Q.Evaluate: $\int \dfrac{1+\cos x}{x+\sin x}\,dx$

Showing the 12 most recent of 32 on this concept.

1. 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 $1-\sin x$:

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

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $u=\sqrt t$, $t=u^2$, $dt=2u\,du$: …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $u=\tan x$, $du=\sec^2 x\,dx$. …

   🔒 Start your 14-day free trial to see the full solution →
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. …

   🔒 Start your 14-day free trial to see the full solution →
5. 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. $\dfrac{dx}{1+x^2}=-du$. …

   🔒 Start your 14-day free trial to see the full solution →
6. KEAM 2025Set eng-2025-04234 marksMCQ
   Q.$\int e^x (x^2 - 2)\cos\left(e^x (x^2 - 2x)\right)\, dx =$ (A) $\sin\left(e^x (x^2 - 2x)\right) + C$ (B) $\sin\left(e^x (x^2 - 2)\right) + C$ (C) $x^2 e^x \sin\left(e^x (x^2 - 2)\right) + C$ (D) $e^x \sin\left(e^x (x^2 - 2)\right) + C$ (E) $e^x \sin\left(x^2 e^x - 2x e^x\right) + C$

   ›Reveal solutionSolution

   Since d/dx[e^x(x^2-2x)] = e^x(x^2-2), the integral is sin(e^x(x^2-2x)) + C.

   Concept and Intuition

   Look for the outer function's argument, g(x) = e^x(x^2-2x), and check whether the rest of the integrand is exactly g'(x); if so the integral is a direct substitution.

   Step-by-Step Solution

   1. Let g(x) = e^x(x^2 - 2x).
   2. g'(x) = e^x(x^2-2x) + e^x(2x-2) = e^x(x^2 - 2), which is exactly the prefactor. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $a=\sqrt2$: …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 …

    🔒 Start your 14-day free trial to see the full solution →
11. 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

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

    🔒 Start your 14-day free trial to see the full solution →
12. 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 …

    🔒 Start your 14-day free trial to see the full solution →