# Evaluate `int sin^6xcos^5xdx` .

### 1 Answer | Add Yours

`int sin^6xcos^5x dx`

Re-write the integrand in such a way that the power of the cosine function result to 1.

`=int sin^6x cosx cos^2x cos^2x dx`

From the Pythagorean identity `sin^2x + cos^2x=1` , replace `cos^2x` with `1-sin^2x` .

`= int sin^6x cosx (1-sin^2x)(1-sin^2x)dx`

Expand `(1-sin^2x)(1-sin^2x)` .

`=int sin^6xcosx(1-2sin^2x+sin^4x)dx`

`=int (sin^6xcosx- 2sin^8xcosx + sin^10xcosx)dx`

`=int sin^6cosxdx - 2int sin^8cosxdx +int sin^10xcosxdx`

Then, apply u-substitution. Let,

`u=sinx`

`du=cosxdx`

Replace u with sinx and cosxdx with du.

`= int u^6du - 2int u^8du + int u^10du`

Use the power formula of integral which is` int u^n du = u^(n+1)/(n+1) + C` .

`= u^7/7 -(2u^9)/9 + u^11/11 + C`

Substitute back u=sinx to return to original variable x.

`= (sin^7x)/7 - (2sin^9x)/9 + (sin^11x)/11 + C`

**Hence, `int sin^6xcos^5x dx = (sin^7x)/7 - (2sin^9x)/9 + (sin^11x)/11 + C` .**