Question

p please solve the problem and give complete solution of it.​

Answer 1

By seeing this u understand properly

Answer 2

$\large\underline{\sf{Given \:Question - }}$

$\sf \: If \: y = {x}^{2} \: sin( {x}^{3}), \: find \: \displaystyle \int \rm \:y \: dx$

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

Given that

$\rm :\longmapsto\: \sf \: \: y = {x}^{2} \: sin( {x}^{3})$

On integrating both sides, w. r. x, we get

$\rm :\longmapsto\: \sf \: \: \displaystyle \int \rm \:y \: dx= \displaystyle \int \rm \:{x}^{2} \: sin( {x}^{3}) \: dx$

So integrate such integral, we use method of Substitution.

So,

Let we substitute,

$\red{\rm :\longmapsto\: {x}^{3} = t}$

$\red{\rm :\longmapsto\: 3{x}^{2}dx = dt}$

$\red{\rm :\longmapsto\: {x}^{2}dx = \dfrac{dt}{3} }$

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

$\rm :\longmapsto\: \sf \: \: \displaystyle \int \rm \:y \: dx= \displaystyle \int \rm \: \: sin( t) \: \frac{dt}{3}$

$\rm :\longmapsto\: \sf \: \: \displaystyle \int \rm \:y \: dx= \frac{1}{3} \displaystyle \int \rm \: \: sin( t) \: dt$

We know that

$\boxed{ \sf \: \displaystyle \int \rm \:sinx \: dx \: = \: - \: cosx \: + \: c}$

So, using this we get

$\rm :\longmapsto\: \sf \: \: \displaystyle \int \rm \:y \: dx= \frac{1}{3}( - \: cost) + c$

$\rm :\longmapsto\: \sf \: \: \displaystyle \int \rm \:y \: dx= \: - \: \frac{1}{3}cost + c$

$\bf :\longmapsto\: \bf \: \: \displaystyle \int \bf \:y \: dx= \: - \: \frac{1}{3}cos {x}^{3} + c$

Additional Information :-

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