Q.Evaluate $\int_{\pi/6}^{\pi/3} \dfrac{dx}{1 + \sqrt{\tan x}}$

Showing the 12 most recent of 48 on this concept.

1. 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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2. 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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3. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/4} \log(1 + \tan x)\,dx =$ (A) $\pi \log 2 + 1$ (B) $\dfrac{\pi}{2}\log 2 + 1$ (C) $\dfrac{\pi}{4}\log 2$ (D) $\dfrac{\pi}{8}\log 2$

   ›Reveal solutionSolution

   A textbook symmetry trick ($x \to \pi/4-x$) doubles the integral into something trivial to evaluate. Answer: $\dfrac{\pi}{8}\log2$.

   Concept and Intuition

   Whenever the limits are $0$ to $a$ and the integrand involves $\tan x$, trying the substitution $x \to a-x$ is a classic move — here it converts $\log(1+\tan x)$ into $\log\!\left(\dfrac{2}{1+\tan x}\right)$, and adding the original and transformed integrals collapses most of the complexity.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/4}\log(1+\tan x)\,dx$.
   2. Substitute $x\to \dfrac{\pi}{4}-x$: $\tan\!\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-\tan x}{1+\tan x}$, so $1+\tan\!\left(\dfrac\pi4-x\right) = \dfrac{2}{1+\tan x}$.
   3. So $I = \displaystyle\int_0^{\pi/4}\log\!\left(\frac{2}{1+\tan x}\right)dx = \int_0^{\pi/4}\log2\,dx - \int_0^{\pi/4}\log(1+\tan x)\,dx = \frac{\pi}{4}\log2 - I$. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{x\sin 2x}{1 + 4\cos^2 2x}\, dx =$ (A) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 2\right)$ (B) $\dfrac{\pi}{4}\left(\mathrm{Tan}^{-1} 2\right)$ (C) $\dfrac{\pi}{2}\left(\mathrm{Tan}^{-1} 2\right)$ (D) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 4\right)$

   ›Reveal solutionSolution

   Use the King's-rule symmetry $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ to eliminate the explicit $x$, then substitute; the result is $\dfrac\pi8\mathrm{Tan}^{-1}2$.

   Concept and Intuition

   Whenever an integrand is $x$ times a function that is itself symmetric (or anti-symmetric) under $x\to a-x$, replacing $x$ by $a-x$ and adding to the original kills the explicit $x$-dependence, leaving a much simpler integral to evaluate.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\frac{x\sin2x}{1+4\cos^22x}\,dx$.
   2. Replace $x\to\frac\pi2-x$: since $\sin(\pi-2x)=\sin2x$ and $\cos(\pi-2x)=-\cos2x$ (so $\cos^2$ is unchanged),

   $I=\int_0^{\pi/2}\frac{\left(\frac\pi2-x\right)\sin2x}{1+4\cos^22x}\,dx=\frac\pi2\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx - I.$

   3. So $2I=\dfrac\pi2 J$ where $J=\displaystyle\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx$.
   4. Let $w=\cos2x$, $dw=-2\sin2x\,dx$. Limits: $x=0\Rightarrow w=1$; $x=\pi/2\Rightarrow w=-1$.

   $J=\int_{1}^{-1}\frac{-dw/2}{1+4w^2}=\int_{-1}^{1}\frac{dw/2}{1+4w^2}=\int_0^1\frac{dw}{1+4w^2}$

   (using the evenness of the integrand over the symmetric interval). …

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5. 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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6. 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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7. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.$\displaystyle\int_{0}^{\pi/2} \log|\tan x + \cot x|\,dx =$ (A) $\pi \log 2$ (B) $-\pi \log 2$ (C) $\dfrac{\pi}{2}\log 2$ (D) $2\pi \log 2$

   ›Reveal solutionSolution

   Simplify $\tan x+\cot x=2/\sin2x$, split the log, and use the classical result $\int_0^\pi \log\sin u\,du=-\pi\log2$. Answer: $\pi\log2$.

   Concept and Intuition

   $\tan x + \cot x$ always simplifies to $\dfrac{2}{\sin 2x}$ via the Pythagorean identity — a very common simplification in definite-integral problems. This converts the problem into the well-known integral of $\log \sin$, whose value over a full "half-period" ($0$ to $\pi$) is a standard memorized result, $-\pi\log 2$.

   Step-by-Step Solution

   1. $\tan x + \cot x = \dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x} = \dfrac{\sin^2x+\cos^2x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x} = \dfrac{2}{\sin 2x}$.
   2. So $\log|\tan x+\cot x| = \log 2 - \log(\sin 2x)$ (since $\sin2x>0$ on $(0,\pi/2)$).
   3. $\displaystyle\int_0^{\pi/2}\log|\tan x+\cot x|\,dx = \frac{\pi}{2}\log 2 - \int_0^{\pi/2}\log(\sin2x)\,dx$.
   4. Let $I=\displaystyle\int_0^{\pi/2}\log(\sin2x)\,dx$. Substitute $u=2x,\,du=2dx$: $I=\dfrac12\displaystyle\int_0^{\pi}\log(\sin u)\,du$. …

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8. 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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9. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.$\int_{-\pi}^{\pi}\dfrac{x\sin^3x}{4-\cos^2x}dx=$ (A) $2\pi(1-\log3)$ (B) $2\pi\left(1-\dfrac{3}{4}\log3\right)$ (C) $\pi\left(1-\dfrac{3}{4}\log3\right)$ (D) $4\pi(1-\log3)$

   ›Reveal solutionSolution

   Use symmetry (odd integrand times $x$, plus $f(\pi-x)=f(x)$) to strip the $x$ out of the integral, then finish with a $u=\cos x$ substitution and partial fractions.

