Question

How do you evaluate the integral `int arc cscx`?

Answer 1

`intarc cscx dx=x*arc cscx+ln|x+sqrt(x^2-1)|+c`

Explanation:

Here,

`I=intarc cscxdx=intarc csc x*1dx`

Using Integration by parts:

`color(blue)(int(u*v)dx=uintvdx-int(u'intvdx)dx`

Let,

`u=arc cscx and v=1to` {Let `color(red)( x > 0=>|x|=x`}

`=>u'=-1/(|x|sqrt(x^2-1))=-1/(xsqrt(x^2-1)) and intvdx`=`x`

So,

`I=arc cscx*(x)-int[(-1)/(xsqrt(x^2-1))*x]dx`

`=x*arc cscx+int1/sqrt(x^2-1)dx`

`=x*arc cscx+ln|x+sqrt(x^2-1)|+c`

Note :

If `color(red)( x < 0 ,then ,|x|=-x`

So,

`I=x*arc cscx-ln|x+sqrt(x^2-1)|+c`