Q.Evaluate: $\int \tan^2 x\,\sec^4 x\,dx$

Showing the 12 most recent of 32 on this concept.

1. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.$\int \frac{\tan^5 x \; \sec^2 x}{\tan^{12} x + 1}\,dx$ is equal to (A) $\frac{1}{6}\tan^{-1}\left[\tan^6 x\right] + C$ (B) $\frac{1}{2}\tan^{-1}\left[\tan^6 x\right] + C$ (C) $\frac{1}{4}\tan^{-1}\left[\tan^4 x\right] + C$ (D) $\frac{1}{3}\tan^{-1}\left[\tan^3 x\right] + C$ (E) $\frac{1}{7}\tan^{-1}\left[\tan^7 x\right] + C$

   ›Reveal solutionSolution

   Substitute $u=\tan^6 x$; the integral reduces to $\tfrac16\int \frac{du}{u^2+1} = \tfrac16\tan^{-1}(\tan^6 x)+C$.

   Let $u = \tan^6 x$. Then $du = 6\tan^5 x \sec^2 x\,dx$, so $\tan^5 x\sec^2 x\,dx = \tfrac{du}{6}$.

   Also $\tan^{12}x = (\tan^6 x)^2 = u^2$. …

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2. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.$\int_{0}^{\pi/4}\left(\tan^{3} x+\tan^{5} x\right)dx =$ (A) $\frac{5}{12}$ (B) $\frac{1}{3}$ (C) $\frac{1}{4}$ (D) $\frac{1}{6}$ (E) $\frac{1}{12}$

   ›Reveal solutionSolution

   Factor out $\tan^3x\sec^2x$ and substitute $u=\tan x$ to get $\frac{1}{4}$.

   $\tan^3x+\tan^5x=\tan^3x(1+\tan^2x)=\tan^3x\sec^2x$.

   Let $u=\tan x$, $du=\sec^2x\,dx$. Limits: $x=0\to u=0$, $x=\frac{\pi}{4}\to u=1$. …

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3. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.$\displaystyle\int \dfrac{\sec x\sqrt{\sec x}}{\sqrt{\sin x} + \sqrt{\cos x}}\, dx =$ (A) $2\left(\sqrt{\tan x} - \log_e\left(\sqrt{\tan x} + 1\right)\right) + C$ (B) $2\left(\sqrt{\tan x} + \log_e\left(\sqrt{\tan x} + 1\right)\right) + C$ (C) $2\sqrt{\tan x} - \log_e\left(\sqrt{\tan x} + 1\right) + C$ (D) $2\sqrt{\tan x} + \log_e\left(\sqrt{\tan x} + 1\right) + C$ (E) $\sqrt{\tan x} - \log_e\left(\sqrt{\tan x} + 1\right) + C$

   ›Reveal solutionSolution

   Factor $\sqrt{\sin x}+\sqrt{\cos x}=\sqrt{\cos x}(\sqrt{\tan x}+1)$ to reduce the integrand to $\dfrac{\sec^2 x}{\sqrt{\tan x}+1}$; substituting $u=\sqrt{\tan x}$ gives option (A).

   Write the denominator as $\sqrt{\sin x}+\sqrt{\cos x}=\sqrt{\cos x}\bigl(\sqrt{\tan x}+1\bigr)$. The numerator is $\sec x\sqrt{\sec x}=\cos^{-3/2}x$, so

   $\dfrac{\cos^{-3/2}x}{\sqrt{\cos x}\,(\sqrt{\tan x}+1)}=\dfrac{\cos^{-2}x}{\sqrt{\tan x}+1}=\dfrac{\sec^2 x}{\sqrt{\tan x}+1}.$

   Let $t=\tan x$, $dt=\sec^2 x\,dx$, then $u=\sqrt t$, $t=u^2$, $dt=2u\,du$: …

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4. KEAM 2024Set eng-2024-06094 marksMCQ
   Q.$\int \dfrac{dx}{x^2 (x^4 + 1)^{3/4}} =$ (A) $-(x^4 + 1)^{1/4} + C$ (B) $(x^4 + 1)^{1/4} + C$ (C) $-\left(\dfrac{x^4 + 1}{x^4}\right)^{1/4} + C$ (D) $\left(\dfrac{x^4 + 1}{x^4}\right) + C$ (E) $\left(\dfrac{x^4 + 1}{x^4}\right)^{3/4} + C$

   ›Reveal solutionSolution

   Factor $x^4$ out of the radical; with $u=1+x^{-4}$, $du=-4x^{-5}dx$, the integral becomes $-u^{1/4}=-\left(\frac{x^4+1}{x^4}\right)^{1/4}+C$.

