Q.Evaluate the integral using substitution $\int_{0}^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi$

Showing the 12 most recent of 16 on this concept.

1. GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
   Q.$\displaystyle\int \dfrac{\sec^{-1}x}{\sqrt{x^4-x^2}}\,dx =$ ____ $+ C$.

   (a) $-\sec^{-1}x$

   (b) $\sec^{-1}x$

   (c) $-\dfrac{1}{2}(\sec^{-1}x)^2$

   (d) $\dfrac{1}{2}(\sec^{-1}x)^2$

   ›Reveal solutionSolution

   Recognize the integrand as $f(x)f'(x)$ where $f(x)=\sec^{-1}x$.

   $\sqrt{x^4-x^2}=|x|\sqrt{x^2-1}$, and $\dfrac{d}{dx}(\sec^{-1}x)=\dfrac{1}{|x|\sqrt{x^2-1}}$.

   So the integrand equals $\sec^{-1}x\cdot\dfrac{d}{dx}(\sec^{-1}x)$, whose integral is $\dfrac12(\sec^{-1}x)^2+C$.

   ✓Final answer

   The correct option is (d) $\dfrac12(\sec^{-1}x)^2$.

2. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.$\displaystyle\int \dfrac{x^9}{(4x^2+1)^6}\,dx$ = ____ $+ C$.

   (a) $\dfrac{1}{5x}\left(4 + \dfrac{1}{x^2}\right)^{-5}$

   (b) $\dfrac{1}{10x}\left(\dfrac{1}{x^2} + 4\right)^{-5}$

   (c) $\dfrac{1}{5}\left(4 + \dfrac{1}{x^2}\right)^{-5}$

   (d) $\dfrac{1}{10}\left(\dfrac{1}{x^2} + 4\right)^{-5}$

   ›Reveal solutionSolution

   Factor $x^{12}$ out of the denominator and substitute $t=4+\dfrac{1}{x^2}$.

   Write $\dfrac{x^9}{(4x^2+1)^6}=\dfrac{x^9}{x^{12}\left(4+\frac{1}{x^2}\right)^6}=x^{-3}\left(4+\dfrac{1}{x^2}\right)^{-6}$.

   Let $t=4+\dfrac{1}{x^2}$, so $dt=-\dfrac{2}{x^3}dx$, i.e. $x^{-3}dx=-\dfrac{dt}{2}$.

   $\int t^{-6}\left(-\dfrac{dt}{2}\right)=-\dfrac12\cdot\dfrac{t^{-5}}{-5}=\dfrac{t^{-5}}{10}=\dfrac{1}{10}\left(\dfrac{1}{x^2}+4\right)^{-5}+C$.

   ✓Final answer

   The correct option is (d) $\dfrac{1}{10}\left(\dfrac{1}{x^2}+4\right)^{-5}$.

3. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin^{24} x}{\cos^{26} x}\,dx$ = ____ $+ C$.

   (a) $\dfrac{\tan^{24} x}{24}$

   (b) $\dfrac{\tan^{26} x}{26}$

   (c) $\dfrac{\tan^{25} x}{25}$

   (d) $\dfrac{\tan^{27} x}{27}$

   ›Reveal solutionSolution

   Split off $\sec^2x$ and write the rest in terms of $\tan x$.

   $\dfrac{\sin^{24}x}{\cos^{26}x}=\tan^{24}x\cdot\sec^2x$.

   With $u=\tan x$, $du=\sec^2x\,dx$: $\int u^{24}du=\dfrac{u^{25}}{25}=\dfrac{\tan^{25}x}{25}+C$.

   ✓Final answer

   The correct option is (c) $\dfrac{\tan^{25}x}{25}$.

4. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin x}{3 + 4\cos^2 x}\,dx$ = ____ $+ C$.

   (a) $\log(3 + 4\cos^2 x)$

   (b) $\dfrac{1}{2\sqrt{3}}\tan^{-1}\left(\dfrac{\sqrt{3}\sec x}{2}\right)$

   (c) $-\dfrac{1}{2\sqrt{3}}\tan^{-1}\left(\dfrac{\cos x}{\sqrt{3}}\right)$

   (d) $\dfrac{1}{2\sqrt{3}}\tan^{-1}\left(\dfrac{2\cos x}{\sqrt{3}}\right)$

   ›Reveal solutionSolution

   Substitute $u=\cos x$ to turn this into a standard $\int\frac{du}{a^2+b^2u^2}$ integral.

   Let $u=\cos x$, so $du=-\sin x\,dx$, i.e. $\sin x\,dx=-du$.

   $\int\dfrac{\sin x}{3+4\cos^2x}dx=-\int\dfrac{du}{3+4u^2}=-\dfrac14\int\dfrac{du}{u^2+3/4}$

   $=-\dfrac14\cdot\dfrac{1}{\sqrt{3}/2}\tan^{-1}\left(\dfrac{u}{\sqrt3/2}\right)=-\dfrac{1}{2\sqrt3}\tan^{-1}\left(\dfrac{2u}{\sqrt3}\right)+C$

   $=-\dfrac{1}{2\sqrt3}\tan^{-1}\left(\dfrac{2\cos x}{\sqrt3}\right)+C$.

