Thread: Trigonometric substitution problem

1. Trigonometric substitution problem

I'm supposed to integrate 8*cos^3(2x)*sin(2x).

I'm whittling it down. I factor out a cosine and use the property cos^2(2x)=1-sin^2(2x).

This gives me the integral of 8*((1-sin^2(2x))*cos(2x)*sin(2x)).

Here I substitute u=sin(2x) and du=2cos(2x)

So I have 1/2 the integral of (1-u^2)*u*du

I suspect that by now I've messed it up. The final answer I SHOULD get is -(cos(2x)^4) What's going on?

2. let $u=\cos(2x)$ and proceed

3. Make $\displaystyle u=\cos 2x$

4. How hard should I hit myself for missing that?

5. Originally Posted by Wolvenmoon
I'm supposed to integrate 8*cos^3(2x)*sin(2x).

I'm whittling it down. I factor out a cosine and use the property cos^2(2x)=1-sin^2(2x).

This gives me the integral of 8*((1-sin^2(2x))*cos(2x)*sin(2x)).

Here I substitute u=sin(2x) and du=2cos(2x)

So I have 1/2 the integral of (1-u^2)*u*du

I suspect that by now I've messed it up. The final answer I SHOULD get is -(cos(2x)^4) What's going on?

Your method, though more involved, was still correct.

Continuing....

$\displaystyle\ 4\int{\left(u-u^3\right)du=4\left(\frac{u^2}{2}-\frac{u^4}{4}\right)+C$

$=2u^2-u^4+C=2sin^2(2x)-sin^4(2x)+C=2\left[1-cos^2(2x)\right]-\left[1-cos^2(2x)\right]^2+C$

$=\displaystyle\ 2-2cos^2(2x)-\left[1-2cos^2(2x)+cos^4(2x)\right]+C$

$=2-2cos^2(2x)-1+2cos^2(2x)-cos^4(2x)+C$

$=1-cos^4(2x)+C=-cos^4(2x)+C_1$

6. Originally Posted by Wolvenmoon
How hard should I hit myself for missing that?
Hahaha, no don't hit yourself yet, you are only at the beginning of your journey into integration, there will be plenty of oppurtunities, I promise...