Question

Integrate cosx/sin^2x+4sinx+5.

Answer 1

put sin x = u
Therefore du = cos(x) dx
Therefore it becomes
$\frac{du}{ {u}^{2} + 4u + 5} \\ = \frac{du}{ {(u + 2)}^{2} + 1}$
If you integrate it you will get
${tan}^{ - 1} (u + 2) = {tan}^{ - 1} ( \sin(x) + 2) + c$
Hope this helps.