Solve for the Anti-Derivative of cosx dx / sin^3x: Expert Answer and Explanation

[laker_gurl3]

was just wondering if this was the right answer..
take the anti-derivative of:

cosx dx / sin^3x

i got
-1/2(sinx)^-2 + C
is that right?

 

[whozum]

Thats what maple says, good job.

 

[whozum]

Just a tip, if your not sure, take the derivative and check if you get the original function.

 

[HallsofIvy]

Maple?? Maple?? ! Let u= sin x. Then du= cos du and the integrand becomes
\frac{cos x}{sin^3 x}dx= \frac{1}{u^3}du= u^{-3}du

The anti-derivative of that is \frac{1}{-3+1}u^{-3+1}+C= \frac{1}{2}u^{-2}+C= \frac{1}{2}\frac{1}{sin^2 t}+C

Good job, laker_gurl3

 

[whozum]

Nothing wrong with being lazy!