What is the antiderivative of #(cos^2x) / sinx#?

Question

1574 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-22T11:03:44

#=cos (x)+\ln |tan (frac{x}{2})|#

#\int \frac{\cos ^2(x)}{\sin \(x)}dx#

We know,
#\cos ^2(x)=1-\sin ^2(x)#

#=\int \frac{1-\sin ^2(x)}{\sin (x)}dx#

Applying sum rule,
#\int f(x)\pm g(x)dx=\int f(x)dx\pm \int g(x)dx#

#=\int \frac{1}{\sin (x)}dx-\int \frac{\sin ^2(x)}{\sin (x)}dx#

Also,we know,
#\int \frac{1}{\sin (x)}dx=\ln |\tan \(\frac{x}{2})\|#

#\int \frac{\sin ^2(x)}{\sin \(x)}dx=-\cos (x)#

now,
#=\ln |\tan (\frac{x}{2})|-\(-\cos (x))#

simplifying it,
#=cos (x)+\ln |tan (frac{x}{2})|#