$\int_{-\pi}^{\pi} (1-x^2) \sin x \cdot \cos^2 x \, dx$ ની કિંમત શોધો.

જો $I = \int\limits_0^{\frac{\pi}{2}} \ln(\sin x) dx$ હોય,તો $\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln(\sin x + \cos x) dx =$

$\int_{\frac{1}{2}}^2 \frac{1}{x} \operatorname{cosec}^{101}\left(x-\frac{1}{x}\right) d x=$

$\int\limits_0^\pi {\frac{{\sin \left( {n + \frac{1}{2}} \right)x}}{{\sin \frac{x}{2}}}} \,dx$,$(n \in N)$ ની કિંમત શોધો.

જો $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^x} dx = \pi(\alpha \pi^2 + \beta)$,જ્યાં $\alpha, \beta \in \mathbb{Z}$,તો $(\alpha + \beta)^2$ ની કિંમત શોધો:

સંકલન $\int_{-1}^{3} \left( \tan^{-1} \frac{x}{x^2+1} + \tan^{-1} \frac{x^2+1}{x} \right) dx = $