Question

If ${x^2} \ne n\pi + 1,n \in N$, then $\int {x\sqrt {\dfrac{{2\sin \left( {{x^2} - 1} \right) - \sin 2\left( {{x^2} - 1} \right)}}{{2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } dx$ is equal to
1. $\ln \left| {\cos \left( {\dfrac{{{x^2} - 1}}{2}} \right)} \right| + c$
2. $\dfrac{1}{2}\ln \left| {\cos \left( {\dfrac{{{x^2} - 1}}{2}} \right)} \right| + c$
3. $\ln \left| {\sec \left( {\dfrac{{{x^2} - 1}}{2}} \right)} \right| + c$
4. $\dfrac{1}{2}\ln \left| {\sec \left( {\dfrac{{{x^2} - 1}}{2}} \right)} \right| + c$

Answer 1

Hint: In this question, we are given the term to integrate $\int {x\sqrt {\dfrac{{2\sin \left( {{x^2} - 1} \right) - \sin 2\left( {{x^2} - 1} \right)}}{{2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } dx$, where ${x^2} \ne n\pi + 1,n \in N$. First, step is to let the angle \[{x^2} - 1 = p\](or any variable). The differentiate it and put the values in given integral. Apply trigonometric identities and you will get the direct trigonometric function for integration. In last, apply the integration formula.

Formula Used:
Trigonometric identity –
$\sin 2A = 2\sin A\cos A$
${\tan ^2}A = \dfrac{{1 - \cos 2A}}{{1 + \cos 2A}}$
Integration formula of trigonometric function –
$\int {\tan xdx = \log \left| {\sec x} \right| + c} $

Complete step by step Solution:
Let, the given integral be equal to $I$
Therefore, $\int {x\sqrt {\dfrac{{2\sin \left( {{x^2} - 1} \right) - \sin 2\left( {{x^2} - 1} \right)}}{{2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } dx - - - - - \left( 1 \right)$
Let, \[{x^2} - 1 = p - - - - - \left( 2 \right)\]
Differentiate equation (2) with respect to $x$,
\[2x = \dfrac{{dp}}{{dx}}\]
\[xdx = \dfrac{{dp}}{2}\]
Put above value in equation (1),
\[I = \int {\sqrt {\dfrac{{2\sin p - \sin 2p}}{{2\sin p + \sin 2p}}} } \dfrac{{dp}}{2}\]
Using trigonometric identity, $\sin 2A = 2\sin A\cos A$
\[ = \dfrac{1}{2}\int {\sqrt {\dfrac{{2\sin p - 2\sin p\cos p}}{{2\sin p + 2\sin p\cos p}}} } dp\]
\[ = \dfrac{1}{2}\int {\sqrt {\dfrac{{2\sin p\left( {1 - \cos p} \right)}}{{2\sin p\left( {1 + \cos p} \right)}}} } dp\]
\[ = \dfrac{1}{2}\int {\sqrt {\dfrac{{\left( {1 - \cos p} \right)}}{{\left( {1 + \cos p} \right)}}} } dp\]
Now, again using a trigonometric identity ${\tan ^2}A = \dfrac{{1 - \cos 2A}}{{1 + \cos 2A}}$
\[ = \dfrac{1}{2}\int {\sqrt {{{\tan }^2}\dfrac{p}{2}} } dp\]
\[ = \dfrac{1}{2}\int {\tan \dfrac{p}{2}} dp\]
Integrate the trigonometric function using integration formula $\int {\tan xdx = \log \left| {\sec x} \right| + c} $
\[ = \dfrac{1}{2}\left[ {\dfrac{{\log \left| {\sec \dfrac{p}{2}} \right|}}{{\left( {\dfrac{1}{2}} \right)}}} \right] + c\]
From equation (2),
\[ = \ln \left| {\sec \left( {\dfrac{{{x^2} - 1}}{2}} \right)} \right| + c\]
It implies that, $\int {x\sqrt {\dfrac{{2\sin \left( {{x^2} - 1} \right) - \sin 2\left( {{x^2} - 1} \right)}}{{2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } dx = \ln \left| {\sec \left( {\dfrac{{{x^2} - 1}}{2}} \right)} \right| + c$

Hence, the correct option is 3.

Note:To solve such problems one should have a good knowledge of trigonometric and integration formulas. Also, when there’s a function inside the function always apply chain rule like we do in differentiation after doing the differentiation of whole function we multiply the differentiation of inner function. But in this we integrate the function and then divide the require term by the differentiation of inner part. It can be written as $\int {f\left( {g\left( x \right)} \right)dx = \dfrac{{\int {f\left( {g\left( x \right)} \right)dx} }}{{\dfrac{d}{{dx}}g\left( x \right)}}} $. Also, if limits would be there in integration then, to solve the limits in integration. First integrate the whole function then subtract the integration using lower limit from the integration using upper limit. For example, the required integration is $2x$ and we have the limit $2$to $4$ then the answer will be $2\left( 4 \right) - 2\left( 2 \right) = 4$.