How do you find the indefinite integral of #int csc2x#?

Question

1701 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-07T08:38:43

# intcscx2xdx=-1/2ln| (csc2x+cot2x)|+C#

start by multiplying by #(csc2x+cot2x)/(csc2x+cot2x)#

#intcscx2xdx=intcsc 2x xx(csc2x+cot2x)/(csc2x+cot2x)dx#

#=int[(csc^2 2x+csc2xcot2x)/(csc2x+cot2x)]dx#

now

#d/(dx)(csc2x)=-2cot2xcsc2x#

and

#d/(dx)(cot2x)=-2csc^2 2x#

also

#int((f'(x))/f(x))dx=ln|f(x)|+C#

using these results we notice that differentiating the denominator:

#d/(dx)(csc2x+cot2x)=-2cot2xcsc2x-2csc^2 2x#

#=-2(csc^2 2x+csc2x+cot2x)=-2 xx" numerator"#

#:. intcscx2xdx=-1/2ln| (csc2x+cot2x)|+C#