Question

Integrate the given function using integration by substitution: 2x sin(x2+ 1) with respect to x:​

Answer 1

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

Given integral is

$\red{\rm :\longmapsto\:\displaystyle\int\rm \: 2x \: sin( {x}^{2} + 1) \: dx \: }$

To evaluate this integral, we use Method of Substitution.

So, Substitute

$\green{\rm :\longmapsto\: {x}^{2} + 1 = y}$

On differentiating both sides, we get

$\green{\rm :\longmapsto\: 2{x}dx = dy}$

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

$\rm \:  =  \: \displaystyle\int\rm \: siny \: dy$

$\rm \:  =  \: - \: cosy \: + \: c$

$\rm \:  =  \: - \: cos( {x}^{2} + 1) \: + \: c$

Therefore,

$\red{\:\boxed{ \tt{ \: \displaystyle\int\rm \: 2x \: sin( {x}^{2} + 1) \: dx = - \: cos( {x}^{2} + 1) + c \: }} }$

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Additional Information :-

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

Answer 2

Answer:

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