calc II problem: need help with steps?

Question

The back of the book has the answer 1/2-ln√2, and I need help with the steps. The problem is ∫ (with the upper bound as (∏/4) and the lower bound is 0) (tan^3) x, dx

Answer 1

int( tan^3(x) )dx
Integrating by parts as tan(x) * tan^2(x) this gives:
int( tan(x) (sec^2(x) - 1) )dx
= int( tan(x)sec^2(x) )dx - int( tan(x) )dx
= int( tan(x) d(tan(x)) - int( d(cos(x) / cos(x) )
= (1/2) tan^2(x) + ln cos(x)
Evaluating from 0 to pi/4, this gives:
(1/2 - 0) + ln (cos(pi/4) / cos(0))
= 1/2 + ln(1/sqrt(2))
= 1/2 - ln( sqrt(2) ).

Answer 2

I'm brand new w/ computers. but just recently I ran into
KINGS LIST on the web. I haven't checked all out. But it has some good things for my G kids. It's wortha try. You're way over my head.

Answer 3

first exchange to cos^3(x) via cos^2(x) cos(x) and to (a million-sin^2(x))cosx Now you have crucial of sin^2(x) [a million-sin^2(x)]cos(x) dx exchange the variable: u = sin(x) and du = cos(x)dx Now you have crucial of u^2[ a million -u^2]du ==> crucial of (u^2 - u^4) du which will result u^3/3 - u^5/5 + C and now exchange lower back u via sin(x) and get the purely suited answer: sin^3(x) /3 - sin^5(x) /5 + C ok