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

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

Let's first evaluate the definite integral,by rewriting the integrand as,

`=intsin^4(x)sin(x)dx`

Now use the identity: `sin^2(x)=1-cos^2(x)`

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

Now apply the integral substitution,

Let `u=cos(x)`

`=>du=-sin(x)dx`

`=int-(1-u^2)^2du`

`=-int(1-2u^2+u^4)du`

`=-int1du+2intu^2du-intu^4du`

`=-u+2(u^3/3)-u^5/5`

Substitute back `u=cos(x)`

and adding a constant C to the solution,

`intsin^5(x)dx=-cos(x)+2/3cos^3(x)-1/5cos^5(x)+C`

Now let's evaluate the definite integral,

`int_0^(pi/2)sin^5(x)dx=[-cos(x)+2/3cos^3(x)-1/5cos^5(x)]_0^(pi/2)`

`=[-cos(pi/2)+2/3cos^3(pi/2)-1/5cos^5(pi/2)]-[-cos(0)+2/3cos^3(0)-1/5cos^5(0)]`

`=[0]-[-1+2/3-1/5]`

`=-[(-15+10-3)/15]`

`=-(-8/15)`

`=8/15`