Question

\[\int\cos^2u\left(\sin u\right)^{-1}\,du\]

Answer 1

1/sinu+sinu^2/sinu

Answer 2

What?

Answer 3

expand cos ... separate denominator, you will get int(cscx) + int (sinx)

Answer 4

oops sorry - at middle

Answer 5

(cos)^2 = 1-(sin)^2
(cos)^2/ sin = [1-(sin)^2] / sin = csc - sin ..

Answer 6

Oh, okay; whooops. That makes sense

Answer 7

\[\begin{array}{l} \text{For the integrand }\cos (u) \cot (u)\text{, write }\cos (u) \cot (u)\text{ as }\csc (u)-\sin (u): \\ \text{}=\int\limits (\csc (u)-\sin (u)) \, du \\ \text{Integrate the sum term by term and factor out constants:} \\ \text{}=\int\limits \csc (u) \, du-\int\limits \sin (u) \, du \\ \text{The integral limits of }\sin (u)\text{ is }-\cos (u): \\ \text{}=\cos (u)+\int\limits \csc (u) \, du \\ \text{The integral limits of }\csc (u)\text{ is }-\log (\cot (u)+\csc (u)): \\ \text{}=\cos (u)-\ln (\cot (u)+\csc (u))+\text{constant} \\ \text{Which is equivalent for restricted }u\text{ values \to:} \\ \text{}=\cos (u)+\ln \left(\sin \left(\frac{u}{2}\right)\right)-\ln \left(\cos \left(\frac{u}{2}\right)\right)+\text{constant} \\\end{array}\]

Answer 8

Wait, how do I prove \(\int\csc u\,du=-\log\left(\cot u+\csc u\right)\)?

Answer 9

differentiate the left hand side .. it's pretty obvious, you can also find the steps

Answer 10

I don't follow.

Answer 11

see the attachment

Answer 12

Any other tricks like that I should know? Like multiplying top and bottom by some trig identity that allows me to evaluate it?

Answer 13

similar for secx

Answer 14

tan -> integration by substitution, similar for cot