Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx$

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\displaystyle\int_{-a}^{a} f(x)\,dx - \int_{0}^{a} f(-x)\,dx =$ (A) $\displaystyle\int_{-a}^{a} f(a-x)\,dx$ (B) $\displaystyle\int_{-a}^{a} f(x) + f(a-x)\,dx$ (C) $\displaystyle\int_{0}^{a} f(x) + f(a-x)\,dx$ (D) $\displaystyle\int_{0}^{a} f(a-x)\,dx$

   ›Reveal solutionSolution

   The expression simplifies to $\int_0^a f(x)dx$, which by the standard reflection property equals $\int_0^a f(a-x)dx$.

   Concept and Intuition

   Split the symmetric integral at the origin and use the substitution $x\to -x$ on the negative half to relate it to $\int_0^a f(-x)dx$ — this is exactly the term being subtracted, so it cancels, leaving a plain $\int_0^a f(x)dx$. Then apply the classic property $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ (true for any integrable $f$, by substituting $x\to a-x$) to match it to the given answer choices.

   Step-by-Step Solution

   1. Split: $\displaystyle\int_{-a}^a f(x)dx = \int_{-a}^0 f(x)dx + \int_0^a f(x)dx$.
   2. In $\int_{-a}^0 f(x)dx$, substitute $x=-t$, $dx=-dt$; limits $x=-a\to t=a$, $x=0\to t=0$: $\displaystyle\int_{-a}^0 f(x)dx = \int_a^0 f(-t)(-dt) = \int_0^a f(-t)dt = \int_0^a f(-x)dx$.
   3. So $\displaystyle\int_{-a}^a f(x)dx = \int_0^a f(-x)dx + \int_0^a f(x)dx$.
   4. Subtracting $\int_0^a f(-x)dx$ from both sides (as required by the question) leaves exactly $\displaystyle\int_0^a f(x)dx$. …

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2. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^x}dx$, ($a > 0$) $=$ (A) $a\pi$ (B) $\frac{\pi}{a}$ (C) $\frac{\pi}{2}$ (D) $2\pi$

   ›Reveal solutionSolution

   This is the classic "King's rule" property $\int_{-a}^{a}\frac{f(x)}{1+b^x}dx=\frac12\int_{-a}^{a}f(x)dx$ for even $f$; here it directly gives $I=\pi/2$, independent of $a$.

   Concept and Intuition

   Whenever an even function $f(x)$ is divided by $1+b^x$ and integrated over a symmetric interval $[-a,a]$, replacing $x\to-x$ swaps $1/(1+b^x)$ with $b^x/(1+b^x)$ — and since these two fractions add to exactly $1$, averaging the original and transformed integral removes the exponential entirely, leaving half of $\int_{-a}^a f(x)dx$. This makes the $a^x$ term a red herring: the answer depends only on $\cos^2x$.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_{-\pi}^{\pi}\frac{\cos^2x}{1+a^x}dx$.
   2. Substitute $x\to -x$ (valid since limits are symmetric): $I=\displaystyle\int_{-\pi}^{\pi}\frac{\cos^2(-x)}{1+a^{-x}}dx=\int_{-\pi}^{\pi}\frac{\cos^2x}{1+a^{-x}}dx$ (using $\cos^2(-x)=\cos^2x$).
   3. Simplify $\dfrac{1}{1+a^{-x}} = \dfrac{a^x}{a^x+1}$, so $I=\displaystyle\int_{-\pi}^\pi\frac{\cos^2x\cdot a^x}{1+a^x}dx$.
   4. Add the two expressions for $I$: $2I=\displaystyle\int_{-\pi}^\pi\cos^2x\left[\frac{1}{1+a^x}+\frac{a^x}{1+a^x}\right]dx=\int_{-\pi}^\pi\cos^2x\,dx$. …

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3. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If $A=\int_0^{\infty}\dfrac{1+x^2}{1+x^4}dx$, $B=\int_0^1\dfrac{1+x^2}{1+x^4}dx$, then (A) $2A=B$ (B) $A=B$ (C) $2B=A$ (D) $2B+A=0$

   ›Reveal solutionSolution

   The reciprocal substitution $x=1/t$ maps $[1,\infty)$ onto $[0,1]$ and leaves this particular integrand invariant, so the tail $\int_1^\infty$ equals $B$ itself, giving $A=2B$.

   Concept and Intuition

   For integrands of the form $\dfrac{1+x^2}{1+x^4}$, the substitution $x\to 1/x$ is special: both the numerator/denominator structure and the $dx=-dt/t^2$ factor combine to reproduce the SAME function of the new variable. Recognizing this self-similarity avoids ever computing the (messy) closed form of the integral.

