$\int_{ - 3}^3 {\frac{{{x^2}\sin 2x}}{{{x^2} + 1}}\,dx = } $

  • A
    $0$
  • B
    $1$
  • C
    $2{\log _e}3$
  • D
    इनमें से कोई नहीं

Explore More

Similar Questions

$\sum_{n=1}^{10} \int_{-2n-1}^{-2n} \sin^{27} x \, dx + \sum_{n=1}^{10} \int_{2n}^{2n+1} \sin^{27} x \, dx$ का मान ज्ञात कीजिए।

माना $u = \int\limits_0^1 {\frac{{\ln (x + 1)}}{{{x^2} + 1}}} \,dx$ और $v = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin 2x)} \,dx$,तो:

यदि $I_{1} = \int_{0}^{1} (1 - x^{50})^{100} dx$ और $I_{2} = \int_{0}^{1} (1 - x^{50})^{101} dx$ इस प्रकार है कि $I_{2} = \alpha I_{1}$,तो $\alpha$ का मान ज्ञात कीजिए:

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} x \tan \left(1+x^2\right) d x$ का मान ज्ञात कीजिए।

मान लीजिए $f(x)$,$(a, b)$ पर समाकलनीय है,जहाँ $b > a > 0$ है। यदि $I_1 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \sec^2 \theta \, d\theta$ और $I_2 = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} f(\tan \theta + \cot \theta) \csc^2 \theta \, d\theta$ है,तो अनुपात $\frac{I_1}{I_2}$ है:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo