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

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Answer 1 · 2018-06-21T03:59:34

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

Here,

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

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

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,

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

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