Trig Integrals

**coolio** · January 30th 2008, 12:50 PM

Could use some help with these:

1. The integral sin^6 x cos^3 x dx
2. The integral sin^2 x cos^4 x dx

Thanks!

**Soroban** · January 30th 2008, 01:24 PM

Hello, coolio!

The first one is quite straight forward . . .

$1)\;\;\int \sin^6\!x\cos^3\!x\, dx$

We have: . $\sin^6\!x\cos^2\!x(\cos x) \;=\;\sin^6\!x(1-\sin^2\!x)(\cos x) \;=\;(\sin^6\!x - \sin^8\!x)(\cos x)$

The integral becomes: . $\int(\sin^6\!x - \sin^8\!x)(\cos x\,dx)$

Let: $u \:=\: \sin x\quad\Rightarrow\quad du \:=\:\cos x\,dx$

Substitute: . $\int(u^6 - u^8)\,du$

. . . Got it?

**coolio** · January 30th 2008, 01:53 PM

Thanks Soroban, that makes a lot more sense. The case where sin and cos both have even powers seems to be giving me the most trouble still. I know it has something to do with the half angle formulas.

**Jhevon** · January 30th 2008, 03:58 PM

Originally Posted by coolio

Thanks Soroban, that makes a lot more sense. The case where sin and cos both have even powers seems to be giving me the most trouble still. I know it has something to do with the half angle formulas.

yes. you would replace them with formulas derived from the half angle formula: $\sin^2 x = \frac {1 - \cos 2x}2$ and $\cos^2 x = \frac {1 + \cos 2x}2$

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