Question

Steps for this integral, please: ∫[sin(x)]/[1+(cos(x))^2] dx =-arctan(cos(x))

Answer 1

so you know the answer but want help with the stepS?

Answer 2

huh? definite integral or indefinite?

Answer 3

\[\int{\sin x\over1+\cos^2x}dx\]

Answer 4

\[\int{\sin x\over1+\cos^2x}dx=-\tan^{-1}(\cos x)+C\]and we want to prove it, right?

Answer 5

right

Answer 6

\[u=\cos x\]\[du=-\sin xdx\]\[\int{\sin x\over1+\cos^2x}dx=-\int{du\over1+u^2}=-\tan^{-1}u+C\]\[=-\tan^{-1}(\cos x)+C\]do you know that last integral?

Answer 7

Thank you for your time
So arctan x=tan^-1x ?

Answer 8

yes, just a different notation

Answer 9

Thank you, once again for your help.

Answer 10

anytime :D
and if you don't know how to show that\[\int{dx\over a^2+x^2}=\frac1a\tan^{-1}(\frac xa)+C\]and you know how to do trig substitution integrals, I could prove that for you, but only if you wanted...

Answer 11

Thanks, I'm good for now :)