Question

If $x^2\ne n\pi +1$, $n\in N$, then $\int x\sqrt{\frac{2\sin \left(x^2-1\right)-\sin 2\left(x^2-1\right)}{2\sin \left(x^2-1\right)+\sin 2\left(x^2-1\right)}}dx$ is equal to:

• A. $\ln \cos \left(\frac{x^2-1}{2}\right)+c$
• B. $\frac{1}{2}\ln \cos \left(\frac{x^2-1}{2}\right)+c$
• C. $\ln \sec \left(\frac{x^2-1}{2}\right)+c$
• D. $\frac{1}{2}\ln \sec \left(\frac{x^2-1}{2}\right)+c$

Answer 1

The correct option is C

$\ln \sec \left(\frac{x^2-1}{2}\right)+c$

Explanation for the correct option:

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

Let $t=x^2-1$

$$\begin{gathered} \Rightarrow \frac{dt}{dx}=2x \\ \Rightarrow \frac{dt}{2}=xdx \end{gathered}$$

So,

$\begin{array}{rcl}\int x\sqrt{\frac{2\sin \left(x^2-1\right)-\sin 2\left(x^2-1\right)}{2\sin \left(x^2-1\right)+\sin 2\left(x^2-1\right)}}dx& =& \frac{1}{2}\int \sqrt{\frac{2\sin t-\sin 2t}{2\sin t+\sin 2t}}dt\\ & =& \frac{1}{2}\int \sqrt{\frac{2\sin t-2\sin t\cos t}{2\sin t+2\sin t\cos t}}dt\left[bysin2x=sinxcosx\right]\\ & =& \frac{1}{2}\int \sqrt{\frac{2\sin t\left(1-\cos t\right)}{2\sin t\left(1+\cos t\right)}}dt\\ & =& \frac{1}{2}\int \sqrt{\frac{1-\cos t}{1+\cos t}}dt\\ & =& \frac{1}{2}\int \sqrt{\frac{2{\sin }^2{\displaystyle \raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2{\cos }^2\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}dt\left[by2si{n}^{2}x=1-cos2xand2co{s}^{2}x=1+cos2x\right]\\ & =& \frac{1}{2}\int \tan \raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.dt\\ & =& \log \left|sec\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right|+c\\ & =& \log \left|sec\frac{\left(x^2-1\right)}{2}\right|+c\end{array}$

Hence, option C is correct.