Evaluate: ∫ (cos 2x)/(sin² xcos² x)dx – Class 12 Mathematics | ConceptFirst

3.1 · Q10

\int \dfrac{\cos 2x}{\sin^2 x\cos^2 x}\,dx

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Use $\cos2x=\cos^2x-\sin^2x$ .

$\dfrac{\cos2x}{\sin^2x\cos^2x}=\dfrac{\cos^2x-\sin^2x}{\sin^2x\cos^2x}=\dfrac{1}{\sin^2x}-\dfrac{1}{\cos^2x}=\csc^2x-\sec^2x$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL2 marks
$\displaystyle\int \sqrt{1+\sin 2x}\cdot dx$
›Reveal solutionSolution

Write $1+\sin2x$ as a perfect square $(\sin x+\cos x)^2$ .

$1+\sin 2x=\sin^2x+\cos^2x+2\sin x\cos x=(\sin x+\cos x)^2$

$\sqrt{1+\sin2x}=\sqrt{(\sin x+\cos x)^2}=|\sin x+\cos x|$

Taking $\sin x+\cos x\ge0$ on the interval considered: …

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CBSE 2023Set ANNUAL2 marksMCQ
$\displaystyle\int \cos^3 x\,dx =$
(a) $\dfrac{1}{12}\sin 3x + \dfrac{3}{4}\sin x + c$
(b) $\dfrac{1}{12}\sin 3x + \dfrac{1}{4}\sin x + c$
(c) $\dfrac{1}{12}\sin 3x - \dfrac{3}{4}\sin x + c$
(d) $\dfrac{1}{12}\sin 3x - \dfrac{1}{4}\sin x + c$
›Reveal solutionSolution

Write $\cos^3x=\cos x(1-\sin^2x)$ and integrate, then convert to the $\sin3x$ form.

$\displaystyle\int\cos^3x\,dx=\int\cos x(1-\sin^2x)\,dx=\sin x-\dfrac{\sin^3x}{3}+c$

Using $\sin3x=3\sin x-4\sin^3x \Rightarrow \sin^3x=\dfrac{3\sin x-\sin3x}{4}$ :

…

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