For x2 [ ] n[ ] + 1, n [ ] N (the set of natural numbers), the integral [ x 2 (x^2 - 1) - …

Question

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)

Accepted Answer

$$\int {x\sqrt {{{2\sin \left( {{x^2} - } \right) - \sin 2\left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } \,\,dx$$

$$ = \int {x\sqrt {{{2\sin \left( {{x^2} - 1} \right) - 2sin\left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + 2\sin \left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)}}} } \,\,dx$$

$$ = \int {x\sqrt {{{1 - \cos \left( {{x^2} - 1} \right)} \over {1 + \cos \left( {{x^2} - 1} \right)}}} } \,dx$$

$$ = \int {x\sqrt {{{2{{\sin }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \over {2{{\cos }^2}\left( {{{{x^2} - 1} \over 2}} \right)}}} } \,dx$$

$$ = \int {x\tan } \left( {{{{x^2} - 1} \over 2}} \right)dx$$

put   $${{{x^2} - 1} \over 2} = t$$

$$ \Rightarrow {x^2} - 1 = 2t$$

$$ \Rightarrow 2xdx = 2dt$$

$$ \Rightarrow xdx = dt$$

$$ \therefore $$  $$\int {\tan t\,dt} $$

$$ = \ln \left| {\sec t} \right| + C$$

$$ = \ln \left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C$$

Solutions · 1