Q.$\int_{0}^{\pi/2} \dfrac{dx}{(a^2\cos^2 x+b^2\sin^2 x)^2}$ (Hint: Divide numerator and denominator by $\cos^4 x$)

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $\displaystyle\int_0^3 x\left(\sqrt[5]{9 - x^2}\right) dx = k\, 3^{1/k}$, then $k =$ (A) $\dfrac{9}{5}$ (B) $\dfrac{5}{9}$ (C) $\dfrac{5}{12}$ (D) $\dfrac{12}{5}$

   ›Reveal solutionSolution

   The substitution $u=9-x^2$ turns the integral into a simple power integral equal to $\frac{5}{12}\cdot3^{12/5}$, which matches $k\cdot3^{1/k}$ exactly when $k=5/12$.

   Concept and Intuition

   Whenever the integrand contains $x$ times a function of $9-x^2$, the substitution $u=9-x^2$ (so $du=-2x\,dx$) removes the awkward fifth root and reduces the problem to integrating a pure power of $u$.

   Step-by-Step Solution

   1. Let $u=9-x^2 \Rightarrow du=-2x\,dx \Rightarrow x\,dx=-\dfrac{du}{2}$. When $x=0,\ u=9$; when $x=3,\ u=0$.
   2. $\int_0^3 x(9-x^2)^{1/5}dx=\int_{9}^{0}u^{1/5}\left(-\frac{du}{2}\right)=\frac12\int_0^9 u^{1/5}du$
   3. $\frac12\left[\frac{u^{6/5}}{6/5}\right]_0^9=\frac12\cdot\frac56\cdot9^{6/5}=\frac{5}{12}\cdot9^{6/5}$
   4. Since $9=3^2$, $9^{6/5}=3^{12/5}$. So the integral equals $\dfrac{5}{12}\cdot3^{12/5}$. …

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2. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.$\displaystyle\int_{-1}^{4} \sqrt{\dfrac{4-x}{x+1}}\,dx =$ (A) 0 (B) $\dfrac{\pi}{2}$ (C) $\dfrac{3\pi}{2}$ (D) $\dfrac{5\pi}{2}$

   ›Reveal solutionSolution

   This integral matches the standard result $\int_a^b\sqrt{\frac{b-x}{x-a}}\,dx=\frac{\pi(b-a)}2$, giving $\frac{5\pi}2$.

   Concept and Intuition

   Integrals of the form $\int_a^b\sqrt{\dfrac{b-x}{x-a}}\,dx$ arise often and have a clean closed form obtained via the substitution $x=a+(b-a)\sin^2\theta$, which converts the square root into $\cot\theta$ and the whole integral into $\int_0^{\pi/2}\cos^2\theta\,d\theta$.

   Step-by-Step Solution

   1. Here $a=-1,\ b=4$ (matching $\sqrt{\frac{4-x}{x+1}}=\sqrt{\frac{b-x}{x-a}}$).
   2. Substitute $x=a+(b-a)\sin^2\theta$: then $b-x=(b-a)\cos^2\theta$, $x-a=(b-a)\sin^2\theta$, so $\sqrt{\frac{b-x}{x-a}}=\cot\theta$, and $dx=2(b-a)\sin\theta\cos\theta\,d\theta$. …

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3. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int_{\pi/6}^{\pi/3}\cos^{-4}x\,dx=$ (A) $\dfrac{64}{9\sqrt3}$ (B) $\dfrac{52\sqrt3}{9}$ (C) $\dfrac{62\sqrt3}{9}$ (D) $\dfrac{44}{9\sqrt3}$

   ›Reveal solutionSolution

   $\cos^{-4}x=\sec^4x$ integrates via the standard $\sec^2x(1+\tan^2x)$ trick to $\tan x+\tan^3x/3$; evaluating between $\pi/6$ and $\pi/3$ gives $\dfrac{44}{9\sqrt3}$.

   Concept and Intuition

   For odd powers of $\sec$ combined this way, peeling off one $\sec^2x$ (to serve as $du$ for $u=\tan x$) and writing the rest in terms of $\tan x$ using $\sec^2x=1+\tan^2x$ is the standard technique.

