JEE MAIN - Mathematics (2019 - 9th January Morning Slot - No. 17)
For x2 $$ \ne $$ n$$\pi $$ + 1, n $$ \in $$ N (the set of natural numbers), the integral
$$\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx} $$ is equal to :
(where c is a constant of integration)
$$\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx} $$ is equal to :
(where c is a constant of integration)
$${\log _e}\left| {{1 \over 2}{{\sec }^2}\left( {{x^2} - 1} \right)} \right| + c$$
$${1 \over 2}{\log _e}\left| {\sec \left( {{x^2} - 1} \right)} \right| + c$$
$${1 \over 2}{\log _e}\left| {{{\sec }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \right| + c$$
$${\log _e}\left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + c$$
Explanation
$$\int {x\sqrt {{{2\sin \left( {{x^2} - } \right) - \sin 2\left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } \,\,dx$$
$$ = \int {x\sqrt {{{2\sin \left( {{x^2} - 1} \right) - 2sin\left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + 2\sin \left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)}}} } \,\,dx$$
$$ = \int {x\sqrt {{{1 - \cos \left( {{x^2} - 1} \right)} \over {1 + \cos \left( {{x^2} - 1} \right)}}} } \,dx$$
$$ = \int {x\sqrt {{{2{{\sin }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \over {2{{\cos }^2}\left( {{{{x^2} - 1} \over 2}} \right)}}} } \,dx$$
$$ = \int {x\tan } \left( {{{{x^2} - 1} \over 2}} \right)dx$$
put $${{{x^2} - 1} \over 2} = t$$
$$ \Rightarrow {x^2} - 1 = 2t$$
$$ \Rightarrow 2xdx = 2dt$$
$$ \Rightarrow xdx = dt$$
$$ \therefore $$ $$\int {\tan t\,dt} $$
$$ = \ln \left| {\sec t} \right| + C$$
$$ = \ln \left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C$$
$$ = \int {x\sqrt {{{2\sin \left( {{x^2} - 1} \right) - 2sin\left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + 2\sin \left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)}}} } \,\,dx$$
$$ = \int {x\sqrt {{{1 - \cos \left( {{x^2} - 1} \right)} \over {1 + \cos \left( {{x^2} - 1} \right)}}} } \,dx$$
$$ = \int {x\sqrt {{{2{{\sin }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \over {2{{\cos }^2}\left( {{{{x^2} - 1} \over 2}} \right)}}} } \,dx$$
$$ = \int {x\tan } \left( {{{{x^2} - 1} \over 2}} \right)dx$$
put $${{{x^2} - 1} \over 2} = t$$
$$ \Rightarrow {x^2} - 1 = 2t$$
$$ \Rightarrow 2xdx = 2dt$$
$$ \Rightarrow xdx = dt$$
$$ \therefore $$ $$\int {\tan t\,dt} $$
$$ = \ln \left| {\sec t} \right| + C$$
$$ = \ln \left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C$$
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