Question

Solve the indefinite integral cosx/ (sinx)^2 dx

Answer 1

Let \(sin(x) = t\)

Answer 2

\[\sin(x) = t\]

\[\cos(x) dx = dt\]

\[\int\limits \frac{dt}{t^2} = ??\]

Answer 3

\[\int\limits_{}^{} \cos x / \sin ^{2}x \]

I have that u=sinx, and du=cosx

So then it looks like \[\int\limits_{}^{} du/u^2\]

Answer 4

Yep..

Answer 5

-u^-1 + C ?

Answer 6

\[\int\limits u^{-2} du \implies \frac{u^{-2 + 1}}{-2 + 1} = \frac{u^{-1}}{-1} + C\]

Answer 7

Yes well done..

Now plug back for u ..

Answer 8

-1/sinx +c

Answer 9

Well Done..

Answer 10

Thanks!

Answer 11

Welcome dear..