How do you find the definite integral of #cot x csc^5 x dx# from #[pi/6, pi/2]#?

Question

2703 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-31T20:44:10

#int_(pi/6)^(pi/2)cotxcsc^5xdx=31/5#

Note that #d/dxcscx=-cotxcscx#.

We can rewrite the integral:

#int_(pi/6)^(pi/2)cotxcsc^5xdx=-int_(pi/6)^(pi/2)csc^4x(-cscxcotxdx)#

Let #u=cscx#. This implies that #du=-cscxcotxdx#. We have both of these in our integral.

When we make the shift from #x# to #u#, we will have to rewrite the bounds:

#x=pi/6=>u=1/sin(pi/6)=2# and #x=pi/2=>u=1/sin(pi/2)=1#. Then:

#=-int_2^1u^4du=int_1^2u^4du=[u^5/5]_1^2=2^5/5-1^5/5=32/5-1/5=31/5#