Q.Integrate the following function: $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$

Showing the 12 most recent of 74 on this concept.

1. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.$\int \left(\frac{\sin 3x}{\sin x} - \frac{\cos 3x}{\cos x}\right)dx$ is equal to (A) $2\cos 2x + C$ (B) $-2\cos 2x + C$ (C) $2\sin 2x + C$ (D) $2x + C$ (E) $x + C$

   ›Reveal solutionSolution

   Simplify each ratio using triple-angle identities; the integrand is the constant $2$, giving $2x+C$.

   Using $\sin3x=3\sin x-4\sin^3x$ and $\cos3x=4\cos^3x-3\cos x$:

   $\frac{\sin3x}{\sin x}=3-4\sin^2x,\qquad\frac{\cos3x}{\cos x}=4\cos^2x-3.$

   Subtract: …

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2. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.$\int \dfrac{1}{(1 + \cot^2 x)\sin^2 x} \, dx =$ (A) $\tan^{-1}(\sin x) + C$ (B) $\tan^{-1}(\cos x) + C$ (C) $\cot^{-1}(\sin x) + C$ (D) $\cot^{-1}(\cos x) + C$ (E) $x + C$

   ›Reveal solutionSolution

   The integrand simplifies to $1$, so the integral is $x+C$.

   Concept and Intuition

   The Pythagorean identity $1+\cot^2 x=\csc^2 x$ makes the whole denominator collapse to $1$.

   Step-by-Step Solution

   1. $1+\cot^2 x=\csc^2 x=\dfrac{1}{\sin^2 x}$.
   2. Denominator $=(1+\cot^2 x)\sin^2 x=\dfrac{1}{\sin^2 x}\cdot\sin^2 x=1$. …

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3. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.$2^{2}\sin\left(\dfrac{x}{2^{2}}\right)\cos\left(\dfrac{x}{2}\right)\cos\left(\dfrac{x}{2^{2}}\right) =$ (A) $\sin 2x$ (B) $\sin x$ (C) $\cos 2x$ (D) $\cos^{2}x$ (E) $\sin\dfrac{x}{2}$

   ›Reveal solutionSolution

   The expression is $4\sin\frac x4\cos\frac x4\cos\frac x2$; apply $2\sin\theta\cos\theta=\sin2\theta$ twice to get $\sin x$.

   Rewrite. $2^{2}\sin\frac{x}{4}\cos\frac{x}{2}\cos\frac{x}{4}=4\sin\frac{x}{4}\cos\frac{x}{4}\cos\frac{x}{2}$. …

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4. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.$\dfrac{1-\cos 2x}{1+\cos 2x} - \sec^2 x =$ (A) $1$ (B) $\tan 2x$ (C) $\sec 2x$ (D) $0$ (E) $-1$

   ›Reveal solutionSolution

   The first term simplifies to $\tan^2 x$, and $\tan^2 x-\sec^2 x=-1$.

   Using $1-\cos 2x = 2\sin^2 x$ and $1+\cos 2x = 2\cos^2 x$:

   $\frac{1-\cos 2x}{1+\cos 2x}=\frac{2\sin^2 x}{2\cos^2 x}=\tan^2 x.$

   Therefore …

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5. KEAM 2024Set eng-2024-06074 marksMCQ
   Q.$\dfrac{\operatorname{cosec}^2(\theta)-1}{\operatorname{cosec}^2(\theta)}-\dfrac{\sec^2(\theta)-1}{\sec^2(\theta)}=$ (A) $2\cos^2\theta$ (B) $2\cos\theta$ (C) $2\sin^2\theta$ (D) $\cos 2\theta$ (E) $2\sin\theta$

   ›Reveal solutionSolution

   $\operatorname{cosec}^2\theta-1=\cot^2\theta$ and $\sec^2\theta-1=\tan^2\theta$. So the expression is $\dfrac{\cot^2\theta}{\operatorname{cosec}^2\theta}-\dfrac{\tan^2\theta}{\sec^2\theta}=\cos^2\theta-\sin^2\theta=\cos 2\theta$.

   For the first fraction: $\dfrac{\operatorname{cosec}^2\theta-1}{\operatorname{cosec}^2\theta}=\dfrac{\cot^2\theta}{\operatorname{cosec}^2\theta}=\dfrac{\cos^2\theta/\sin^2\theta}{1/\sin^2\theta}=\cos^2\theta.$ …

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6. KEAM 2024Set eng-2024-06074 marksMCQ
   Q.If $7\cos^2 x + 3\sin^2 x = 6$, then the value of $\cos 2x$ is equal to (A) $\dfrac{1}{2}$ (B) $\dfrac{3}{2}$ (C) $\dfrac{5}{2}$ (D) $1$ (E) $2$

   ›Reveal solutionSolution

   Write $7\cos^2x+3\sin^2x=3(\cos^2x+\sin^2x)+4\cos^2x=3+4\cos^2x=6$, so $\cos^2x=\tfrac34$ and $\cos 2x=2\cos^2x-1=\tfrac12$.

   Group terms using $\sin^2x+\cos^2x=1$: …

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7. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.If $\theta = \cot^{-1}\sqrt{\dfrac{1-x}{1+x}}$, then $\sec^2\theta$ (A) $\dfrac{1+x}{2}$ (B) $\dfrac{1-x}{2}$ (C) $\dfrac{2}{1-x}$ (D) $x$ (E) $2x$

   ›Reveal solutionSolution

   $\sec^2\theta = \dfrac{2}{1-x}$.

