How to evaluate this integral with $\sin$ and $\cos$?

Question

$$ \int \sin^6 x\cos^4x\ dx$$

I have absolutely no idea how to solve it. I tried to use formula of degree reduction but failed. Help me please. Thank you.

score 3 · Answer 1 · answered May 22 '14 at 16:01

3

As $\displaystyle\sin2A=2\sin A\cos A,\cos2B=2\cos^2B-1=1-2\sin^2B,$

$$\sin^6x\cos^4x=\sin^2x(\sin x\cos x)^4=\frac{1-\cos2x}2\cdot\frac{\sin^42x}{16}$$

Now, $\displaystyle\sin^4(2x)=(\sin^22x)^2=\frac{(1-\cos4x)^2}4=\frac{1-2\cos4x+\cos^24x}4$

Again, $\displaystyle\cos^24x=\frac{1+\cos8x}2$

answered May 22 '14 at 16:01

lab bhattacharjee

It is easier to note that the derivative of $\sin^5 2x$ is $10 \cos 2x \sin^4 2x$ - then you don't need to use Werner – Mark Bennet May 22 '14 at 16:08
@MarkBennet, Good Observation – lab bhattacharjee May 22 '14 at 16:11

score 3 · Answer 2 · answered May 22 '14 at 16:04

3

HINT:

If you are allowed to use Euler Formula,

$\displaystyle\sin x=\frac{e^{ix}-e^{-ix}}{2i}$ and $\displaystyle\cos x=\frac{e^{ix}+e^{-ix}}2$

and use Binomial Expansion

answered May 22 '14 at 16:04

lab bhattacharjee

2 Answers2