   Step-by-Step Solution

   1. Write $A=\displaystyle\int_0^1\frac{1+x^2}{1+x^4}dx+\int_1^\infty\frac{1+x^2}{1+x^4}dx = B+\int_1^\infty\frac{1+x^2}{1+x^4}dx$.
   2. In the second integral, substitute $x=\dfrac1t$, so $dx=-\dfrac{dt}{t^2}$; as $x:1\to\infty$, $t:1\to0$.
   3. $1+x^2=1+\dfrac1{t^2}=\dfrac{t^2+1}{t^2}$, and $1+x^4=1+\dfrac1{t^4}=\dfrac{t^4+1}{t^4}$, so $\dfrac{1+x^2}{1+x^4}=\dfrac{(t^2+1)t^2}{t^4+1}$.
   4. Multiplying by $dx=-dt/t^2$: the integrand becomes $\dfrac{(t^2+1)t^2}{t^4+1}\cdot\left(-\dfrac{1}{t^2}\right)dt=-\dfrac{t^2+1}{t^4+1}dt$. …

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4. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.$\displaystyle\int_{\pi/6}^{\pi/3} \dfrac{1}{1+\sqrt{\cot x}}\,dx=$ (A) $\pi/4$ (B) $\pi/2$ (C) $\pi/6$ (D) $\pi/12$

   ›Reveal solutionSolution

   Using the reflection property $\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$ with $a+b=\pi/2$ makes the two forms of the integrand add up to exactly $1$, giving $I=\pi/12$.

   Concept and Intuition

   Integrals of the shape $\int \dfrac{dx}{1+\sqrt{h(x)}}$ over symmetric limits around $x=\pi/4$-type reflections often pair up with their "co-function" version to sum to a constant — a classic trick that avoids ever actually antidifferentiating $\cot$ or $\tan$ raised to a half power.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\cot x}}$.
   2. Since $\pi/6+\pi/3=\pi/2$, substitute $x\to\pi/2-x$: $\cot(\pi/2-x)=\tan x$, so $I=\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{dx}{1+\sqrt{\tan x}}$ as well.
   3. Add the two expressions for $I$: $2I=\displaystyle\int_{\pi/6}^{\pi/3}\left[\dfrac1{1+\sqrt{\cot x}}+\dfrac1{1+\sqrt{\tan x}}\right]dx$. …

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5. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin(\pi/4+x)+\sin(3\pi/4+x)}{\cos x+\sin x}dx =$ (A) $\dfrac{\pi}{\sqrt{2}}$ (B) $\dfrac{\pi}{2\sqrt{2}}$ (C) $\dfrac{\pi}{3\sqrt{2}}$ (D) $\dfrac{\pi}{4\sqrt{2}}$

   ›Reveal solutionSolution

   Simplify the numerator using angle-addition identities, then use the classic $x\to\pi/2-x$ symmetry trick (King's rule) to evaluate the definite integral without direct antidifferentiation.

   Concept and Intuition

   Symmetric definite integrals over $[0,\pi/2]$ often simplify beautifully using the substitution $x\to\pi/2-x$, which swaps $\sin$ and $\cos$. Adding the original and transformed integrals frequently collapses to something trivial.

   Step-by-Step Solution

   1. Expand: $\sin(\pi/4+x)=\dfrac{\sqrt2}{2}(\cos x+\sin x)$ and $\sin(3\pi/4+x)=\dfrac{\sqrt2}{2}(\cos x-\sin x)$.
   2. Sum $= \dfrac{\sqrt2}{2}\big[(\cos x+\sin x)+(\cos x-\sin x)\big] = \sqrt2\cos x$.
   3. So $I=\displaystyle\int_0^{\pi/2}\dfrac{\sqrt2\cos x}{\cos x+\sin x}dx$.
   4. Substitute $x\to\pi/2-x$: $I=\displaystyle\int_0^{\pi/2}\dfrac{\sqrt2\sin x}{\sin x+\cos x}dx = J$ (same value, by the King's-rule symmetry). …

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6. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int_{-1}^{1} \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx =$ (A) $2$ (B) $4$ (C) $0$ (D) $8$

   ›Reveal solutionSolution

   The integrand is an odd function of $x$, so its integral over the symmetric interval $[-1,1]$ vanishes.

   Concept and Intuition

   $\int_{-a}^a f(x)dx=0$ whenever $f$ is odd, i.e. $f(-x)=-f(x)$. Recognising this symmetry avoids a painful direct integration of the square roots.