   Concept and Intuition

   With $\theta=\cot^{-1}\sqrt{(1-x)/(1+x)}$ we know $\cot^2\theta$ directly, and $\sec^2\theta=1+\tan^2\theta$ can be built from $\cot^2\theta$ via reciprocals.

   Step-by-Step Solution

   1. $\cot\theta=\sqrt{\dfrac{1-x}{1+x}}$ so $\cot^2\theta=\dfrac{1-x}{1+x}$.
   2. $\tan^2\theta=\dfrac{1}{\cot^2\theta}=\dfrac{1+x}{1-x}$. …

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8. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.$x = \frac{1+cos2\theta}{tan\theta - sec\theta}$ and $y = \frac{tan\theta + sec\theta}{sec^2\theta}$, then $\frac{y}{x} =$ (A) $\frac{1}{2}$ (B) $2$ (C) $-2$ (D) $\frac{-1}{2}$ (E) $1$

   ›Reveal solutionSolution

   Reduce $x$ and $y$ separately using double-angle and reciprocal identities, then form the ratio; it collapses to a constant $-\tfrac12$.

   Given $x = \dfrac{1+\cos 2\theta}{\tan\theta - \sec\theta}$ and $y = \dfrac{\tan\theta + \sec\theta}{\sec^2\theta}$.

   Simplify $x$: $1+\cos 2\theta = 2\cos^2\theta$ and $\tan\theta - \sec\theta = \dfrac{\sin\theta - 1}{\cos\theta}$.

   $x = \frac{2\cos^2\theta}{(\sin\theta-1)/\cos\theta} = \frac{2\cos^3\theta}{\sin\theta - 1}.$

   Simplify $y$: $\tan\theta + \sec\theta = \dfrac{\sin\theta + 1}{\cos\theta}$ and dividing by $\sec^2\theta$ multiplies by $\cos^2\theta$: …

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9. KEAM 2025Set eng-2025-04294 marksMCQ
   Q.$\dfrac{1 - \sin^6\theta - \cos^6\theta}{\cos^2 2\theta} =$ (A) $\dfrac{1}{4}\tan^2 2\theta$ (B) $\dfrac{1}{2}\tan^2 2\theta$ (C) $\dfrac{3}{2}\tan^2 2\theta$ (D) $\dfrac{3}{4}\tan^2 2\theta$ (E) $\tan^2 2\theta$

   ›Reveal solutionSolution

   Since $\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$, the numerator is $3\sin^2\theta\cos^2\theta=\frac34\sin^2 2\theta$; dividing by $\cos^2 2\theta$ gives $\frac34\tan^2 2\theta$. …

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10. KEAM 2026Set eng-2026-04184 marksMCQ
    Q.The value of $\sin\left(x+\dfrac{7\pi}{4}\right)+\sin\left(x-\dfrac{7\pi}{4}\right)$ is equal to (A) $\sqrt{2}\sin x$ (B) $-\sqrt{2}\cos x$ (C) $-\sqrt{2}\sin x$ (D) $\sqrt{2}\cos x$ (E) $-2\cos x$

    ›Reveal solutionSolution

    Sum-to-product gives $2\sin x\cos\tfrac{7\pi}{4}=\sqrt2\sin x$.

    Using $\sin(A+B)+\sin(A-B)=2\sin A\cos B$ with $A=x$, $B=\tfrac{7\pi}{4}$:

    $2\sin x\cos\tfrac{7\pi}{4}.$ …

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11. KEAM 2024Set eng-2024-06054 marksMCQ
    Q.$\left(1+\cos\dfrac{\pi}{8}\right)\left(1+\cos\dfrac{7\pi}{8}\right)=$ (A) $\dfrac{1}{\sqrt{2}}\left(1+\sqrt{2}\right)$ (B) $\dfrac{1}{2\sqrt{2}}\left(1+\sqrt{2}\right)$ (C) $\dfrac{1}{2\sqrt{2}}\left(\sqrt{2}-1\right)$ (D) $\dfrac{1}{\sqrt{2}}\left(\sqrt{2}-1\right)$ (E) $\dfrac{1}{2}\left(\sqrt{2}+1\right)$

    ›Reveal solutionSolution

    The product becomes $\sin^2\frac{\pi}{8}=\frac{2-\sqrt2}{4}$, which equals $\frac{1}{2\sqrt2}(\sqrt2-1)$.

    Since $\cos\frac{7\pi}{8}=\cos(\pi-\frac{\pi}{8})=-\cos\frac{\pi}{8}$,

    $\left(1+\cos\tfrac{\pi}{8}\right)\left(1-\cos\tfrac{\pi}{8}\right)=1-\cos^2\tfrac{\pi}{8}=\sin^2\tfrac{\pi}{8}.$ …

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12. KEAM 2024Set eng-2024-06074 marksMCQ
    Q.$\operatorname{cosec} x + \cot x =$ (A) $\tan\left(\dfrac{x}{2}\right)$ (B) $\sec\left(\dfrac{x}{2}\right)$ (C) $\cot\left(\dfrac{x}{2}\right)$ (D) $\cos\left(\dfrac{x}{2}\right)$ (E) $\sin\left(\dfrac{x}{2}\right)$

    ›Reveal solutionSolution

    $\operatorname{cosec}x+\cot x=\dfrac{1+\cos x}{\sin x}$. Using $1+\cos x=2\cos^2(x/2)$ and $\sin x=2\sin(x/2)\cos(x/2)$ gives $\cot(x/2)$.

    $\operatorname{cosec}x+\cot x=\dfrac{1}{\sin x}+\dfrac{\cos x}{\sin x}=\dfrac{1+\cos x}{\sin x}$. …

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