Let x2≠nπ−1,n∈$$N.The$$ value of ∫x2sin ⁡$$x2+1$$−sin⁡ $$2x2+12sin$$ ⁡$$x2+1+sin$$ ⁡$$2x2+1dx$$ is equal to

log ⁡ 1 2 sec ⁡ x 2 + 1 + C

Questions

Let $x^{2} \neq nπ - 1, n \in N$ .The value of $\int x \sqrt{\frac{2 \sin (x^{2} + 1) - \sin 2 (x^{2} + 1)}{2 \sin (x^{2} + 1) + \sin 2 (x^{2} + 1)}} dx$ is equal to

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\log |\frac{1}{2} \sec (x^{2} + 1)| + C

\log |\sec (\frac{x^{2} + 1}{2})| + C

\frac{1}{2} \log |\sec (x^{2} + 1)| + C

None of the above

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detailed solution

Correct option is B

We have, $\int x \sqrt{\frac{2 \sin (x^{2} + 1) - \sin 2 (x^{2} + 1)}{2 \sin (x^{2} + 1) + \sin 2 (x^{2} + 1)}} dx$

$\begin{array}{l} = \int x \sqrt{\frac{2 \sin (x^{2} + 1) - 2 \sin (x^{2} + 1) \cdot \cos (x^{2} + 1)}{2 \sin (x^{2} + 1) + 2 \sin (x^{2} + 1) \cdot \cos (x^{2} + 1)}} dx \\ = \int x \sqrt{\frac{1 - \cos (x^{2} + 1)}{1 + \cos (x^{2} + 1)}} dx = \int xtan (\frac{x^{2} + 1}{2}) dx \\ = \int \tan (\frac{x^{2} + 1}{2}) d (\frac{x^{2} + 1}{2}) \\ = \log |\sec (\frac{x^{2} + 1}{2})| + C \end{array}$