Question

evaluate the definite integral of (x sin(x))/cos^3(x) dx, on the interval [0, pi/4]

Answer 1

Hmm looks like `Integration by Parts` for this problem.

\[\Large\rm \int\limits \frac{x \sin x}{\cos^3x}~dx\]
\[\Large\rm u=x,\qquad\qquad dv=\frac{\sin x}{\cos^3x}~dx\]Understand how to find your v?

Answer 2

eek - not from sinx/cos^3x dx :(

Answer 3

u-sub if you need to.
u=cos x

You end up with an integral like 1/u^3,
leading to 1/(2cos^2x) or something similiar.

Answer 4

Err actually, I don't want to use u for our parts AND here.
Let's call it something else,
\[\Large\rm m=\cos x, \qquad\qquad -dm=\sin x~dx\]
\[\Large\rm \int\limits \frac{-dm}{m^3}=\frac{1}{2m^2}=\frac{1}{2cos^2x}\]Like uhhh that? Yah?

Answer 5

OK! Now, the integration by parts formula that I know is
\[\int\limits u dv=uv - \int\limits vdu\]
Is this the formula to use even though I have \[\frac{ u }{ dv }\]

Answer 6

You have u dv not u/dv
Recall he called x=u and the other part of the fraction dv

Answer 7

Oh - got it!

Answer 8

Thank you so much!! I think I have an answer. :)