$-\int (1-u^2)^2u^8(x)du$

FOIL:
$u^8(1-u^2)(1-u^2)$

$(1-u^2-u^2+u^4)$

$u^8(1-2u^2+u^4)$

$(u^8-2u^{10}+u^{12})$

Enter in Newly FOIL'ed equation:
$-\int (u^8-2u^{10}+u^{12})du$

Integrate:
$-(\frac{{u}^{9}}{9}-2\frac{{u}^{11}}{11}+\frac{{u}^{13}}{13})+C$

Revert 'u' back to cos(x):
$(-\frac{1}{9}cos^{9}(x)+\frac{2}{11}cos^{11}(x)-\frac{1}{13}cos^{13}(x))+C$

Now in Short from Wolfram Alpha (the correct answer):

$\frac{-(cos^9(x) (-540 cos(2 x)+99 cos(4 x)+505))}{10296} +constant$

Where did I go wrong? How do I do this integral without referring to reduction formula?

2. You have not done anything wrong. All you need is a little more simplification to get to the desired result.

$(-\frac{1}{9}cos^{9}(x)+\frac{2}{11}cos^{11}(x)-\frac{1}{13}cos^{13}(x))+C$

$= \dfrac{-143cos^{9}x+234 cos^{11}x-99cos^{13}x}{1287}+C$

Taking $-cos^{9}x$ as a common term,if you simplify the terms further, you will get the value you have stated at the bottom.

3. Originally Posted by Zanderist
$(-\frac{1}{9}cos^{9}(x)+\frac{2}{11}cos^{11}(x)-\frac{1}{13}cos^{13}(x))+C$

Now in Short from Wolfram Alpha (the correct answer):

$\frac{-(cos^9(x) (-540 cos(2 x)+99 cos(4 x)+505))}{10296} + \text{Constant}$
You will see that if you type in $-\frac{1}{9} \cos^{9}x+\frac{2}{11} \cos^{11}x-\frac{1}{13} \cos^{13}x+C$ into Wolfram|Alpha and scroll down to the "alternate forms" section, you will see that there will be the equation in the form that came from Wolfram|Alpha's integration.