   Step-by-Step Solution

   1. $\sec^4x=\sec^2x\cdot\sec^2x=\sec^2x(1+\tan^2x)$.
   2. With $u=\tan x$, $du=\sec^2x\,dx$: $\int\sec^4x\,dx=\int(1+u^2)du=u+\frac{u^3}{3}+c=\tan x+\frac{\tan^3x}{3}+c$.
   3. At $x=\pi/3$: $\tan x=\sqrt3$, so value $=\sqrt3+\frac{3\sqrt3}{3}=\sqrt3+\sqrt3=2\sqrt3$.
   4. At $x=\pi/6$: $\tan x=\frac{1}{\sqrt3}$, so value $=\frac{1}{\sqrt3}+\frac{1}{3\cdot3\sqrt3}=\frac{1}{\sqrt3}+\frac{1}{9\sqrt3}=\frac{9+1}{9\sqrt3}=\frac{10}{9\sqrt3}$. …

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4. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.$\displaystyle\int_{1/5}^{1/2} \dfrac{\sqrt{x-x^2}}{x^3}\,dx =$ (A) $\dfrac{21}{2}$ (B) $\dfrac{14}{3}$ (C) $\dfrac{7}{3}$ (D) $\dfrac{7}{2}$

   ›Reveal solutionSolution

   This tests factoring $x-x^2=x^2(1/x-1)$ to expose a clean substitution $t=1/x-1$. The definite integral evaluates to $\dfrac{14}{3}$, option (B).

   Concept and Intuition

   The integrand $\dfrac{\sqrt{x-x^2}}{x^3}$ looks like it needs a trig substitution for $\sqrt{x-x^2}$, but factoring out $x^2$ from under the root — valid since $x>0$ on $[1/5,1/2]$ — turns it into $\dfrac{\sqrt{1/x-1}}{x^2}$, a form whose derivative-friendly piece $1/x^2\,dx$ is exactly what appears when differentiating $1/x$. This makes $t=1/x-1$ the natural substitution.

   Step-by-Step Solution

   1. Since $x>0$: $\sqrt{x-x^2}=\sqrt{x^2\left(\dfrac1x-1\right)}=x\sqrt{\dfrac1x-1}$.
   2. So the integrand is $\dfrac{x\sqrt{1/x-1}}{x^3}=\dfrac{\sqrt{1/x-1}}{x^2}$.
   3. Let $t=\dfrac1x-1$; then $dt=-\dfrac{1}{x^2}dx$, i.e. $\dfrac{dx}{x^2}=-dt$.
   4. The integral becomes $\int\sqrt t\,(-dt)=-\dfrac23t^{3/2}+c$.
   5. Limits: at $x=\dfrac15$, $t=5-1=4$; at $x=\dfrac12$, $t=2-1=1$. …

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5. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.$\displaystyle\int_0^2 x^8\left(\dfrac{4}{x^2} - 1\right)^{5/2} dx =$ (A) $\dfrac{2^{15}}{63}$ (B) $\dfrac{2^{16}}{315}$ (C) $\dfrac{2^{16}}{189}$ (D) $\dfrac{2^{10}}{63}$

   ›Reveal solutionSolution

   Simplify the fractional-power term algebraically first (removing the ugly $5/2$ power on a fraction), then a plain $u=4-x^2$ substitution gives $\frac{2^{10}}{63}$.

   Concept and Intuition

   A term like $\left(\frac{4}{x^2}-1\right)^{5/2}$ looks like it needs a trig substitution, but multiplying it out against the accompanying $x^8$ first often collapses the whole thing into a much simpler polynomial-times-power form that a basic $u$-substitution can handle — always simplify algebraically before reaching for a substitution.

