# Thread: how can solve this integral

1. ## how can solve this integral

hi!

tanx/cos x^2 dx

thanks

2. A couple options. I will do one to completion (a way that I suspect they do not want you to do), and will get you started on the way I'd imagine they want you to do this:

1st Method -

$\int \frac{tan(x)dx}{cos^{2}(x)}\Rightarrow$

$=\int \frac{sin(x)dx}{cos^{3}(x)}$

Make the substitution:

$u=cos(x); -du=sin(x)dx$

$=\int \frac{sin(x)dx}{cos^{3}(x)} \Rightarrow -\int \frac{du}{u^{3}}$

$=\frac{1}2{u^2}$

$=\frac{1}{2cos^{2}(x)}$

The second method begins much in the same way by rewritting the integral:

$\int \frac{tan(x)dx}{cos^{2}(x)}\Rightarrow$

$\int \frac{tan(x)sec(x)dx}{cos(x)}$

See if you can take it from there.

3. I would think that

$\int \frac{\tan x}{\cos^2x}dx=\int \tan x\cdot\sec^2xdx$

And if $u=\tan x$, then $du=\sec^2xdx$ and...

$\int u\cdot du=\frac{u^2}{2}+C=\frac{\tan^2x}{2}+C$

4. Originally Posted by hoger
hi!

Is it $\int{\frac{\tan{x}}{\cos^2{x}}\,dx}$ or $\int{\frac{\tan{x}}{\cos{(x^2)}}\,dx}$?