Question

For ${x^2} \ne n\pi + 1$ , $n \in \mathbb{N}$ (the set of natural numbers), the integral
$\int {x\sqrt {\dfrac{{2\sin \left( {{x^2} - 1} \right) - \sin 2\left( {{x^2} - 1} \right)}}{{2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} dx}$ is equal to:
(where c is a constant of integration)
A) ${\log _e}\left| {\dfrac{1}{2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c$
B) $\dfrac{1}{2}{\log _e}\left| {{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c$
C) $\dfrac{1}{2}{\log _e}\left| {{{\sec }^2}\left( {\dfrac{{{x^2} - 1}}{2}} \right)} \right| + c$
D) ${\log _e}\left| {{{\sec }^2}\left( {\dfrac{{{x^2} - 1}}{2}} \right)} \right| + c$

Answer 1

We can give a substitution to the term inside the bracket. Then we can also find its derivative and give substitutions to make the integral in terms of the new variable. Then we can use trigonometric identities to simplify the term inside the integral. Then we can integrate and re-substitute the variable to get the required solution.

Complete step by step solution:
We need to find the integral $\int {x\sqrt {\dfrac{{2\sin \left( {{x^2} - 1} \right) - \sin 2\left( {{x^2} - 1} \right)}}{{2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} dx}$ .
Let $I = \int {x\sqrt {\dfrac{{2\sin \left( {{x^2} - 1} \right) - \sin 2\left( {{x^2} - 1} \right)}}{{2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} dx}$
We can give the substitution $t = \left( {{x^2} - 1} \right)$ … (1)
Then its derivative is given by,
$\Rightarrow \dfrac{{dt}}{{dx}} = 2x$
On rearranging, we get,
$\Rightarrow \dfrac{{dt}}{2} = xdx$ .. (2)
On substituting (1) and (2), It will become,
$\Rightarrow I = \int {\dfrac{1}{2}\sqrt {\dfrac{{2\sin t - \sin 2t}}{{2\sin t + \sin 2t}}} dt}$
We know that $\sin 2A = 2\sin A\cos A$ . On applying this relation on both numerator and denominator of I, we get,
$\Rightarrow I = \int {\dfrac{1}{2}\sqrt {\dfrac{{2\sin t - 2\sin t\cos t}}{{2\sin t + 2\sin t\cos t}}} dt}$
On taking the common factors from both numerator and denominator, we get,
$\Rightarrow I = \int {\dfrac{1}{2}\sqrt {\dfrac{{2\sin t\left( {1 - \cos t} \right)}}{{2\sin t\left( {1 + \cos t} \right)}}} dt}$
On cancelling the common terms, we get,
$\Rightarrow I = \int {\dfrac{1}{2}\sqrt {\dfrac{{\left( {1 - \cos t} \right)}}{{\left( {1 + \cos t} \right)}}} dt}$
We know that $1 - \cos A = 2{\sin ^2}\dfrac{A}{2}$ and $1 + \cos A = 2{\cos ^2}\dfrac{A}{2}$ . So, it will become,
$\Rightarrow I = \int {\dfrac{1}{2}\sqrt {\dfrac{{2{{\sin }^2}\dfrac{t}{2}}}{{2{{\cos }^2}\dfrac{t}{2}}}} dt}$
We know that $\dfrac{{\sin A}}{{\cos A}} = \tan A$ . On applying this relation and cancelling the common terms, we get,
$\Rightarrow I = \int {\dfrac{1}{2}\sqrt {{{\tan }^2}\dfrac{t}{2}} dt}$
On taking the square root, we get,
$\Rightarrow I = \int {\dfrac{1}{2}\tan \dfrac{t}{2}dt}$
We can take the constant outside the integral.
$\Rightarrow I = \dfrac{1}{2}\int {\tan \dfrac{t}{2}dt}$
We know that $\int {\tan x} = \log \left| {\sec x} \right|$
$\Rightarrow I = \dfrac{1}{2}\dfrac{{\log \left| {\sec \dfrac{t}{2}} \right|}}{{\dfrac{1}{2}}}$
On cancelling the common terms, we get,
$\Rightarrow I = \log \left| {\sec \dfrac{t}{2}} \right|$
Now we can re-substitute the value of t. So, we get,
$\Rightarrow I = \log \left| {\sec \dfrac{{\left( {{x^2} - 1} \right)}}{2}} \right|$
On adding the constant of integration, we get,
$\Rightarrow I = \log \left| {\sec \dfrac{{\left( {{x^2} - 1} \right)}}{2}} \right| + c$
Hence the required value of the integral is $\log \left| {\sec \dfrac{{\left( {{x^2} - 1} \right)}}{2}} \right| + c$ .

So, the correct answer is option A.

Note:
We must give the substitution to the whole of the term inside the trigonometric function. The condition ${x^2} \ne n\pi + 1$ is given for the domain because, when ${x^2} = n\pi + 1$ , the denominator of the function will become zero. After integration, we must re-substitute for the variable we substituted. We must also add the constant of integration as it is an indefinite integration.
The trigonometric identities used in this problem are,
$\sin 2A = 2\sin A\cos A$
$1 - \cos A = 2{\sin ^2}\dfrac{A}{2}$
$1 + \cos A = 2{\cos ^2}\dfrac{A}{2}$