Q.Evaluate the definite integral: $\int_{0}^{2} \frac{6x+3}{x^2+4} \, dx$

Showing the 12 most recent of 16 on this concept.

1. 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

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

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2. 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 $(t-1)(t+1)=t^2-1$. …

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3. 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

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

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4. 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 $u=\sin x$, $du=\cos x\,dx$: …

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5. GUJCET 2019Set 171 markMCQ
   Q.If $\int \sin^{13}x\cos^3 x\,dx = A\sin^{14}x + B\sin^{16}x + C$, then $A + B = \underline{\hspace{2cm}}$ (A) $\dfrac{17}{112}$ (B) $\dfrac{15}{112}$ (C) $\dfrac{1}{110}$ (D) $\dfrac{1}{112}$

   ›Reveal solutionSolution

   Substituting $u=\sin x$ gives $\int u^{13}(1-u^2)\,du=\dfrac{\sin^{14}x}{14}-\dfrac{\sin^{16}x}{16}$, so $A+B=\dfrac{1}{112}$.

   Concept: $\cos^3x=(1-\sin^2x)\cos x$. Let $u=\sin x$, $du=\cos x\,dx$:

   $\int u^{13}(1-u^2)\,du=\frac{u^{14}}{14}-\frac{u^{16}}{16}$ …

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6. 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. …

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

   …

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8. 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. $x^{-3}dx=-\dfrac{dt}{2}$.

   …

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9. 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 $\tan x$.

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

   …

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10. 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. $\sin x\,dx=-du$.

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

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

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

    …

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11. 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 $t=xe^x$.

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

    …

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12. 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.

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

    …

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