# `int x^2/sqrt(36-x^2) dx` Find the indefinite integral

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Given

`int x^2/sqrt(36-x^2) dx`

This can be solved by using the Trigonometric substitutions  (Trig substitutions)

when the integral contains `sqrt(a-bx^2)` then we have to take

`x=sqrt(a/b) sin(t)` in order to solve the integral easily

so here , For

`int x^2/sqrt(36-x^2) dx`

`x` is given as

`x= sqrt(36/1) sin(t) = 6sin(t) `

=> `dx = 6 cos(t) dt`

so ,

`int x^2/sqrt(36-x^2) dx`

=`int...

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