Question #f550a

1 Answer
Feb 10, 2018

#int\ (1-sin^2(x))/sin^2(x)\ dx=-cot(x)-x+C#

Explanation:

We can first split the fraction into two:
#int\ (1-sin^2(x))/sin^2(x)\ dx=int\ 1/sin^2(x)-sin^2(x)/sin^2(x)\ dx=#

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

We can now use the following identity:
#1/sin(theta)=csc(theta)#

#int\ csc^2(x)\ dx-x#

We know that the derivative of #cot(x)# is #-csc^2(x)#, so we can add a minus sign both outside and inside the integral (so they cancel) to work it out:
#-int\ -csc^2(x)\ dx-x=-cot(x)-x+C#