Q.Integrate the function $\sqrt{4-x^2}$

Showing the 12 most recent of 32 on this concept.

1. KEAM 2024Set eng-2024-06064 marksMCQ
   Q.$\int\frac{4x\cos\sqrt{4x^{2}+7}}{\sqrt{4x^{2}+7}}\,dx =$ (A) $\frac{1}{2}\sin\sqrt{4x^{2}+7}+C$ (B) $\frac{7}{2}\sin\sqrt{4x^{2}+7}+C$ (C) $\sin\sqrt{4x^{2}+7}+C$ (D) $\frac{1}{4}\sin\sqrt{4x^{2}+7}+C$ (E) $\frac{7}{4}\sin\sqrt{4x^{2}+7}+C$

   ›Reveal solutionSolution

   With $u=\sqrt{4x^2+7}$ the integrand is exactly $\cos u\,du$, giving $\sin\sqrt{4x^2+7}+C$.

   Let $u=\sqrt{4x^2+7}$. Then $\frac{du}{dx}=\frac{8x}{2\sqrt{4x^2+7}}=\frac{4x}{\sqrt{4x^2+7}}$, so $du=\frac{4x}{\sqrt{4x^2+7}}\,dx$.

   The integral becomes $\int\cos u\,du=\sin u+C=\sin\sqrt{4x^2+7}+C$.

   ✓Final answer

   The correct option is (C).

2. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.$\int \frac{\sin(\cot^{-1}x)}{1 + x^2} dx$ is equal to (A) $-\cos(\cot^{-1}x) + C$ (B) $\cos(\cot^{-1}x) + C$ (C) $\frac{\cos(\cot^{-1}x)}{1 + x^2} + C$ (D) $\frac{\cos(\cot^{-1}x)}{2} + C$ (E) $\frac{-\cos(\cot^{-1}x)}{1 + x^2} + C$

   ›Reveal solutionSolution

   Substitute $u=\cot^{-1}x$ so $du=-\tfrac{dx}{1+x^2}$.

   Let $u=\cot^{-1}x\Rightarrow du=-\dfrac{dx}{1+x^2}$, i.e. $\dfrac{dx}{1+x^2}=-du$.

   $\int\frac{\sin(\cot^{-1}x)}{1+x^2}\,dx=\int\sin u\,(-du)=-\left(-\cos u\right)+C=\cos u+C.$

   Back-substituting, $=\cos(\cot^{-1}x)+C$.

   ✓Final answer

   The correct option is (B).

3. 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

   $\frac{1}{\cos x\sqrt{2\sin 2x}}=\frac{1}{2\cos x\sqrt{\sin x\cos x}}=\frac{1}{2\cos^2 x\sqrt{\tan x}}=\frac{\sec^2 x}{2\sqrt{\tan x}}.$

   Let $u=\tan x$, $du=\sec^2 x\,dx$:

   $\frac12\int u^{-1/2}\,du=\frac12\cdot2u^{1/2}=\sqrt{u}=\sqrt{\tan x}+C.$

   ✓Final answer

   The correct option is (B).

4. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.$\int \dfrac{\sin^{-1} x}{\sqrt{1 - x^2}}\, dx =$ (A) $\dfrac{1}{2}(\sin^{-1} x)^2 + C$ (B) $-(\sin^{-1} x)\sqrt{1 - x^2} + C$ (C) $(\sin^{-1} x)\sqrt{1 - x^2} + x + C$ (D) $(\sin^{-1} x)\sqrt{1 - x^2} - x + C$ (E) $(\sin^{-1} x)^2 + C$

   ›Reveal solutionSolution

   Substituting u=sin^{-1}x turns the integral into u du = (1/2)(sin^{-1}x)^2 + C.

   Concept and Intuition

   The factor 1/sqrt(1-x^2) is exactly the derivative of sin^{-1}x, so the substitution u = sin^{-1}x collapses the integral.

   Step-by-Step Solution

   1. Let u = sin^{-1}x, then du = dx/sqrt(1-x^2).
   2. The integral becomes integral of u du.
   3. That equals u^2/2 + C = (1/2)(sin^{-1}x)^2 + C.

   Common Mistakes

   • Forgetting the factor 1/2 from integrating u.

   ✓Final answer

   The correct option is (A) — (1/2)(sin^{-1}x)^2 + C.

   ANSWER: A

5. KEAM 2022Set eng-2022-P2-B14 marksMCQ
   Q.$\int \frac{1}{x^3}\sqrt{1 - \frac{1}{x^2}} \, dx =$ (A) $\frac{-1}{6}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (B) $\frac{1}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (C) $\frac{-1}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (D) $\frac{4}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$ (E) $\frac{-4}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$

   ›Reveal solutionSolution

   The integral equals $\frac13\left(1-\frac1{x^2}\right)^{3/2}+C$.

   Concept and Intuition

   The derivative of $1-\frac1{x^2}$ is $\frac2{x^3}$, which matches the $\frac1{x^3}$ factor outside the root, so a substitution linearizes the integral.

   Step-by-Step Solution

   1. Let $u=1-\frac1{x^2}$.
   2. $du=\frac{2}{x^3}\,dx\Rightarrow \frac{dx}{x^3}=\frac{du}{2}$.
   3. $\int \frac{1}{x^3}\sqrt{1-\frac1{x^2}}\,dx=\int \sqrt u\,\frac{du}{2}=\frac12\cdot\frac{u^{3/2}}{3/2}$.
   4. $=\frac12\cdot\frac23 u^{3/2}=\frac13 u^{3/2}=\frac13\left(1-\frac1{x^2}\right)^{3/2}+C$.

