How do you find the antiderivative of #cosx/cscx#?

Question

Ezra Adams · Accepted Answer

#-1/4cos2x+C.# Since, #1/cscx=sinx, and, 2sinxcosx=sin2x#, we have, #intcosx/cscxdx=intcosx*sinxdx=1/2int(2sinxcosx)dx# #=1/2intsin2xdx=1/2((-cos2x)/2)=-1/4cos2x+C.#

Christopher Allers · Answer

Ratnaker M. has given the correct answer. Here are two other ways to get (and to write) the answer.

Once we realize that we need to evaluate #int cosx sinx dx# we have choices for how to evaluate this. One is shown in Ratnaker's answer. OR let #u=cosx#, so that #du = -sinx dx# and the integral becomes #- int u du = -u^2/2+C = -cos^2x +C#. OR let #u=sinx#, so that #du = cosx dx# and the integral becomes #int u du = u^2/2+C = sin^2x +C#. Challenge show that all of these answers are "the same". (Hint: find the difference. -- it's in the #C#.)

Samantha Adkison · Answer

undefined To find the antiderivative of $\frac{\cos x}{\csc x}$, you can rewrite it in terms of sine and cosine functions. Using the identity $\csc x = \frac{1}{\sin x}$, the expression becomes $\frac{\cos x \cdot \sin x}{1}$. Then, using the quotient rule for integration, integrate each term separately. The antiderivative of $\cos x \cdot \sin x$ can be found using integration by parts. This yields the final result of $-\cos x + C$, where $C$ is the constant of integration.