   Step-by-Step Solution

   1. Let $g(x)=\sqrt{1+x+x^2}$. Then $g(-x)=\sqrt{1-x+x^2}$.
   2. The integrand is $f(x)=g(x)-g(-x)$.
   3. Check parity: $f(-x) = g(-x)-g(x) = -\big(g(x)-g(-x)\big) = -f(x)$ — so $f$ is odd.
   4. Hence $\int_{-1}^1 f(x)\,dx = 0$. …

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7. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi} \dfrac{x\tan x}{\sec x + \tan x}\, dx =$ (A) $\dfrac{\pi(\pi-2)}{2}$ (B) $\dfrac{\pi+2}{2}$ (C) $\dfrac{\pi(\pi+2)}{2}$ (D) $\dfrac{\pi-2}{2}$

   ›Reveal solutionSolution

   The integrand times $x$ over $[0,\pi]$ is handled with the symmetry trick $\int_0^a xf(x)dx=\frac{a}{2}\int_0^af(x)dx$ when $f(a-x)=f(x)$.

   Concept and Intuition

   $\tan x/(\sec x+\tan x)$ simplifies to $\sin x/(1+\sin x)$, a function symmetric under $x\to\pi-x$. This symmetry lets us replace the troublesome factor of $x$ with a constant $\pi/2$, reducing the problem to a standard definite integral.

   Step-by-Step Solution

   1. Simplify: $\dfrac{\tan x}{\sec x+\tan x}=\dfrac{\sin x/\cos x}{(1+\sin x)/\cos x}=\dfrac{\sin x}{1+\sin x}=f(x)$.
   2. Check symmetry: $f(\pi-x)=\dfrac{\sin(\pi-x)}{1+\sin(\pi-x)}=\dfrac{\sin x}{1+\sin x}=f(x)$. ✓
   3. By the King's rule, $I=\displaystyle\int_0^\pi xf(x)dx=\dfrac{\pi}{2}\int_0^\pi f(x)dx$.
   4. Compute $J=\displaystyle\int_0^\pi\dfrac{\sin x}{1+\sin x}dx=\int_0^\pi\left[1-\dfrac{1}{1+\sin x}\right]dx=\pi-\int_0^\pi\dfrac{dx}{1+\sin x}$. …

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8. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin^3 x}{\sin x + \cos x}\, dx =$ (A) $\dfrac{\pi + 1}{4}$ (B) $\dfrac{\pi - 2}{2}$ (C) $\dfrac{\pi - 2}{4}$ (D) $\dfrac{\pi - 1}{4}$

   ›Reveal solutionSolution

   By symmetry $I=J$ (the $\cos^3$ analogue) and $I+J=\pi/2-1/2$ via a sum-of-cubes factorisation, giving $I=(\pi-1)/4$.

   Concept and Intuition

   Define $J$ as the same integral with $\sin^3x$ replaced by $\cos^3x$. The substitution $x\to\pi/2-x$ shows $I=J$ exactly (not just numerically), because it swaps sine and cosine while leaving $\sin x+\cos x$ and the interval unchanged. Then the sum-of-cubes identity $\sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x)$ lets $I+J$ collapse to an elementary integral, and since $I=J$, each equals half of that sum.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\dfrac{\sin^3x}{\sin x+\cos x}dx$ and $J=\displaystyle\int_0^{\pi/2}\dfrac{\cos^3x}{\sin x+\cos x}dx$.
   2. Substituting $u=\pi/2-x$ in $I$: $\sin x\to\cos u$, $\cos x\to\sin u$, limits unchanged, giving $I=\displaystyle\int_0^{\pi/2}\dfrac{\cos^3u}{\cos u+\sin u}du = J$. So $I=J$.
   3. $I+J = \displaystyle\int_0^{\pi/2}\dfrac{\sin^3x+\cos^3x}{\sin x+\cos x}dx$. Using $\sin^3x+\cos^3x=(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=(\sin x+\cos x)(1-\sin x\cos x)$, this simplifies to $\displaystyle\int_0^{\pi/2}(1-\sin x\cos x)\,dx$. …

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9. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int_0^{3\pi/2} \dfrac{\cos^3 x}{\cos^3 x+\sin^3 x}\,dx=$ (A) 0 (B) 1 (C) $\dfrac{\pi}{4}$ (D) $\dfrac{3\pi}{4}$

   ›Reveal solutionSolution

   Using the substitution $x\to\frac{3\pi}{2}-x$ swaps the roles of $\cos^3x$ and $\sin^3x$ in the integrand, showing the given integral equals its "sine" counterpart; since the two together give the full interval length, each equals $\frac{3\pi}{4}$.

   Concept and Intuition

   This is a King's-rule-style symmetry trick: $\int_0^{a}f(x)dx=\int_0^af(a-x)dx$. Applying it with $a=3\pi/2$ swaps $\cos x\leftrightarrow-\sin x$ and $\sin x\leftrightarrow -\cos x$, which exactly interchanges the roles of $\cos^3x$ and $\sin^3x$ in the fraction (the minus signs cancel through the cube and the overall ratio), letting us equate the two "complementary" integrals.