   Checking by differentiation confirms this is correct: $\dfrac{d}{dx}\left[\dfrac{1}{2\sqrt3}\tan^{-1}\left(\dfrac{2\cos x}{\sqrt3}\right)\right]=-\dfrac{\sin x}{3+4\cos^2x}$, i.e. option (d) as literally printed is the negative of the true antiderivative -- the argument $\frac{2\cos x}{\sqrt3}$ is exactly right, only the overall sign in the option is off (likely a printing slip, since options (b) and (c) both have the wrong argument form entirely).

   ✓Final answer

   $\displaystyle\int\frac{\sin x}{3+4\cos^2x}\,dx=-\dfrac{1}{2\sqrt3}\tan^{-1}\left(\dfrac{2\cos x}{\sqrt3}\right)+C$ -- matching option (d) in every respect except a sign that appears mistyped in the option.

5. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.$\displaystyle\int \dfrac{e^x(1+x)}{\cos^2(x \cdot e^x)}\,dx$ = ____ $+ C$.

   (a) $-\cot(x \cdot e^x)$

   (b) $\tan(e^x)$

   (c) $\tan(x \cdot e^x)$

   (d) $\cot(e^x)$

   ›Reveal solutionSolution

   Recognize that $e^x(1+x)$ is the derivative of $x\cdot e^x$, so substitute $t=xe^x$.

   $\dfrac{d}{dx}(xe^x)=e^x+xe^x=e^x(1+x)$.

   Let $t=xe^x$, $dt=e^x(1+x)dx$: $\displaystyle\int\dfrac{dt}{\cos^2t}=\int\sec^2t\,dt=\tan t+C=\tan(xe^x)+C$.

   ✓Final answer

   The correct option is (c) $\tan(x\cdot e^x)$.

6. GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{e^x + e^{-x}}$ = ____ $+ C$.

   (a) $\tan^{-1}(e^x)$

   (b) $\log(e^x - e^{-x})$

   (c) $\tan^{-1}(e^{-x})$

   (d) $\log(e^x + e^{-x})$

   ›Reveal solutionSolution

   Multiply through by $e^x$ to turn the denominator into $1+e^{2x}$, a standard arctan form.

   $\dfrac{1}{e^x+e^{-x}}=\dfrac{e^x}{e^{2x}+1}$.

   Let $u=e^x$, $du=e^x dx$: $\displaystyle\int\dfrac{du}{u^2+1}=\tan^{-1}u+C=\tan^{-1}(e^x)+C$.

   ✓Final answer

   The correct option is (a) $\tan^{-1}(e^x)$.

7. GUJCET 2023Set 091 markMCQ
   Q.$\int x^{2019}\cdot e^{x^{2020}}\,dx =$ ______ $+ C$. (A) $\frac{1}{2019}e^{x^{2019}}$ (B) $\frac{1}{2020}e^{x^{2019}}$ (C) $e^{x^{2020}}$ (D) $\frac{1}{2020}e^{x^{2020}}$

   ›Reveal solutionSolution

   The x²⁰¹⁹ factor is (up to a constant) the derivative of the exponent x²⁰²⁰.

   Concept. Substitution $u=x^{2020}$, $du=2020\,x^{2019}\,dx$.

   Solution.

   $\int x^{2019}e^{x^{2020}}\,dx=\frac{1}{2020}\int e^{u}\,du=\frac{1}{2020}e^{x^{2020}}+C.$

   ✓Final answer

   (D) $\frac{1}{2020}e^{x^{2020}}$

   ANSWER: (D)

8. GUJCET 2022Set 081 markMCQ
   Q.$\int e^{\sin x}\sin 2x\,dx=$ ______ $+C$. (A) $e^{\sin x}(\sin x+1)$ (B) $2e^{\sin x}(\sin x-1)$ (C) $2e^{\sin x}(\sin x+1)$ (D) $e^{\sin x}(\sin x-1)$

   ›Reveal solutionSolution

   $\sin2x=2\sin x\cos x$ turns the integral into $2\int u e^u\,du=2e^u(u-1)$.

   Concept. $\int e^{\sin x}\sin2x\,dx=\int e^{\sin x}\,2\sin x\cos x\,dx$.

