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

Question

For $x^2
e n\pi+1$, $n\in N$ (the set of natural numbers), the integral $\displaystyle\int\frac{x\left[2\sin(x^2-1)-\sin2(x^2-1)ight]}{2\sin(x^2-1)+\sin2(x^2-1)}\,dx$ is equal to:

Accepted Answer

Using $\sin2\phi=2\sin\phi\cos\phi$ with $\phi=x^2-1$:
$$I=\int\frac{x\left[1-\cos(x^2-1)\right]}{1+\cos(x^2-1)}dx$$

Let $\dfrac{x^2-1}2=t\Rightarrow x\,dx=dt$ (using half-angle: $1-\cos\phi=2\sin^2(\phi/2)$, $1+\cos\phi=2\cos^2(\phi/2)$):
$$I=\int\tan t\,dt=\ln|\sec t|+C=\ln\left|\sec\frac{x^2-1}2\right|+C=\frac12\ln\left|\sec^2\frac{x^2-1}2\right|+C$$

Solutions · 1