Solve the Integral
I = xsec^{2}x^{2} dx
tan^{x}2
PLEASE HELP THANKS!!
If you meant to integrate:
$\displaystyle I = \frac{x \sec^2{(x^2)}}{\tan{(x^2)}}$
Then let $\displaystyle u = \tan{(x^2)} \therefore\,\, du = 2 x \sec^2{(x^2)}\,\,dx$
Now:
$\displaystyle \begin{align*}\int \frac{x \sec^2{(x^2)}}{\tan{(x^2)}}\,\,dx =& \int \frac{2x \sec^2{(x^2)}}{2 \tan{(x^2)}}\,\,dx\\ =& \int \frac{1}{2u}\,\,du....\text{[by substituting values of } u \text{ and } du\text{]}\\ =& \frac{1}{2}\ln{(u)} + C\\=& \frac{1}{2}\ln{(\tan{(x^2)})} + C....\text{[by substituting } u\text{]}\end{align*}$