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Question

Question: How do you find the Integral \[x\sin {x^2}dx\] ?...

How do you find the Integral xsinx2dxx\sin {x^2}dx ?

Explanation

Solution

Indefinite integral simply represents the area under a given curve without any boundary conditions. So here by using this basic definition we can integrate xsinx2dxx\sin {x^2}dx.Also we know one of the basic identity: sinxdx=cosx+C\int {\sin xdx = - \cos x + C} .The above expression and equation can be used to integrate xsinx2dxx\sin {x^2}dx.

Complete step by step answer:
Given, xsinx2dx.........................(i)x\sin {x^2}dx.........................\left( i \right)
Also by the basic definition of indefinite integral we can write that:
Indefinite integral is given by: f(x)dx\int {f\left( x \right)dx}
Such to integrate xsinx2dxx\sin {x^2}dx we can write
xsinx2dx..........................(ii)\int {x\sin {x^2}dx} ..........................\left( {ii} \right)
Now on observing (i) we can say that the term xsinx2dxx\sin {x^2}dx cannot be integrated directly such that let’s assume:
x2=t................................(iii){x^2} = t................................\left( {iii} \right)
Now let’s differentiate equation (ii) and find the value of xdxxdx such that we can substitute it in (i):
So we get:

\Rightarrow 2xdx = dt \\\ \Rightarrow xdx = \dfrac{{dt}}{2}......................\left( {iv} \right) \\\ $$ Now let’s substitute all the above values in (i): So we can write: $$\int {x\sin {x^2}dx} = \int {\sin \left( t \right)\dfrac{{dt}}{2}} ..........................\left( v \right)$$ Now to integrate (v) we can use one of the basic trigonometric property to integrate sine function which is: $\int {\sin xdx = - \cos x + C} $ Here but instead of $x$ we have$t$. So we can write: $$\int {x\sin {x^2}dx} = \int {\sin \left( t \right)\dfrac{{dt}}{2}} \\\ \Rightarrow\int {x\sin {x^2}dx}= - \dfrac{1}{2}\cos t + C...............\left( {vi} \right) \\\ $$ Now let’s substitute the value of $t$ back in the equation (vi) such that we can write: $$\int {x\sin {x^2}dx} = - \dfrac{1}{2}\cos t + C \\\ \Rightarrow\int {x\sin {x^2}dx}= - \dfrac{1}{2}\cos {x^2} + C \\\ \therefore\int {x\sin {x^2}dx}= - \dfrac{{\cos {x^2}}}{2} + C.......................\left( {vii} \right) \\\ $$ **Therefore from (vii) we can write that on integrating $$x\sin {x^2}dx$$ we get $$ - \dfrac{{\cos {x^2}}}{2} + C$$.** **Note:** Since the basic definition indefinite integral simply implies the area under a curve such that the value of an integral must be finite or else the integral doesn’t exist. Also can’t integrate an expression directly then we have to use the formula for integration by parts which is given by: $\int {udv = uv - \int {vdu} } $