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

  • A
    $x \sin ^{-1} x+\sqrt{1-x^2}+c$
  • B
    $\sin ^{-1} x+\sqrt{1-x^2}+c$
  • C
    $x \sin ^{-1} x+c$
  • D
    $\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+c$

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$\int \frac{\tan^{-1} x - \cot^{-1} x}{\tan^{-1} x + \cot^{-1} x} \, dx$ का मान ज्ञात कीजिए।

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$\int(\sqrt{1+\sin (2 x)}) d x=$

List-$I$ की निम्नलिखित वस्तुओं को List-$II$ में सुमेलित कीजिए। सही विकल्प चुनिए।
List-$I$List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$$A. \frac{\tan^2 x}{2} + \ln|\cos x| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
$E. \cos x - \sec x + c$

यदि $f(x) = \int x^2 \cos^2 x (2x \tan^2 x - 2x - 6 \tan x) dx$ और $f(0) = \pi$ है,तो $f(x) =$

समाकलन $\int \frac{\sin \theta \cdot \sin 2 \theta \left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$ का मान ज्ञात कीजिए (जहाँ $c$ समाकलन का एक स्थिरांक है)।

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