therefore $\int_{}^{} {\frac{{{x^2}}}{{{{(9 - {x^2})}^{32}}}}\,dx} = \int_{}^{} {\frac{{9{{\sin }^2}\theta }}{{{{(9 - 9{{\sin }^2}\theta )}^{32}}.\,3\cos \theta }}\,d\theta } $
$ = \int_{}^{} {\frac{{27{{\sin }^2}\theta \cos \theta }}{{27{{\cos }^3}\theta }}} \,d\theta = \int_{}^{} {{{\tan }^2}\theta \,d\theta } = \int_{}^{} {({{\sec }^2}\theta - 1)\,d\theta } $
$ = \tan \theta - \theta + c = \tan \left\{ {{{\sin }^{ - 1}}\left( {\frac{x}{3}} \right)} \right\} - {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + c$
$ = \tan {\tan ^{ - 1}}\left( {\frac{{\left( {\frac{x}{3}} \right)}}{{\sqrt {1 - ({x^2}9)} }}} \right) - {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + c$
$ = \frac{x}{{\sqrt {9 - {x^2}} }} - {\sin ^{ - 1}}\left( {\frac{x}{3}} \right) + c.$
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$\text{If}\ \text{P}(\text{A})=\frac{1}{2},\ \text{P}(\text{B})=0,\ \text{then}\ \text{P}(\text{A}|\text{B})\ \text{is}:$