Calculus: Early Transcendentals, Chapter 7, 7.2, Section 7.2, Problem 26

int_0^(pi/4)sec^4(theta)tan^4(theta)d theta
Let's evaluate the indefinite integral by rewriting the integrand as,
intsec^3(theta)tan^4(theta)d theta=intsec^2(theta)sec^2(theta)tan^4(theta)d theta
Now use the identity:1+tan^2(x)=sec^2(x)
=int(1+tan^2(theta))sec^2(theta)tan^4(theta)d theta
Now apply integral substitution,
Let u=tan(theta)
=>du=sec^2(theta)d theta
=int(1+u^2)u^4du
=int(u^4+u^6)du
=intu^4du+intu^6du
=u^5/5+u^7/7
Substitute back u=tan(theta)
=1/5tan^5(theta)+1/7tan^7(theta)
Add a constant to the solution,
=1/5tan^5(theta)+1/7tan^7(theta)+C
Now let's evaluate the definite integral,
int_0^(pi/4)sec^4(theta)tan^4(theta)d theta=[1/5tan^5(theta)+1/7tan^7(theta)]_0^(pi/4)
=[1/5tan^5(pi/4)+1/7tan^7(pi/4)]-[1/5tan^5(0)+1/7tan^7(0)]
=[1/5+1/7]-[0]
=[(7+5)/35]
=12/35