Q.Find $\int \cos 6x \sqrt{1 + \sin 6x}\, dx$

Showing the 12 most recent of 32 on this concept.

1. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.$\int x^5 e^{1-x^6} \, dx =$ (A) $\dfrac{1}{6} e^{1-x^6} + C$ (B) $-e^{1-x^6} + C$ (C) $\dfrac{-1}{6} e^{1-x^6} + C$ (D) $\dfrac{x^5}{5} e^{1-x^6} + C$ (E) $\dfrac{x^6}{6} e^{1-x^6} + C$

   ›Reveal solutionSolution

   $\int x^5 e^{1-x^6}\,dx=-\dfrac16 e^{1-x^6}+C$.

   Concept and Intuition

   The exponent's derivative $\dfrac{d}{dx}(1-x^6)=-6x^5$ matches the algebraic factor $x^5$, so a $u$-substitution collapses the integral.

   Step-by-Step Solution

   1. Let $u=1-x^6$, then $du=-6x^5\,dx$, i.e. $x^5\,dx=-\dfrac16\,du$.
   2. Integral $=\displaystyle\int e^{u}\left(-\tfrac16\right)du=-\tfrac16 e^{u}+C$.
   3. Back-substitute: $=-\dfrac16 e^{1-x^6}+C$.

   Common Mistakes

   • Dropping the negative sign from $du=-6x^5\,dx$.

   ✓Final answer

   The correct option is (C) — $\dfrac{-1}{6} e^{1-x^6} + C$.

   ANSWER: C

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

   $\int \dfrac{\tan^5 x\sec^2 x}{\tan^{12}x+1}\,dx = \dfrac{1}{6}\int \dfrac{du}{u^2+1} = \dfrac{1}{6}\tan^{-1}u + C = \dfrac{1}{6}\tan^{-1}(\tan^6 x)+C$.

   ✓Final answer

   The correct option is (A).

3. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.If $\displaystyle\int \dfrac{1}{x^7 \left(\dfrac{1}{x^6}+1\right)^{2/3}} \, dx = -\dfrac{1}{2}\left(\dfrac{1}{\dfrac{1}{x^6}+1}\right)^{p} + c$, then $p =$ (A) $\dfrac{2}{3}$ (B) $\dfrac{-1}{3}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{-2}{3}$ (E) $\dfrac{1}{6}$

   ›Reveal solutionSolution

   Substitute $u=\tfrac1{x^6}+1$; the result is $-\tfrac12u^{1/3}$, matched to the given form $-\tfrac12(1/u)^{p}$ gives $p=-1/3$.

   Let $u=\dfrac{1}{x^6}+1$, so $du=-\dfrac{6}{x^7}\,dx$, i.e. $\dfrac{dx}{x^7}=-\dfrac{du}{6}$.

   $\int \frac{u^{-2/3}}{x^7}\,dx=\int u^{-2/3}\left(-\frac{du}{6}\right)=-\frac16\cdot\frac{u^{1/3}}{1/3}=-\frac12u^{1/3}+c.$

   The given answer is $-\dfrac12\left(\dfrac{1}{u}\right)^{p}=-\dfrac12 u^{-p}$. Matching exponents, $-p=\dfrac13$, so $p=-\dfrac13$.

   ✓Final answer

   The correct option is (B).

4. 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

5. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.$\int \left(\frac{\sin t + \cos t}{13 + 36\sin 2t}\right) dt$ is equal to (A) $\frac{1}{84}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$ (B) $\frac{1}{81}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$ (C) $\frac{1}{84}\log\left|\frac{7 - 6(\sin t - \cos t)}{7 + 6(\sin t - \cos t)}\right| + C$ (D) $\frac{1}{48}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$ (E) $\frac{1}{64}\log\left|\frac{7 + 6(\sin t - \cos t)}{7 - 6(\sin t - \cos t)}\right| + C$

   ›Reveal solutionSolution

   Let $u=\sin t-\cos t$; the numerator becomes $du$ and $13+36\sin 2t=49-36u^2$, giving a standard $\tfrac{du}{a^2-k^2u^2}$ integral.

   Set $u=\sin t-\cos t$. Then

   $du=(\cos t+\sin t)\,dt,$

   which is exactly the numerator, and

   $u^2=1-2\sin t\cos t=1-\sin 2t\;\Rightarrow\;\sin 2t=1-u^2.$

   So the denominator is

   $13+36\sin 2t=13+36(1-u^2)=49-36u^2.$

   Hence

   $\int\frac{du}{49-36u^2}=\int\frac{du}{7^2-(6u)^2}.$

   Using $\int\tfrac{du}{a^2-k^2u^2}=\tfrac{1}{2ak}\log\left|\tfrac{a+ku}{a-ku}\right|$ with $a=7,\;k=6$,

   $=\frac{1}{2\cdot 7\cdot 6}\log\left|\frac{7+6u}{7-6u}\right|=\frac{1}{84}\log\left|\frac{7+6(\sin t-\cos t)}{7-6(\sin t-\cos t)}\right|+C.$

   ✓Final answer

   The correct option is (A).

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 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

   • Miscounting powers so the $\sec^2$ factor is not isolated.
   • Forgetting the $\frac1{26}$ from integrating $u^{25}$.

   ✓Final answer

   The correct option is (D) — $\frac{\tan^{26}(x)}{26} + C$.

   ANSWER: D

8. 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).

9. 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 $u=\sqrt t$, $t=u^2$, $dt=2u\,du$:

   $\int\dfrac{dt}{\sqrt t+1}=\int\dfrac{2u\,du}{u+1}=2\int\left(1-\dfrac{1}{u+1}\right)du=2\bigl(u-\log(u+1)\bigr)+C.$

   Thus the integral is $2\bigl(\sqrt{\tan x}-\log(\sqrt{\tan x}+1)\bigr)+C$.

   ✓Final answer

   The correct option is (A).

10. 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

    $\int\frac{\sec x}{(\sec x+\tan x)^2}\,dx=\int\frac{1}{u^2}\cdot\frac{du}{u}=\int u^{-3}\,du=-\frac{1}{2u^2}+C=\frac{-1}{2(\sec x+\tan x)^2}+C.$

    ✓Final answer

    The correct option is (B).

11. 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).

12. 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 $\cos^2x$):

    $\frac{1}{\cos^{2/3}x\,\sin^{4/3}x}=\sec^2x\cdot\frac{\cos^{4/3}x}{\sin^{4/3}x}=\sec^2x\,\cot^{4/3}x=\frac{\sec^2x}{\tan^{4/3}x}.$

    Let $t=\tan x$, $dt=\sec^2x\,dx$:

    $\int t^{-4/3}\,dt=\frac{t^{-1/3}}{-1/3}=-3\,t^{-1/3}+C=-3\tan^{-1/3}x+C.$

    ✓Final answer

    The correct option is (D).