Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 4

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