   Step-by-Step Solution

   1. Rewrite the bracket: $\dfrac{4}{x^2}-1=\dfrac{4-x^2}{x^2}$, so

   $x^8\left(\frac{4-x^2}{x^2}\right)^{5/2}=x^8\cdot\frac{(4-x^2)^{5/2}}{(x^2)^{5/2}}=x^8\cdot\frac{(4-x^2)^{5/2}}{x^5}=x^3(4-x^2)^{5/2}.$

   2. The integral becomes $\displaystyle I=\int_0^2 x^3(4-x^2)^{5/2}\,dx$.
   3. Substitute $u=4-x^2$, so $du=-2x\,dx$ and $x^2=4-u$. Write $x^3\,dx=x^2(x\,dx)=(4-u)\left(-\frac{du}{2}\right)$.
   4. Limits: $x=0\Rightarrow u=4$; $x=2\Rightarrow u=0$. So

   $I=\int_{u=4}^{0}(4-u)u^{5/2}\left(-\frac12\right)du=\frac12\int_0^4(4-u)u^{5/2}\,du.$

   5. Expand: $\displaystyle I=\frac12\left[4\int_0^4 u^{5/2}du-\int_0^4 u^{7/2}du\right]$.
   6. $\displaystyle\int_0^4 u^{5/2}du=\frac{2}{7}u^{7/2}\Big|_0^4=\frac27\cdot4^{7/2}=\frac27\cdot128=\frac{256}{7}$ (using $4^{7/2}=(2^2)^{7/2}=2^7=128$). …

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6. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$\int_{1/\sqrt[5]{31}}^{1/\sqrt[5]{242}}\dfrac{1}{\sqrt[5]{x^{30}+x^{25}}}\,dx=$ (A) $\dfrac{65}{4}$ (B) $\dfrac{-75}{4}$ (C) $\dfrac{75}{4}$ (D) $\dfrac{-65}{4}$

   ›Reveal solutionSolution

   Factor out the highest power of $x$ inside the fifth root to expose a clean substitution $w=1+x^{-5}$, then evaluate at the given nasty-looking but designed-to-simplify limits.

   Concept and Intuition

   The limits $31^{-1/5}$ and $242^{-1/5}$ look intimidating, but they're chosen precisely so that $1+x^{-5}$ becomes the perfect fifth powers $32=2^5$ and $243=3^5$ at the two ends — a strong hint to substitute $w=1+x^{-5}$.

   Step-by-Step Solution

   1. $x^{30}+x^{25}=x^{25}(x^5+1)$, so $(x^{30}+x^{25})^{1/5}=x^5(x^5+1)^{1/5}=x^5\cdot x(1+x^{-5})^{1/5}=x^6(1+x^{-5})^{1/5}$ (for $x>0$).
   2. So the integrand is $\dfrac{1}{x^6(1+x^{-5})^{1/5}}=x^{-6}(1+x^{-5})^{-1/5}$.
   3. Let $w=1+x^{-5}$, so $dw=-5x^{-6}dx\Rightarrow x^{-6}dx=-\dfrac{dw}{5}$.
   4. $\displaystyle\int x^{-6}(1+x^{-5})^{-1/5}dx=-\frac15\int w^{-1/5}dw=-\frac15\cdot\frac{w^{4/5}}{4/5}+c=-\frac14w^{4/5}+c$.
   5. Lower limit $x_1=31^{-1/5}\Rightarrow x_1^{-5}=31\Rightarrow w_1=32=2^5\Rightarrow w_1^{4/5}=2^4=16$. …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{1}{1+\sqrt{\tan x}} \, dx =$ (A) $0$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   The classic King's-rule trick ($x\to a-x$) makes the integrand add to $1$; the integral evaluates to $\dfrac\pi4$ — (D).