   Let $u=\sin x$, $du=\cos x\,dx$:

   $2\int u e^{u}\,du=2\big(u e^{u}-e^{u}\big)+C=2e^{\sin x}(\sin x-1)+C.$

   ✓Final answer

   (B) $2e^{\sin x}(\sin x-1)$

   ANSWER: (B)

9. GUJCET 2022Set 081 markMCQ
   Q.$\int\sqrt{\frac{\cos x-\cos^3 x}{1-\cos^3 x}}\,dx=$ ______ $+C$. (A) $-\frac{3}{2}\cos^{-1}\left(\cos^{3/2}x\right)$ (B) $-\frac{2}{3}\cos^{-1}\left(\cos^{3/2}x\right)$ (C) $\frac{3}{2}\cos^{-1}\left(\cos^{3/2}x\right)$ (D) $\frac{2}{3}\cos^{-1}\left(\cos^{3/2}x\right)$

   ›Reveal solutionSolution

   Simplify $\cos x-\cos^3x=\cos x\sin^2x$ and recognise the derivative of $\cos^{-1}(\cos^{3/2}x)$.

   Concept. Numerator $=\cos x(1-\cos^2x)=\cos x\sin^2x$, so the integrand is

   $\sqrt{\frac{\cos x\sin^2x}{1-\cos^3x}}=\frac{\sin x\,\cos^{1/2}x}{\sqrt{1-\cos^3x}}.$

   Now with $u=\cos^{3/2}x$, $\dfrac{d}{dx}\cos^{-1}u=\dfrac{-u'}{\sqrt{1-u^2}}=\dfrac{\tfrac32\sin x\cos^{1/2}x}{\sqrt{1-\cos^3x}}$. Thus the integrand $=\tfrac23\dfrac{d}{dx}\cos^{-1}(\cos^{3/2}x)$, and

   $\int\dots dx=\frac23\cos^{-1}\!\left(\cos^{3/2}x\right)+C.$

   ✓Final answer

   (D) $\frac{2}{3}\cos^{-1}\left(\cos^{3/2}x\right)$

   ANSWER: (D)

10. GUJCET 2022Set 081 markMCQ
    Q.$\int(x+1)(x+3)(x+2)^7\,dx=$ ______ $+C$. (A) $\frac{(x+3)^{10}}{10}+\frac{(x+3)^8}{8}$ (B) $\frac{(x+2)^{10}}{10}+\frac{(x+2)^8}{8}$ (C) $\frac{(x+3)^{10}}{10}-\frac{(x+3)^8}{8}$ (D) $\frac{(x+2)^{10}}{10}-\frac{(x+2)^8}{8}$

    ›Reveal solutionSolution

    Centering at $t=x+2$ turns $(x+1)(x+3)$ into $t^2-1$.

    Concept. Let $t=x+2$, so $x+1=t-1$, $x+3=t+1$, and $(t-1)(t+1)=t^2-1$.

    $\int(t^2-1)t^7\,dt=\int(t^9-t^7)\,dt=\frac{t^{10}}{10}-\frac{t^8}{8}=\frac{(x+2)^{10}}{10}-\frac{(x+2)^8}{8}+C.$

    ✓Final answer

    (D) $\frac{(x+2)^{10}}{10}-\frac{(x+2)^8}{8}$

    ANSWER: (D)

11. GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
    Q.$\int x^2 e^{x^3} \, dx$ equals ___.

    (a) $\frac{1}{3} e^{x^2} + C$

    (b) $\frac{1}{3} e^{x^3} + C$

    (c) $\frac{1}{2} e^{x^3} + C$

    (d) $\frac{1}{2} e^{x^2} + C$

    ›Reveal solutionSolution

    Substitution $u=x^3$ turns the integral into $\tfrac13\int e^u\,du$.

    Let $u = x^3 \Rightarrow du = 3x^2\,dx \Rightarrow x^2\,dx = \tfrac13\,du.$

    $\int x^2 e^{x^3}\,dx = \tfrac13\int e^u\,du = \tfrac13 e^{x^3}+C.$

    ✓Final answer

    (b) $\tfrac13 e^{x^3}+C$.

12. GUJCET 2020Set 071 markMCQ
    Q.$\int \frac{\sqrt{\cot x}}{\cos x \sin x}\,dx =$ ________ $+ C$. (A) $-2\sqrt{\cot x}$ (B) $-2\sqrt{\tan x}$ (C) $2\sqrt{\cot x}$ (D) $\frac{1}{\sqrt{\cot x}}$

    ›Reveal solutionSolution

    Substitute $t=\cot x$.

    Concept: Rewrite $\dfrac{1}{\sin x\cos x}=\dfrac{\csc^2 x}{\cot x}$ and substitute.

    Since $\cos x=\cot x\,\sin x$, we have $\sin x\cos x=\sin^2x\,\cot x$, so

    $\frac{\sqrt{\cot x}}{\sin x\cos x}=\frac{\sqrt{\cot x}\,\csc^2 x}{\cot x}=(\cot x)^{-1/2}\csc^2 x$

    Let $t=\cot x\Rightarrow dt=-\csc^2x\,dx$:

    $\int t^{-1/2}\csc^2x\,dx=-\int t^{-1/2}\,dt=-2\,t^{1/2}=-2\sqrt{\cot x}+C$

    ✓Final answer

    (A) $-2\sqrt{\cot x}$

    ANSWER: (A)