   Write $(x^4+1)^{3/4}=x^3(1+x^{-4})^{3/4}$, so

   $\int\frac{dx}{x^2(x^4+1)^{3/4}}=\int\frac{dx}{x^5(1+x^{-4})^{3/4}}.$ …

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5. KEAM 2025Set eng-2025-04294 marksMCQ
   Q.$\int \dfrac{dx}{\cos x \sqrt{2 \sin 2x}} =$ (A) $\dfrac{1}{2}\sqrt{\tan x} + C$ (B) $\sqrt{\tan x} + C$ (C) $2\sqrt{\tan x} + C$ (D) $4\sqrt{\tan x} + C$ (E) $3\sqrt{\tan x} + C$

   ›Reveal solutionSolution

   Writing $2\sin2x=4\sin x\cos x$ reduces the integrand to $\frac12\sec^2x\,(\tan x)^{-1/2}$; with $u=\tan x$ this integrates to $\sqrt{\tan x}+C$.

   Use $\sin 2x=2\sin x\cos x$, so $2\sin 2x=4\sin x\cos x$ and $\sqrt{2\sin 2x}=2\sqrt{\sin x\cos x}$. Then …

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6. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.$\int\dfrac{\sec^2\left(\sqrt{2x+5}\right)}{\sqrt{2x+5}}\,dx=$ (A) $2\tan\left(\sqrt{2x+5}\right)+C$ (B) $\dfrac{1}{2}\tan\left(\sqrt{2x+5}\right)+C$ (C) $\tan\left(2x+5\right)+C$ (D) $\tan\left(\sqrt{2x+5}\right)+C$ (E) $2\tan\left(2x+5\right)+C$

   ›Reveal solutionSolution

   Substitute $u=\sqrt{2x+5}$; the differential cancels the $\dfrac{1}{\sqrt{2x+5}}$ factor.

   Let $u=\sqrt{2x+5}$. Then

   $du=\dfrac{1}{2\sqrt{2x+5}}\cdot 2\,dx=\dfrac{dx}{\sqrt{2x+5}}.$

   So the integral becomes …

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7. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.$\displaystyle\int \dfrac{\sec x}{(\sec x + \tan x)^9} \, dx =$ (A) $\dfrac{1}{9}(\sec x + \tan x)^9 + C$ (B) $\dfrac{-1}{9}(\sec x + \tan x)^9 + C$ (C) $\dfrac{-1}{9}(\sec x + \tan x)^{-9} + C$ (D) $\dfrac{1}{9}(\sec x + \tan x)^{-9} + C$ (E) $(\sec x + \tan x)^{-9} + C$

   ›Reveal solutionSolution

   Substitute $u=\sec x+\tan x$; then $\sec x\,dx=du/u$.

   Let $u=\sec x+\tan x$. Then $du=(\sec x\tan x+\sec^2x)\,dx=\sec x(\tan x+\sec x)\,dx=\sec x\,u\,dx$, so $\sec x\,dx=\dfrac{du}{u}$. …

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8. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.$\int \frac{\sec x}{(\sec x + \tan x)^2}\,dx =$ (A) $\frac{2}{5(\sec x + \tan x)^4} + C$ (B) $\frac{-1}{2(\sec x + \tan x)^2} + C$ (C) $\frac{2}{3(\sec x + \tan x)^{3/2}} + C$ (D) $\frac{-2}{3(\sec x + \tan x)^3} + C$ (E) $(\sec x + \tan x)^2 + C$

   ›Reveal solutionSolution

   The substitution $u=\sec x+\tan x$ turns it into $\int u^{-3}\,du$.

