Evaluate the following: #int_(pi/6)^(pi/2)(cscxcotx)dx# I know that we are suppose to find the anti-derivative of cscxcotx, but i dont know how to?

Question

Evaluate the following: #int_(pi/6)^(pi/2)(cscxcotx)dx#  I know that we are suppose to find the anti-derivative of cscxcotx, but i dont know how to?

Charles Arocha · Accepted Answer

Recall that #d/dxcscx=-cscxcotx#. Reversing this shows that #int(-cscxcotx)dx=cscx+C#.

So, we have

#int_(pi//6)^(pi//2)(cscxcotx)dx=-int_(pi//6)^(pi//2)(-cscxcotx)dx=[-cscx]_(pi//6)^(pi//2)#

So, the anti-derivative you were looking for was #-cscx#. The actual evaluation of the integral gives:

#-csc(pi/2)-(-csc(pi/6))=-1+2=1#

Another way to find the integral is to use #cscx=1/sinx# and #cotx=cosx/sinx#:

#int(cscxcotx)dx=int1/sinxcosx/sinxdx=intcosx/sin^2xdx#

This can be found with the substitution #u=sinx#, which implies that #du=cosxdx#.

#=int1/u^2du=intu^-2du#

Using #intu^ndu=u^(n+1)/(n+1)#:

#=u^-1/(-1)=-1/u=-1/sinx=-cscx+C#

As we determined above!

Ezekiel Alexander · Answer

undefined To evaluate the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \csc(x) \cot(x) \, dx$, we can use the fact that $\csc(x) \cot(x) = \frac{\cos(x)}{\sin^2(x)}$. Then, perform a u-substitution letting $u = \sin(x)$. This will transform the integral into a more manageable form. After integrating with respect to $u$, you can then substitute back in terms of $x$ and evaluate the definite integral over the given interval.