Question

A Challenge !

$\quad \qquad { \bigstar { \underline { \boxed { \bf { \red { \int \dfrac{x²}{( x Sinx + Cos x )²} dx }}}}}}{\bigstar}$ ​

Answer 1

Answer:

refer to the above attachment....

I hope it helpful to you...

Answer 2

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\sf \frac{ {x}^{2} }{(xsinx + cosx) {}^{2} } \: dx$

Let consider,

$\red{\rm :\longmapsto\:\dfrac{d}{dx}(xsinx + cosx)}$

$\red{\rm \: = \: x\dfrac{d}{dx}sinx + sinx\dfrac{d}{dx}x + \dfrac{d}{dx}cosx}$

$\red{\rm \: = \: xcosx + sinx - sinx}$

$\red{\rm \: = \: xcosx }$

So, we have to make this term in numerator,

So, given integral can be rewritten as

$\rm :\longmapsto\:\displaystyle\int\sf \frac{ {x}^{2} \times cosx}{(xsinx + cosx) {}^{2} } \times \frac{1}{cosx} \: dx$

$\rm \: = \: \:\displaystyle\int\sf \frac{ (xcosx)(xsecx)}{(xsinx + cosx) {}^{2} } \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\sf \frac{(xcosx)}{ {(xsinx + cosx)}^{2} } \times (xsecx) \: dx$

Now, using integration by parts, we get

$\rm= xsecx\displaystyle\int\sf \frac{(xcosx)}{ {(xsinx + cosx)}^{2} }dx - \displaystyle\int\sf \bigg[ \dfrac{d}{dx}xsecx\displaystyle\int\sf \frac{xcosx}{ {(xsinx + cosx)}^{2} }dx\bigg]dx$

Now, Consider

$\red{\rm :\longmapsto\:\dfrac{d}{dx}xsecx}$

$\red{\rm \: = \: x\dfrac{d}{dx}secx + secx\dfrac{d}{dx}x}$

$\red{\rm \: = \: xsecx \: tanx + secx}$

$\red{\rm \: = \: \dfrac{xsinx}{ {cos}^{2} x} + \dfrac{1}{cosx} }$

$\red{\rm \: = \: \dfrac{xsinx + cosx}{ {cos}^{2} x} }$

Now, Consider

$\red{\rm :\longmapsto\:\displaystyle\int\sf \frac{xcosx}{ {(xsinx + cosx)}^{2} }}$

To evaluate this integral, Substitute

$\red{\rm :\longmapsto\:xsinx + cosx = y}$

$\red{\rm :\longmapsto\:\dfrac{d}{dx}(xsinx + cosx) = \dfrac{d}{dx}y}$

$\red{\rm :\longmapsto\:xcosx \: dx \: = \: dy}$

So, above integral can be rewritten as

$\red{\rm \: = \: \displaystyle\int\sf \frac{dy}{ {y}^{2} } }$

$\red{\rm \: = \: \dfrac{ {y}^{ - 2 + 1} }{ - 2 + 1} }$

$\red{\rm \: = \: \dfrac{ {y}^{ - 1} }{ - 1}}$

$\red{\rm \: = \: - \dfrac{1}{y} }$

$\red{\rm \: = \: - \dfrac{1}{xsinx + cosx} }$

So, Now given integral can be reduced to

$\rm =\dfrac{ - xsecx}{xsinx + cosx} + \displaystyle\int\sf \dfrac{xsinx + cosx}{ {cos}^{2} x} \times \dfrac{1}{xsinx + cosx} \: dx$

$\rm =\dfrac{ - xsecx}{xsinx + cosx} + \displaystyle\int\sf {sec}^{2} x \: dx$

$\rm =\dfrac{ - xsecx}{xsinx + cosx} + tanx + c$

Hence,

$\boxed{ \tt{ \: \displaystyle\int\sf \frac{ {x}^{2} }{ {(xsinx + cosx)}^{2} }dx = \dfrac{ - xsecx}{xsinx + cosx} + tanx + c \: }}$

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Additional Information :-

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

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$