   Step-by-Step Solution

   1. Define $I=\displaystyle\int_0^{3\pi/2}\frac{\cos^3x}{\cos^3x+\sin^3x}dx$ and $J=\displaystyle\int_0^{3\pi/2}\frac{\sin^3x}{\cos^3x+\sin^3x}dx$.
   2. Adding: $I+J=\displaystyle\int_0^{3\pi/2}1\,dx=\frac{3\pi}{2}$.
   3. Substitute $x\to\frac{3\pi}{2}-x$ in $I$: $\cos\left(\frac{3\pi}{2}-x\right)=-\sin x$ and $\sin\left(\frac{3\pi}{2}-x\right)=-\cos x$. …

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10. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.$\int_{\pi/11}^{9\pi/22} \dfrac{dx}{1 + \sqrt{\tan x}} =$ (A) $\pi/4$ (B) $\pi/22$ (C) $\pi/11$ (D) $7\pi/44$

    ›Reveal solutionSolution

    This tests the classic King's-rule substitution $x\to a+b-x$ for definite integrals whose limits sum to $\pi/2$, applied to a $1/(1+\sqrt{\tan x})$ integrand.

    Concept and Intuition

    Whenever the limits of integration sum to a value that makes $\tan x \to \cot x$ under the substitution $x\to a+b-x$ (here $a+b=\pi/2$, so $\tan(\pi/2-x)=\cot x$), adding the original and transformed integrals often produces a simple constant integrand.

    Step-by-Step Solution

    1. Check $a+b$: $\dfrac{\pi}{11}+\dfrac{9\pi}{22} = \dfrac{2\pi}{22}+\dfrac{9\pi}{22}=\dfrac{11\pi}{22}=\dfrac{\pi}{2}$.
    2. Substitute $x\to \dfrac\pi2-x$: $\tan\left(\dfrac\pi2-x\right)=\cot x = \dfrac1{\tan x}$, so the transformed integrand is $\dfrac{1}{1+1/\sqrt{\tan x}}=\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}$.
    3. So $I=\displaystyle\int_a^b\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$ as well (same value as the original, by the substitution property). …

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

    ›Reveal solutionSolution

    This is the classic "$x\,f(\sin x)$" King's-rule setup; it reduces to $\pi/2$ times a standard arctan integral, giving $\dfrac{\pi^2}4$ — (D).

    Concept and Intuition

    Whenever the integrand has the shape $x\cdot g(\sin x,\cos^2x)$ over $[0,\pi]$ and $g$ is invariant under $x\to\pi-x$ (true here since $\sin(\pi-x)=\sin x$ and $\cos^2(\pi-x)=\cos^2x$), the King's-rule substitution $x\to\pi-x$ lets us replace the "$x$" prefactor with "$\pi-x$", and adding the two forms eliminates the awkward $x$ entirely.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_0^\pi\dfrac{x\sin x}{1+\cos^2x}dx$ and $f(x)=\dfrac{\sin x}{1+\cos^2x}$.
    2. Check $f(\pi-x)=\dfrac{\sin(\pi-x)}{1+\cos^2(\pi-x)}=\dfrac{\sin x}{1+\cos^2x}=f(x)$ — confirmed invariant.
    3. By King's rule: $I=\displaystyle\int_0^\pi(\pi-x)f(x)\,dx=\pi\int_0^\pi f(x)dx - I$.
    4. So $2I=\pi\displaystyle\int_0^\pi f(x)\,dx=\pi J$, where $J=\displaystyle\int_0^\pi\dfrac{\sin x}{1+\cos^2x}dx$. …

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12. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.$\displaystyle\int_0^{\pi} \dfrac{x\sin x}{1+\cos^2 x}\,dx =$ (A) $\dfrac{\pi^2}{4}$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi^2}{2}$ (D) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    This is the classic $\int_0^\pi x f(\sin x,\cos x)\,dx$ trick using the substitution $x\to \pi-x$; the answer is (A).

    Concept and Intuition

    For integrals of the form $\int_0^\pi x\,g(\cos x)\sin x\,dx$, substituting $x\to\pi-x$ (using $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$, but here $\cos^2$ is unaffected) converts the $x$ factor into $\pi - x$, letting you solve for the integral algebraically in terms of a simpler integral without the $x$ weight.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_0^\pi\frac{x\sin x}{1+\cos^2 x}dx$.
    2. Substitute $x\to\pi-x$: $\sin(\pi-x)=\sin x$, $\cos^2(\pi-x)=\cos^2 x$, so

    $I=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx-I.$

    3. So $2I=\pi\displaystyle\int_0^\pi\frac{\sin x}{1+\cos^2 x}dx$. …

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