   Concept and Intuition

   Integrals of the form $\int_{-a}^a x\,g(x)\,dx$ where $g$ is odd become $2\int_0^a x\,g(x)\,dx$ (since $x\,g(x)$ is even). If additionally $g(\pi-x)=g(x)$ on $[0,\pi]$, the classic "King's Rule" trick $\int_0^\pi x\,g(x)\,dx=\dfrac\pi2\int_0^\pi g(x)\,dx$ removes the $x$ factor entirely.

   Step-by-Step Solution

   1. Let $g(x)=\dfrac{\sin^3x}{4-\cos^2x}$. Since $\sin^3(-x)=-\sin^3x$ and $\cos^2(-x)=\cos^2x$, $g$ is odd, so $x\,g(x)$ is even: $\int_{-\pi}^\pi xg(x)dx=2\int_0^\pi xg(x)dx$.
   2. Also $g(\pi-x)=\dfrac{\sin^3(\pi-x)}{4-\cos^2(\pi-x)}=\dfrac{\sin^3x}{4-\cos^2x}=g(x)$ (since $\sin(\pi-x)=\sin x$, $\cos(\pi-x)=-\cos x$), so King's Rule gives $\int_0^\pi xg(x)dx=\dfrac\pi2\int_0^\pi g(x)dx$.
   3. Combining: original integral $=2\cdot\dfrac\pi2\int_0^\pi g(x)dx=\pi\int_0^\pi g(x)dx=\pi J$.
   4. Compute $J=\int_0^\pi \dfrac{\sin^3x}{4-\cos^2x}dx$ via $u=\cos x$: $J=\int_{-1}^{1}\dfrac{1-u^2}{4-u^2}du=2\int_0^1\left[1-\dfrac{3}{4-u^2}\right]du$ (using $1-u^2=(4-u^2)-3$). …

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10. 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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11. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
    Q.$\displaystyle\int_{-\pi/4}^{\pi/4} x\tan(1+x^2)\,dx=$ (A) 0 (B) $\dfrac{\pi}{4}$ (C) $\dfrac{-\pi}{4}$ (D) 1

    ›Reveal solutionSolution

    This is a symmetric-limits definite integral where checking odd/even parity instantly gives the answer without doing any actual antiderivative work: it is $0$.

    Concept and Intuition

    For $\displaystyle\int_{-a}^{a} f(x)\,dx$: if $f$ is odd ($f(-x)=-f(x)$) the integral is $0$; if $f$ is even it equals $2\int_0^a f(x)dx$. Any function built as (odd function of $x$) $\times$ (function of $x^2$) is automatically odd, because a function of $x^2$ never changes sign under $x\to -x$.

    Step-by-Step Solution

    1. Let $f(x) = x\tan(1+x^2)$.
    2. Replace $x$ by $-x$: $f(-x) = (-x)\tan(1+(-x)^2) = -x\tan(1+x^2) = -f(x)$.
    3. So $f$ is odd on the symmetric interval $\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]$. …

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12. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.$\int_{-4\pi}^{4\pi}\tan^9 x\sin^6 x\cos^3 x\,dx=$ (A) $16\times\frac{\pi}{2}$ (B) $8\times\frac{2}{3}$ (C) $16\times\frac{14}{17}\times\frac{12}{15}\times\ldots\times\frac{2}{3}$ (D) $0$

    ›Reveal solutionSolution

    Odd × even × even = odd, and the integral of any odd function over a symmetric interval is zero — no actual antiderivative work is needed.

    Concept and Intuition

    Before grinding through a nasty trig integral, always check parity. $\tan(-x)=-\tan x$ (odd), and raising an odd function to an odd power ($9$) keeps it odd. $\sin(-x)=-\sin x$ raised to an even power ($6$) becomes even, and $\cos(-x)=\cos x$ raised to any power stays even. Odd times even times even is odd, and an odd function's graph is antisymmetric about the origin, so equal positive and negative area cancels exactly over any interval symmetric about $0$.

    Step-by-Step Solution

    1. Let $g(x)=\tan^9x\sin^6x\cos^3x$.
    2. $g(-x)=\tan^9(-x)\sin^6(-x)\cos^3(-x)=(-\tan x)^9(\sin x)^6(\cos x)^3=-\tan^9x\sin^6x\cos^3x=-g(x)$.
    3. So $g$ is an odd function. …

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