   Common Mistakes

   • Getting the sign or factor of $du$ wrong (it is $+\frac2{x^3}$).
   • Forgetting to halve after substituting.

   ✓Final answer

   The correct option is (B) — $\frac{1}{3}\left(1 - \frac{1}{x^2}\right)^{\frac{3}{2}} + C$.

   ANSWER: B

6. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.$\displaystyle\int \dfrac{\sqrt{\sqrt{x} + 1}}{\sqrt{x}}\, dx =$ (A) $\dfrac{4}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C$ (B) $\dfrac{2}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C$ (C) $\dfrac{4}{3}(\sqrt{x} + 1)^{\frac{3}{4}} + C$ (D) $\dfrac{1}{3}(\sqrt{x} + 1)^{\frac{3}{2}} + C$ (E) $\dfrac{3}{4}(\sqrt{x} + 1)^{\frac{3}{2}} + C$

   ›Reveal solutionSolution

   Substitute $u=\sqrt{x}+1$.

   With $u=\sqrt{x}+1$, $du=\dfrac{1}{2\sqrt{x}}\,dx$, i.e. $\dfrac{dx}{\sqrt{x}}=2\,du$. Then

   $\int\frac{\sqrt{\sqrt{x}+1}}{\sqrt{x}}\,dx=\int\sqrt{u}\cdot 2\,du=2\cdot\frac{2}{3}u^{3/2}+C=\frac{4}{3}(\sqrt{x}+1)^{3/2}+C.$

   ✓Final answer

   The correct option is (A).

7. 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

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

   Let $u=1+x^{-4}$, then $du=-4x^{-5}\,dx$, i.e. $x^{-5}dx=-\tfrac14\,du$:

   $\int-\tfrac14 u^{-3/4}\,du=-u^{1/4}+C=-(1+x^{-4})^{1/4}+C=-\left(\frac{x^4+1}{x^4}\right)^{1/4}+C.$

   ✓Final answer

   The correct option is (C).

8. KEAM 2025Set eng-2025-04274 marksMCQ
   Q.$\int \dfrac{\cos\theta}{2 - \sin^2\theta}\,d\theta =$ (A) $\dfrac{1}{2}\log\left|\dfrac{\sqrt{2} - \sin\theta}{\sqrt{2} + \sin\theta}\right| + C$ (B) $\dfrac{1}{2}\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$ (C) $\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$ (D) $\dfrac{1}{\sqrt{2}}\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$ (E) $\dfrac{1}{2\sqrt{2}}\log\left|\dfrac{\sqrt{2} + \sin\theta}{\sqrt{2} - \sin\theta}\right| + C$

   ›Reveal solutionSolution

   Substitution $u=\sin\theta$ gives $\int\dfrac{du}{2-u^2}=\dfrac{1}{2\sqrt2}\log\left|\dfrac{\sqrt2+\sin\theta}{\sqrt2-\sin\theta}\right|+C$.

   Let $u=\sin\theta$, so $du=\cos\theta\,d\theta$. The integral becomes

   $\int\frac{du}{2-u^2}=\int\frac{du}{(\sqrt2)^2-u^2}.$

   Using the standard result $\displaystyle\int\frac{du}{a^2-u^2}=\frac{1}{2a}\log\left|\frac{a+u}{a-u}\right|+C$ with $a=\sqrt2$:

   $=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+u}{\sqrt2-u}\right|+C=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+\sin\theta}{\sqrt2-\sin\theta}\right|+C.$

   ✓Final answer

   The correct option is (E).

9. 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: $x=0\to u=0$, $x=\frac{\pi}{4}\to u=1$.

   $\int_0^{\pi/4}\tan^3x\sec^2x\,dx=\int_0^1u^3\,du=\left[\frac{u^4}{4}\right]_0^1=\frac{1}{4}$.

   ✓Final answer

   The correct option is (C).

10. 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

    $\int \sec^{2}u\,du=\tan u+C=\tan\left(\sqrt{2x+5}\right)+C.$

    ✓Final answer

    The correct option is (D).

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

    ›Reveal solutionSolution

    $\int\dfrac{\cos(\tan x)}{\cos^2 x}\,dx=\sin(\tan x)+C$.

    Concept and Intuition

    The factor $1/\cos^2 x=\sec^2 x$ is exactly the derivative of $\tan x$, so substituting $u=\tan x$ linearizes the integral.

    Step-by-Step Solution

    1. Rewrite as $\int\cos(\tan x)\sec^2 x\,dx$.
    2. Let $u=\tan x$, $du=\sec^2 x\,dx$.
    3. Integral $=\int\cos u\,du=\sin u+C=\sin(\tan x)+C$.

    Common Mistakes

    • Failing to recognize $1/\cos^2 x=\sec^2 x$ as $d(\tan x)$.

    ✓Final answer

    The correct option is (B) — $\sin(\tan x) + C$.

    ANSWER: B

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

    ›Reveal solutionSolution

    The substitution $u=x+\frac{1}{x}$ turns it into $\int\frac{du}{u^2+1}=\tan^{-1}\left(x+\frac{1}{x}\right)+C$.

    Divide numerator and denominator by $x^2$:

    $\frac{x^2-1}{x^4+3x^2+1}=\frac{1-\frac{1}{x^2}}{x^2+3+\frac{1}{x^2}}$.

    Let $u=x+\frac{1}{x}$, so $du=\left(1-\frac{1}{x^2}\right)dx$ and $x^2+\frac{1}{x^2}=u^2-2$. The denominator becomes $u^2-2+3=u^2+1$.

    Thus $\int\frac{du}{u^2+1}=\tan^{-1}u+C=\tan^{-1}\left(x+\frac{1}{x}\right)+C$.

    ✓Final answer

    The correct option is (E).