Question

Any way to solve integral of (cosx)^4 dx using integration by parts?

Answer 1

let u = cos^3x --> du = -3cos^2 sin dx
dv = cos dx ---> v = sinx
it turns \[sinx cos^3x +3\int cos^2sin^2dx\]
just replace cos^2 sin^2 = sin^2(2x) and take integral again

Answer 2

I'm trying it out, I know how to do it using sourwig's identity.

Answer 3

Sometimes, profs ask us apply that way to solve the problem. That's why

Answer 4

Ibelive you meant 2cos^2 sin^2 = sin^2(2x)

Answer 5

I believe*

Answer 6

heehehe... I underestimated it so that I typed a wrong thing twice.:)

Answer 7

no problem! awesome

Answer 8

\[\int\limits \left( \cos ^{2}x \right)^{2}dx=\int\limits \left( \frac{ 1+\cos 2x }{ 2 } \right)^2 dx\]
\[=\frac{ 1 }{ 4 }\int\limits \left( 1+2\cos 2x+\cos ^{2}2x \right)dx\]
\[=\frac{ 1 }{ 4 }\int\limits \left\{ 1+2\cos 2x+\frac{ 1+\cos 4x }{ 2 } \right\}dx\]
\[=\frac{ 1 }{ 4 }\int\limits \left( \frac{ 3 }{ 2 }+2\cos 2x+\cos 4x \right)dx\]
complete it.