## Help with Trigonometric Integral

Limits, differentiation, related rates, integration, trig integrals, etc.

### Help with Trigonometric Integral

Hey everyone. This is my first post. I just need help with this problem in Calculus II. The problem states: Integrate the trigonometric integral: Integral of sin ^ 2 (5x) * cos (2x) dx. I wasn't sure how to approach it. Thanks for any help. BTW, this is what I have so far:

I've moved it around a little using trig identities:

Integral of (1 - cos ^ 2 (5x)) cos (2x) dx

= Integral of (cos (2x) - cos ^ 2 (5x) * cos (2x)) dx

= [1/2 * sin (2x)] - Integral of (cos ^ 2 (5x) * cos (2x)) dx
paulh428

Posts: 7
Joined: Wed Apr 27, 2011 11:17 pm

### Re: Help with Trigonometric Integral

paulh428 wrote:Hey everyone. This is my first post. I just need help with this problem in Calculus II. The problem states: Integrate the trigonometric integral: Integral of sin ^ 2 (5x) * cos (2x) dx. I wasn't sure how to approach it. Thanks for any help. BTW, this is what I have so far:

I've moved it around a little using trig identities:

Integral of (1 - cos ^ 2 (5x)) cos (2x) dx

= Integral of (cos (2x) - cos ^ 2 (5x) * cos (2x)) dx

= [1/2 * sin (2x)] - Integral of (cos ^ 2 (5x) * cos (2x)) dx

use the following

$\sin(A)\cos(B)=\frac{1}{2}\left[\sin(A-B)+\sin(A+B)\right]$

$\sin(A)\sin(B)=\frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right]$

$\cos(A)\cos(B)=\frac{1}{2}\left[\cos(A-B)+\cos(A+B)\right]$

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA

### Re: Help with Trigonometric Integral

Thank you. I actually used it that way after this post. Not sure if what I have now is the correct answer but it was a long one nonetheless. Thank you very much.
paulh428

Posts: 7
Joined: Wed Apr 27, 2011 11:17 pm

### Re: Help with Trigonometric Integral

Martingale wrote:
$\sin(A)\cos(B)=\frac{1}{2}\left[\sin(A-B)+\sin(A+B)\right]$

$\sin(A)\sin(B)=\frac{1}{2}\left[\cos(A-B)-\cos(A+B)\right]$

$\cos(A)\cos(B)=\frac{1}{2}\left[\cos(A-B)+\cos(A+B)\right]$

paulh428 wrote:Thank you. I actually used it that way after this post. Not sure if what I have now is the correct answer but it was a long one nonetheless. Thank you very much.

$sin ^ 2 (5x) \cdot cos (2x)$

$=\frac{1}{2}(1-\cos(10x))\cos(2x)$

$=\frac{1}{2}\cos(2x)-\frac{1}{2}\cos(10x)\cos(2x)$

$=\frac{1}{2}\cos(2x)- \frac{1}{4}\left[\cos(10x-2x)+\cos(10x+2x)\right]$

$=\frac{1}{2}\cos(2x)- \frac{1}{4}\cos(8x)-\frac{1}{4}\cos(12x)$

$\int\left[\frac{1}{2}\cos(2x)- \frac{1}{4}\cos(8x)-\frac{1}{4}\cos(12x)\right]dx$

$\frac{1}{4}\sin(2x)-\frac{1}{32}\sin(8x)-\frac{1}{48}\sin(12x)+c$

Martingale

Posts: 363
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA