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

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`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 (6sin(t))^2/sqrt(36-(6sin(t))^2) (6 cos(t) dt)`

= `int 36(sin(t))^2/sqrt(36-(6sin(t))^2) (6 cos(t) dt)`

=` int ((36)*(6)(sin(t))^2 *cos(t)) /sqrt(36-(6sin(t))^2) dt`

=`int (216(sin(t))^2 *cos(t)) /sqrt(36-36(sin(t))^2) dt`

= `int (216(sin(t))^2 *cos(t)) /sqrt(36(1-(sin(t))^2)) dt`

=`int (216(sin(t))^2 *cos(t)) /sqrt(36(cos(t))^2) dt`

=`int (216(sin(t))^2 *cos(t)) /(6(cos(t))) dt`

= `int (216/6) sin^2(t) dt`

= `int 36 sin^2(t) dt`

= `36 int sin^2(t) dt`

= `36 int (1-cos(2t))/2 dt`

= `(36/2) int (1-cos(2t)) dt`

= `18 [int 1 dt - int cos(2t) dt]+c`

= `18[t- (1/2)sin(2t)]+c`

but we know that

`x= 6sin(t)`

=> `x/6 = sin (t)`

=> `t= sin^(-1) (x/6) or arcsin(x/6)`


`18[t- (1/2)sin(2t)]+c`

= `18[(arcsin(x/6))- (1/2)sin(2(arcsin(x/6)))]+c`


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

=`18arcsin(x/6)- 9sin(2(arcsin(x/6)))+c`

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