# Evaluate the integral `int_0^(pi/2) sin^5x dx`

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### 1 Answer

The integral `int_0^(pi/2) sin^5x dx` has to be determined.

`int sin^5x dx`

=> `int sinx*sin^4x dx`

=> `int sinx*(1 - cos^2x)^2 dx`

let `cos x = y` , `dy = -sin x dx`

=> `-int (1 - y^2)^2 dy`

=> `-int 1 + y^4 - 2y^2 dy`

=> `-y - y^5/5 + (2y^3)/3`

substitute y = cos x

=> `-cos x - (cos^5x)/5 + (2/3)*cos^3x`

`int_0^(pi/2) sin^5x dx`

=> ` (-cos x - cos^5x/5 + (2/3)*cos^3x)_0^(pi/2)`

=> `-cos(pi/2) + cos 0 - cos^5(pi/2)/5 + cos 0/5 + (2/3)*cos^3(pi/2) - (2/3)*cos^3 0`

=> `0 + 1 - 0 + 1/5 + 0 - 2/3`

=> `8/15`

**The integral `int_0^(pi/2) sin^5x dx = 8/15` **