# Evaluate the following integral. `int cos^4xsin^3xdx`

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### 1 Answer

`int cos^4xsin^3xdx`

To evaluate, let's use u-substitution. To be able to apply this, re-write the integrand in way that the power of sine function would result to 1.

`= int cos^4x sinx sin^2x dx `

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

`= int cos^4xsinx(1-cos^2x)dx`

`=int (cos^4xsinx - cos^6xsinx)dx`

`=int cos^4xsinxdx - int cos^6xsinx dx`

Then let,

`u=cos x`

`du=-sinxdx ` ===> `-du= sinxdx`

Replace th x variable of the integral with u.

`=int u^4(-du) - int u^6(-du) = -int u^4du + int u^6du `

`= int u^6du - int u^4 du`

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

`= u^7/7 - u^5/5 + C`

Substitute back u=cos x to return to lvariable x.

`= (cos^7x)/7 - (cos^5x)/5 + C`

**Hence, `int cos^4xsin^3dx = (cos^7x)/7 - (cos^5x)/5 + C` .**