Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 3

int_0^(pi/2)sin^7(theta)cos^5(theta)d theta
Let's first compute the indefinite integral by rewriting the integrand,
intsin^7(theta)cos^5(theta)d theta=intsin^6(theta)sin(theta)cos^5(theta)d theta
Now use the identity:sin^2(x)=1-cos^2(x)
=int(1-cos^2(theta))^3sin(theta)cos^5(theta)d theta
Now apply integral substitution,
Let u=cos(theta)
=>du=-sin(theta)d theta
=int(1-u^2)^3u^5(-du)
=-int(1-u^2)^3u^5du
=-int(1-u^6-3u^2+3u^4)u^5du
=-int(u^5-u^11-3u^7+3u^9)du
=-intu^5du+intu^11du+3intu^7du-3intu^9du
=-u^6/6+u^12/12+3(u^8/8)-3(u^10/10)
=u^12/12-3/10u^10+3/8u^8-1/6u^6
Substitute back u=cos(x)
=1/12cos^12(x)-3/10cos^10(x)+3/8cos^8(x)-1/6cos^6(x)
Add a constant C to the solution,
=1/12cos^12(x)-3/10cos^10(x)+3/8cos^8(x)-1/6cos^6(x)+C
Now let's evaluate the definite integral,
int_0^(pi/2)sin^7(theta)cos^5(theta)d theta=[1/12cos^12(x)-3/10cos^10(x)+3/8cos^8(x)-1/6cos^6(x)]_0^(pi/2)
=[1/12cos^12(pi/2)-3/10cos^10(pi/2)+3/8cos^8(pi/2)-1/6cos^6(x)]-[1/12cos^12(0)-3/10cos^10(0)+3/8cos^8(0)-1/6cos^6(0)]
Plug in the values of cos(pi/2)=0.cos(0)=1
=[0]-[1/12-3/10+3/8-1/6]
=-[(10-36+45-20)/120]
=-[-1/120]
=1/120