   Concept and Intuition

   For $\displaystyle\int_0^a f(x)dx$, substituting $x\to a-x$ gives an equal integral $\displaystyle\int_0^a f(a-x)dx$. When $f(x)+f(a-x)$ simplifies to a constant, adding the two versions of the integral collapses everything to a trivial computation.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\dfrac{dx}{1+\sqrt{\tan x}}$.
   2. By the property $\displaystyle\int_0^a f(x)dx=\int_0^a f(a-x)dx$: $I=\displaystyle\int_0^{\pi/2}\dfrac{dx}{1+\sqrt{\tan(\pi/2-x)}}=\int_0^{\pi/2}\dfrac{dx}{1+\sqrt{\cot x}}$.
   3. Simplify: $\dfrac1{1+\sqrt{\cot x}}=\dfrac1{1+1/\sqrt{\tan x}}=\dfrac{\sqrt{\tan x}}{\sqrt{\tan x}+1}$. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int_0^1 \dfrac{x}{(1-x)^{3/4}}\,dx =$ (A) $\dfrac{4}{5}$ (B) $\dfrac{8}{15}$ (C) $\dfrac{14}{5}$ (D) $\dfrac{16}{5}$

   ›Reveal solutionSolution

   With $u=1-x$, the integral becomes $\int_0^1(u^{-3/4}-u^{1/4})du = \dfrac{16}{5}$.

   Concept and Intuition

   Integrals of the form $\int x^m(1-x)^n dx$ over $[0,1]$ are cleanly handled by substituting $u=1-x$ so the fractional power becomes a simple power of $u$.

   Step-by-Step Solution

   1. Let $u=1-x \Rightarrow x=1-u,\ dx=-du$; limits $x:0\to1$ becomes $u:1\to0$.
   2. $\displaystyle\int_0^1\dfrac{x}{(1-x)^{3/4}}dx = \int_0^1\dfrac{1-u}{u^{3/4}}du = \int_0^1\left(u^{-3/4}-u^{1/4}\right)du$.
   3. $= \left[4u^{1/4} - \dfrac{4}{5}u^{5/4}\right]_0^1 = 4-\dfrac{4}{5} = \dfrac{16}{5}$.

   Common Mistakes …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$\displaystyle\int_0^1\sqrt{\dfrac{2+x}{2-x}}\,dx=$ (A) $\pi+2$ (B) $\dfrac{1}{2}(\pi+2)$ (C) $\dfrac{\pi}{2}+2+\sqrt{3}$ (D) $\dfrac{\pi}{3}+2-\sqrt{3}$

   ›Reveal solutionSolution

   Splitting the integrand as $\frac{2+x}{\sqrt{4-x^2}}$ into two standard forms gives $\frac{\pi}{3}+2-\sqrt3$, option (D).

   Concept and Intuition

   Multiplying numerator and denominator by $\sqrt{2+x}$ turns the awkward square root of a ratio into $\frac{2+x}{\sqrt{4-x^2}}$, which splits cleanly into an arcsine-type term and a simple power-rule term.

   Step-by-Step Solution

   1. $\sqrt{\dfrac{2+x}{2-x}}=\dfrac{2+x}{\sqrt{(2-x)(2+x)}}=\dfrac{2+x}{\sqrt{4-x^2}}$.
   2. Split: $\displaystyle\int_0^1\frac{2}{\sqrt{4-x^2}}dx+\int_0^1\frac{x}{\sqrt{4-x^2}}dx$.
   3. First integral: $2\left[\arcsin\frac{x}{2}\right]_0^1=2\arcsin\frac12=2\cdot\frac{\pi}{6}=\frac{\pi}{3}$.
   4. Second integral: $\left[-\sqrt{4-x^2}\right]_0^1=-\sqrt3-(-2)=2-\sqrt3$. …

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10. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.If $5f(x)+3f\left(\dfrac{1}{x}\right)=2-\dfrac{1}{x}$, $x\ne0$, then $\displaystyle\int_1^2 f\left(\dfrac{1}{x}\right)dx=$ (A) $\dfrac{6\log2-7}{32}$ (B) $\dfrac{6\log2-17}{32}$ (C) $\dfrac{6\log2-1}{32}$ (D) $\dfrac{6\log2-7}{16}$

    ›Reveal solutionSolution

    Replacing $x$ by $1/x$ in the functional equation gives a second equation; solving the linear system for $f(1/x)$ and integrating over $[1,2]$ gives $\dfrac{6\log2-7}{32}$.