   Let $u=\sec x+\tan x$. Then $du=(\sec x\tan x+\sec^2 x)\,dx=\sec x(\tan x+\sec x)\,dx=\sec x\,u\,dx$, so $\sec x\,dx=\dfrac{du}{u}$. Hence …

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9. KEAM 2025Set eng-2025-04284 marksMCQ
   Q.$\int \frac{dx}{\cos^{2/3}x \, \sin^{4/3}x}$ is (A) $3\tan^3 x + C$ (B) $3\tan^{1/3} x + C$ (C) $-3\tan^{1/3} x + C$ (D) $-3\tan^{-1/3} x + C$ (E) $3\tan^{-1/3} x + C$

   ›Reveal solutionSolution

   Rewrite as $\sec^2x\,\tan^{-4/3}x\,dx$; sub $t=\tan x$ to get $\int t^{-4/3}dt=-3t^{-1/3}=-3\tan^{-1/3}x+C$.

   The integrand $\dfrac{1}{\cos^{2/3}x\,\sin^{4/3}x}$ has denominator powers summing to $2$, so factor out $\cos^2x$. Multiplying numerator and denominator by $\cos^{4/3}x$ (equivalently dividing top and bottom by $\cos^2x$): …

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10. KEAM 2025Set eng-2025-04264 marksMCQ
    Q.$\displaystyle\int_0^{\pi/2} \dfrac{1}{1+\sin x}\, dx =$ (A) $2$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{4}$ (D) $1$ (E) $0$

    ›Reveal solutionSolution

    Rationalise the denominator, integrate $\sec^2x-\sec x\tan x$.

    Multiply numerator and denominator by $1-\sin x$:

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

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11. KEAM 2025Set eng-2025-04234 marksMCQ
    Q.$\int x^7 (x^8 + 1)^{-3/4}\, dx =$ (A) $\dfrac{1}{2}\left(1 + \dfrac{1}{x^8}\right)^{1/4} + C$ (B) $4\left(1 + \dfrac{1}{x^8}\right)^{1/4} + C$ (C) $(x^8 + 1)^{1/4} + C$ (D) $4(x^8 + 1)^{1/4} + C$ (E) $\dfrac{1}{2}(x^8 + 1)^{1/4} + C$

    ›Reveal solutionSolution

    Substituting u=x^8+1 gives (1/2)(x^8+1)^{1/4} + C.

    Concept and Intuition

    The presence of x^7 alongside x^8 signals the substitution u = x^8 + 1, whose differential absorbs x^7 dx.

    Step-by-Step Solution

    1. Let u = x^8 + 1, then du = 8 x^7 dx, so x^7 dx = du/8.
    2. Integral = (1/8) integral of u^{-3/4} du.
    3. = (1/8) * u^{1/4}/(1/4) = (1/8)4u^{1/4} = (1/2)u^{1/4}. …

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12. KEAM 2022Set eng-2022-P2-B14 marksMCQ
    Q.$\int \frac{\sin^{25} x}{\cos^{27} x} \, dx$ is equal to (A) $\frac{\sin^{26}(x)}{26} + C$ (B) $\frac{\cos^{26}(x)}{26} + C$ (C) $\tan^{26}(x) + C$ (D) $\frac{\tan^{26}(x)}{26} + C$ (E) $26\tan^{26}(x) + C$

    ›Reveal solutionSolution

    The integral equals $\frac{\tan^{26}x}{26}+C$.

    Concept and Intuition

    Split off a $\sec^2x$ factor so the rest becomes a power of $\tan x$, then substitute $u=\tan x$.

    Step-by-Step Solution

    1. $\frac{\sin^{25}x}{\cos^{27}x}=\tan^{25}x\cdot\frac{1}{\cos^2x}=\tan^{25}x\,\sec^2x$.
    2. Let $u=\tan x\Rightarrow du=\sec^2x\,dx$.
    3. $\int \tan^{25}x\,\sec^2x\,dx=\int u^{25}\,du=\frac{u^{26}}{26}+C$.
    4. $=\frac{\tan^{26}x}{26}+C$.

    Common Mistakes …

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