$\int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ का मान ज्ञात कीजिए।

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}} d x=$

$\int_{\pi / 5}^{3 \pi / 10} \frac{d x}{\sec ^2 x+\left(\tan ^{2022} x-1\right)\left(\sec ^2 x-1\right)}=$

$\int_{-1}^{1} \log(x + \sqrt{x^2 + 1}) \, dx = $

मान लीजिए $f(x) = 7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$ सभी $x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ के लिए है। तो सही व्यंजक है/हैं:
$(A) \int_0^{\pi/4} x f(x) dx = \frac{1}{12}$
$(B) \int_0^{\pi/4} f(x) dx = 0$
$(C) \int_0^{\pi/4} x f(x) dx = \frac{1}{6}$
$(D) \int_0^{\pi/4} f(x) dx = 1$

यदि $I_n = \int_{-\pi}^{\pi} \frac{\sin(nx)}{(1+\pi^x) \sin x} dx$,$n=0, 1, 2, \ldots$,तो
$(A)$ $I_n = I_{n+2}$
$(B)$ $\sum_{m=1}^{10} I_{2m+1} = 10\pi$
$(C)$ $\sum_{m=1}^{10} I_{2m} = 0$
$(D)$ $I_n = I_{n+1}$