How do you find the antiderivative of Cosx/Sin^2xcosxsin2x?

2 Answers
Mar 11, 2018

-cosecx+Ccosecx+C

Explanation:

I=intcosx/sin^2xdx=int1/sinx*cosx/sinxdxI=cosxsin2xdx=1sinxcosxsinxdx
I=intcscx*cotxdx=-cscx+CI=cscxcotxdx=cscx+C

Mar 11, 2018

int\ cos(x)/sin^2(x)\ dx=-csc(x)+C

Explanation:

int\ cos(x)/sin^2(x)\ dx

The trick to this integral is a u-substitution with u=sin(x). We can see this is the right way to go because we've got the derivative of u, cos(x) in the denominator.

To integrate with respect to u, we need to divide by the derivative, cos(x):

int\ cos(x)/sin^2(x)\ dx=int\ cancel(cos(x))/(cancel(cos(x))u^2)\ du=int\ 1/u^2\ du=int\ u^-2\ du

We can evaluate this integral using the reverse power rule:

int\ x^n\ dx=x^(n+1)/(n+1)

int\ u^-2\ du=u^-1/(-1)+C=-1/u+C

Now we resubstitute u=sin(x) to get the answer in terms of x:

-1/u+C=-1/sin(x)+C=-csc(x)+C