How do you find the antiderivative of #(cosx)^2/(sinx)#?

Question

Emma Aldridge · Accepted Answer

#-ln(abs(cscx+cotx))+cosx+C#

We have:

#intcos^2x/sinxdx#

Through the Pythagorean identity:

#=int(1-sin^2x)/sinxdx#

Split the integral:

#=intcscxdx-intsinxdx#

These are both well-known integrals:

#=-ln(abs(cscx+cotx))+cosx+C#

William Abdalla · Answer

undefined To find the antiderivative of $ \frac{{\cos(x)^2}}{{\sin(x)}} $, we can use trigonometric identities. One approach is to rewrite the expression using the Pythagorean identity, $ \sin^2(x) + \cos^2(x) = 1 $. After rewriting, we integrate term by term using the substitution method. The steps involved are as follows:

1. Rewrite the expression using the Pythagorean identity: $ \frac{{\cos(x)^2}}{{\sin(x)}} = \frac{{1 - \sin^2(x)}}{{\sin(x)}} $.

2. Expand the expression: $ \frac{{1}}{{\sin(x)}} - \frac{{\sin^2(x)}}{{\sin(x)}} $.

3. Rewrite each term separately: $ \csc(x) - \sin(x) $.

4. Integrate each term separately:
   - The antiderivative of $ \csc(x) $ is $ -\ln|\csc(x) + \cot(x)| $.
   - The antiderivative of $ \sin(x) $ is $ -\cos(x) $.

Therefore, the antiderivative of $ \frac{{\cos(x)^2}}{{\sin(x)}} $ is $ -\ln|\csc(x) + \cot(x)| - \cos(x) + C $, where $ C $ is the constant of integration.