જો $f(x) = \int x^2 \cos^2 x (2x \tan^2 x - 2x - 6 \tan x) dx$ અને $f(0) = \pi$ હોય,તો $f(x) =$

  • A
    $x^2 \sin x + \pi$
  • B
    $\cos x + \pi - 1$
  • C
    $-x^3 \sin 2x + \pi$
  • D
    $x^3 \cos 2x + \pi \cos x$

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$\int x^{2} [ \sqrt{2} \sin ( \frac{\pi}{4} + x ) + e^{x} ] dx =$

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$\int \frac{\sqrt{1-x^2} \sin ^{-1} x+x}{\sqrt{1-x^2}} d x=$

સંકલન $\int \frac{\sin \theta \cdot \sin 2 \theta \left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$ નું મૂલ્ય શોધો (જ્યાં $c$ એ સંકલનનો અચળાંક છે).

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