$\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx = $

  • A
    $\frac{2}{15}$
  • B
    $\frac{4}{15}$
  • C
    $\frac{6}{15}$
  • D
    $\frac{8}{15}$

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मान लीजिए $f(x) = \int_{1}^{x} \sqrt{2 - t^2} dt$ है। तो समीकरण $x^2 - f'(x) = 0$ के वास्तविक मूल हैं

$\int_0^\pi x \sin^7 x \cos^6 x \, dx =$

$\mathop {Limit}\limits_{x \to {x_1}} \,\,\frac{x}{{x - {x_1}}}\,\,\int\limits_{{x_1}}^x {f(t)} \, dt$ का मान ज्ञात कीजिए:

यदि $\int\limits_e^x {t\,f(t)\,dt = \sin x - x\cos x - \frac{{{x^2}}}{2}}$ सभी $x \in R - \{0\}$ के लिए सत्य है,तो $f(\frac{\pi}{6})$ का मान ज्ञात कीजिए।

$\int_0^2 x^{\frac{5}{2}} \sqrt{2-x} \, dx =$

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