Question

integral x tan^2 x dx

Answer 1

i know its integration by parts

Answer 2

Lol, I was just about to say that.

Answer 3

\[\int\limits u dv= uv - \int\limits v du\]

Answer 4

is u = \[\tan ^{2}x\]

Answer 5

Let's do it then.
\[u=x \implies du= dx, dv=\tan^2(x) \implies v=\tan(x)-x\]
The integral becomes:
\[x(\tan(x)-x)-\int\limits (\tan(x)-x)dx\]

Answer 6

\[\int\limits_{}^{}fg' dx=fg-\int\limits_{}^{}f'g dx\]
\[f=x ; g'=\tan^2(x)\]'
\[f'=1; g=\int\limits_{}^{}\tan^2(x) dx=\int\limits_{}^{}\frac{\sin^2(x)}{\cos^2(x)} dx=\int\limits_{}^{}\frac{1-\cos^2(x)}{\cos^2(x)} dx\]
\[\int\limits_{}^{}\frac{1-\cos^2(x)}{\cos^2(x)} dx=\int\limits_{}^{}(\sec^2(x)-1) dx=\tan(x)-x\]

Answer 7

\[\int\limits_{}^{}x \tan^2(x) dx=x \cdot g-\int\limits_{}^{} 1 \cdot(\tan(x)-x) dx\]

Answer 8

\[\int\limits (\tan(x)-x)dx=-\log(\cos{x})-{x^2 \over 2}+c\]

Answer 9

There you have it!

Answer 10

Thank you guys