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Let $x^2+1 \neq n \pi$ and $2 x^2+1 \neq n \pi \,\forall n \,\in \mathbb{N}$, then $$ \int x\sqrt{\left|\frac{2\sin(x^2+1)-\sin(2x^2+1)}{2\sin(x^2+1)+\sin(2x^2+1)}\right|} dx=? $$ This problem is inspired from a really easy problem from JEE mains standard problems. That problem is as follows:

Let $x^2+1 \neq n \pi$ ,$\,\forall n \,\in \mathbb{N}$, then $$ \int x\sqrt{\left\{\frac{2\sin(x^2+1)-\sin(2(x^2+1))}{2\sin(x^2+1)+\sin(2(x^2+1))}\right\}} dx=? $$ I first mistakenly forgot to apply the set of parentheses in the second term in both numerator and denominator. Later after solving the original problem successfully, I decided to embrace my mistake and explore this new integral. However I tried all possible substitutions I could think of, but I couldn't evaluate with ease. Please assist.

Aditya Naskar
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  • Welcome to MSE. I learned the hard way that you can't just ask people on this site to answer your questions for you without you at least trying it out for yourself first. – Gordon Oct 19 '23 at 01:42
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    Alright sir, I have provided as much context I could. What more should I show? Please assist. – Aditya Naskar Oct 19 '23 at 07:20
  • This question got upvoted which means it is a good enough question already. I'd help you but I'm as lost as you here. – Gordon Oct 19 '23 at 11:06
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    You may simplify a bit by setting $y=x^2+1$. But there is no reason for a general function with sin, cos and square root to have an explicit antiderivative. In the initial problem, the function could be simplified a lot by using trigonometric identities. This situation is exceptional. – Christophe Leuridan Dec 06 '23 at 19:45

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$$ I=\int x\sqrt{\frac{2\sin(x^2+1)-\sin(2(x^2+1))}{2\sin(x^2+1)+\sin(2(x^2+1))}} dx $$ Let $t=x^2+1 \quad\implies\quad dt=2x\,dx$ $$ I=\int \frac12\sqrt{{\frac{2\sin(t)-\sin(2t)}{2\sin(t)+\sin(2t)}}} dx $$ $$ I=\int \frac12\sqrt{{\frac{1-\cos(t)}{1+\cos(t)}}} dx $$ $$ I=\int \frac12\tan\left(\frac{t}{2}\right) dx $$ $$ I=\frac12\ln\left(1+\tan^2 \left(\frac{t}{2}\right)\right)+C $$ $$\boxed{ \int x\sqrt{\frac{2\sin(x^2+1)-\sin(2(x^2+1))}{2\sin(x^2+1)+\sin(2(x^2+1))}} dx = \frac12\ln\left(1+\tan^2 \left(\frac{x^2+1}{2}\right)\right)+C} $$

JJacquelin
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    As mentioned, OP managed to solve the integral you've computed. They are interested in computing the modified integral with sin(2$x^2$+1) instead of sin(2($x^2$+1)). – XXXMATHEMATICION Dec 11 '23 at 19:46