    Concept and Intuition

    A functional equation relating $f(x)$ and $f(1/x)$ is solved by generating a second equation (via the substitution $x\to1/x$) and treating $f(x), f(1/x)$ as two unknowns in a linear system.

    Step-by-Step Solution

    1. Given: $5f(x)+3f(1/x)=2-1/x$ … (1)
    2. Replace $x\to1/x$: $5f(1/x)+3f(x)=2-x$ … (2)
    3. Compute $5\times(1)-3\times(2)$: wait — instead eliminate $f(x)$: multiply (1) by $3$ and (2) by $5$: $15f(x)+9f(1/x)=6-3/x$ and $15f(x)+25f(1/x)=10-5x$.
    4. Subtract: $16f(1/x)=4-5x+3/x\Rightarrow f(1/x)=\dfrac{4-5x+3/x}{16}$.
    5. $\displaystyle\int_1^2 f(1/x)\,dx=\dfrac1{16}\int_1^2\left(4-5x+\dfrac3x\right)dx$. …

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11. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.$\displaystyle\int_{1/25}^{3} \dfrac{e^{3/x}}{x^2}\,dx=$ (A) $-\dfrac{1}{3}(e^{75}-e)$ (B) $\dfrac{1}{3}(e^{50}-e^{25})$ (C) $-\dfrac{1}{3}(e^{50}-e)$ (D) $\dfrac{1}{3}(e^{75}-e)$

    ›Reveal solutionSolution

    The substitution $u=3/x$ converts the awkward $e^{3/x}/x^2$ integrand into a plain exponential, evaluating to $\dfrac13(e^{75}-e)$.

    Concept and Intuition

    Whenever you see $e^{g(x)}\cdot g'(x)$-like structure — here $\dfrac{1}{x^2}$ is (up to a constant) the derivative of $3/x$ — substituting $u=3/x$ turns the whole integral into $\int e^u\,du$, the simplest possible exponential integral.

    Step-by-Step Solution

    1. Let $u=\dfrac{3}{x}$. Then $\dfrac{du}{dx}=-\dfrac{3}{x^2}$, so $\dfrac{dx}{x^2}=-\dfrac{du}{3}$.
    2. Change the limits: at $x=\dfrac1{25}$, $u=3\cdot25=75$; at $x=3$, $u=3/3=1$.
    3. The integral becomes $\displaystyle\int_{x=1/25}^{3}\dfrac{e^{3/x}}{x^2}\,dx=\int_{u=75}^{1}e^u\left(-\dfrac13\right)du=-\dfrac13\int_{75}^{1}e^u\,du$.
    4. Flip the limits (introducing a sign change): $-\dfrac13\displaystyle\int_{75}^1 e^u\,du=\dfrac13\displaystyle\int_1^{75}e^u\,du$. …

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12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.Let $T > 0$ be a fixed number. $f: \mathbb{R} \to \mathbb{R}$ is a continuous function such that $f(x+T) = f(x)$ $\forall x \in \mathbb{R}$. If $I = \displaystyle\int_0^T f(x)\,dx$, then $\displaystyle\int_0^{5T} f(2x)\,dx =$ (A) $10\,I$ (B) $\dfrac{5}{2}\,I$ (C) $5\,I$ (D) $2\,I$

    ›Reveal solutionSolution

    A substitution $u=2x$ turns the integral into one over $10$ full periods of $f$, giving $5I$.

    Concept and Intuition

    If $f$ is periodic with period $T$, the integral of $f$ over any interval of length $nT$ (a whole number of periods) equals $n$ times the integral over one period. Stretching the variable via $u=2x$ effectively doubles the length of the interval in $u$-space.

    Step-by-Step Solution

    1. Let $u=2x \Rightarrow du = 2\,dx \Rightarrow dx = \dfrac{du}{2}$. When $x=0$, $u=0$; when $x=5T$, $u=10T$.
    2. $\displaystyle\int_0^{5T}f(2x)\,dx = \int_0^{10T}f(u)\cdot\frac{du}{2} = \frac12\int_0^{10T}f(u)\,du$. …

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