Question

Evaluate
$\\ \tt \color{blu} \boxed{ {{\tt\int \frac{3 x^{2}}{\sqrt{9-x^{6}}} d x}}}$

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Answer 1

Answer:

$\rm (-1)·\bigg(atan \bigg(\dfrac{\sqrt{-x^6+9}}{x^3} \bigg)\bigg)+C$

Step-by-step explanation:

${{ \displaystyle \: \rm \: \int \frac{3 x^{2}}{\sqrt{9-x^{6}}} d x}}$

$3\displaystyle \: \rm \: \int \frac{ x^{2}}{\sqrt{-x^{6} + 9}} dx$

$\rm let \: -1+\dfrac{9}{x^6}=u^2$

$\rm \dfrac{d}{dx} \bigg(-1+\dfrac{9}{x^6}\bigg)=\dfrac{d}{du}(u^2)$

$\rm -\dfrac{54dx}{x^7}=2udu$

$\displaystyle \: \rm \: \int \frac{ x^{2}}{\sqrt{-x^{6} + 9}} dx = \displaystyle \rm \int -\dfrac{1}{3u^2+3}du$

$3 \displaystyle \rm \int -\dfrac{1}{3u^2+3}du$

$-3 \displaystyle \rm \int \dfrac{1}{3u^2+3}du$

$\rm let \: u = tan(v)$

$\rm \dfrac{d}{du} u = \dfrac{d}{dv}tan(v)$

$\rm du = (1+ tan(v)^2)dv$

$\displaystyle \rm \int \dfrac{1}{3u^2+3}du = \displaystyle \rm \int \dfrac{1}{3} dv$

$-3 \displaystyle \rm \int \dfrac{1}{3} dv$

$\displaystyle \rm \int \dfrac{1}{3} dv = \dfrac{v}{3}= -v$

$\rm (-1)·(atan(u))$

$\rm (-1)·\bigg(atan \bigg(\dfrac{\sqrt{-x^6+9}}{x^3} \bigg)\bigg)$

$\rm (-1)·\bigg(atan \bigg(\dfrac{\sqrt{-x^6+9}}{x^3} \bigg)\bigg)+C$

Answer 2

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm \frac{3 x^{2}}{\sqrt{9-x^{6}}} d x$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm \frac{3 x^{2}}{\sqrt{9- {( {x}^{3}) }^{2} }} d x$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\rm \: {x}^{3} = y$

$\rm\implies \: {3x}^{2} \: dx \: = \: dy \:$

So, on substituting these values in above integral, we get

$\rm \: =  \: \displaystyle\int\rm \frac{dy}{ \sqrt{9 - {y}^{2} } }$

can be further rewritten as

$\rm \: =  \: \displaystyle\int\rm \frac{dy}{ \sqrt{ {3}^{2} - {y}^{2} } }$

We know,

$\boxed{\sf{  \: \: \displaystyle\int\rm \frac{dx}{ \sqrt{ {a}^{2} - {x}^{2} } } \: = \: {sin}^{ - 1} \frac{x}{a} \: + \: c \: }}$

So, using this result, we get

$\rm \: =  \: {sin}^{ - 1}\dfrac{y}{3} + c \:$

$\rm \: =  \: {sin}^{ - 1}\dfrac{ {x}^{3} }{3} + c \:$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \displaystyle\int\rm \frac{3 x^{2}}{\sqrt{9-x^{6}}} d x=  \: {sin}^{ - 1}\dfrac{ {x}^{3} }{3} + c \: }}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$