How to find $\int_0^{\pi/4}\sec(x)^3dx$ How do I find $$\int_0^{\pi/4}\sec(x)^3dx$$
I arrive at $\sec x\tan x + 1/3 \ln| (\cos^3(x))| dx$
where $u = \sec x$ and $v' = \sec^2(x)$
What's my error? Could I get a step by step solution?
 A: In your last step, you did 
$$\int v \, du = \int \tan^2 x \sec x \, dx $$
and you somehow got to $\frac{1}{3} \ln |\cos^3 x \, dx$.  First, the $dx$ there does not make any sense; you are adding that infinitessimal to the ordinary quantity $\sec x \tan x$, which is always a sign of something wrong.  But assuming that that was just a typo, your mistake was in integrating 
$$\int \tan^2 x \sec x \, dx = \int u^2 du$$
with $u = \tan x$. which is 
$$\frac{1}{3} tan^3 x$$
So the indefinite integral is 
$$
\tan x \sec x + \frac{1}{3} \tan^3 x
$$
The last  part of the problem is evaluating this at the lower endpoint: 
$$
1 \cdot \sqrt{2} + \frac{1}{3} \cdot 1 - 0 \cdot 1 -\frac{1}{3}\cdot 0 = \sqrt{2}+\frac{1}{3}
$$
A: So you're integrating by parts with $u = \sec x$ and $dv = \sec^2 x = (\tan x)'$. Then:
$$
I := \int_{0}^{\frac{\pi}{4}} \sec^3 x dx = \sec \left(\frac{\pi}{4}\right)\tan \left(\frac{\pi}{4}\right)  - \int_{0}^{\frac{\pi}{4}}\tan^2 x \sec x dx
$$
Where the last integral is
$$
\int_{0}^{\frac{\pi}{4}}\tan^2 x \sec x dx = \int_{0}^{\frac{\pi}{4}} \frac{1-\cos^2 x}{\cos^3 x} = I - \int_{0}^{\frac{\pi}{4}} \sec xdx 
$$
Plugging this in the first expression yields:
$$
2I = \sec \left(\frac{\pi}{4}\right)\tan \left(\frac{\pi}{4}\right)  + \int_{0}^{\frac{\pi}{4}} \sec xdx  = \sec \left(\frac{\pi}{4}\right)\tan \left(\frac{\pi}{4}\right)  + \log\left\lvert \sec \left(\frac{\pi}{4}\right) + \tan \left(\frac{\pi}{4}\right) \right\rvert
$$
So $I = \frac{1}{\sqrt{2}} + \log \sqrt{1+\sqrt{2}}$.
