Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx$

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1. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The value of $\int_{0}^{\pi/2} \frac{\cos^{11} x}{\cos^{11} x + \sin^{11} x}\, dx$ is equal to (A) $\pi$ (B) $\frac{3\pi}{2}$ (C) $\frac{\pi}{2}$ (D) $\frac{\pi}{4}$ (E) $2\pi$

   ›Reveal solutionSolution

   Apply $\int_0^{a}f(x)dx=\int_0^{a}f(a-x)dx$ with $a=\pi/2$; adding gives twice the value $=\pi/2$.

   Let $I=\int_0^{\pi/2}\frac{\cos^{11}x}{\cos^{11}x+\sin^{11}x}dx$.

   Replacing $x\to\frac\pi2-x$ swaps sine and cosine: $I=\int_0^{\pi/2}\frac{\sin^{11}x}{\sin^{11}x+\cos^{11}x}dx$. …

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2. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.The value of $\int_{0}^{\pi/2} \frac{\cos^{2024} x}{\sin^{2024} x + \cos^{2024} x}\,dx$ is equal to (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{2}$ (C) $2\pi$ (D) $\pi$ (E) $\frac{\pi}{3}$

   ›Reveal solutionSolution

   Apply $x\to\frac{\pi}{2}-x$ and add to get $2I=\frac{\pi}{2}$.

   Let $I=\displaystyle\int_0^{\pi/2}\frac{\cos^{2024}x}{\sin^{2024}x+\cos^{2024}x}\,dx$. Using $x\to\frac{\pi}{2}-x$ swaps sine and cosine:

   $I=\int_0^{\pi/2}\frac{\sin^{2024}x}{\sin^{2024}x+\cos^{2024}x}\,dx.$ …

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3. KEAM 2024Set eng-2024-06084 marksMCQ
   Q.$\displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{\tan x+\sin x}{1+\cos^2 x}\,dx$ is equal to (A) $0$ (B) $2$ (C) $\sqrt{2}$ (D) $2\sqrt{2}$ (E) $-2\sqrt{2}$

   ›Reveal solutionSolution

   Numerator $\tan x+\sin x$ is odd and denominator $1+\cos^2x$ is even, so the whole integrand is odd; integrating an odd function over a symmetric interval gives $0$.

   Let $g(x)=\dfrac{\tan x+\sin x}{1+\cos^2 x}$. Replace $x$ by $-x$:

   • $\tan(-x)=-\tan x$ and $\sin(-x)=-\sin x$, so the numerator changes sign.
   • $\cos(-x)=\cos x$, so $1+\cos^2 x$ is unchanged. …

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4. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.$\int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx$ is equal to (A) $0$ (B) $\frac{\pi}{6}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{12}$ (E) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   Apply $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$ with $a+b=\frac{\pi}{2}$, which swaps $\sin$ and $\cos$.

   Let $I=\displaystyle\int_{\pi/6}^{\pi/3}\dfrac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx$. …

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5. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The value of $\displaystyle\int_0^{2\pi} \dfrac{\sin 2x}{x(2\pi - x)}\, dx$ is equal to (A) $\pi$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{8}$ (E) $0$

   ›Reveal solutionSolution

   The substitution $x\to 2\pi-x$ flips the sign of $\sin 2x$ while leaving $x(2\pi-x)$ unchanged, so the integrand is antisymmetric about $x=\pi$ and the integral vanishes. …

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6. KEAM 2024Set eng-2024-06054 marksMCQ
   Q.$\int_{-\pi/2}^{\pi/2} \frac{\cos^2 x}{1 + 2^{-x}}\,dx$ is equal to (A) $\frac{\pi}{3}$ (B) $\frac{\pi}{4}$ (C) $1$ (D) $\frac{1}{2}$ (E) $\frac{\pi}{2}$

   ›Reveal solutionSolution

   The $\frac{1}{1+2^{-x}}$ factor with an even numerator halves the plain integral of $\cos^2 x$: $I=\tfrac12\int_{-\pi/2}^{\pi/2}\cos^2x\,dx=\tfrac{\pi}{4}$.

   Let $I=\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^2 x}{1+2^{-x}}\,dx$. Replace $x\to -x$ (limits symmetric, $\cos^2$ even):

   $I=\displaystyle\int_{-\pi/2}^{\pi/2}\frac{\cos^2 x}{1+2^{x}}\,dx$.

   Adding the two forms: …

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7. KEAM 2021Set eng-2021-P2-B14 marksMCQ
   Q.The value of $\displaystyle\int_{\pi/8}^{3\pi/8} \dfrac{\sin^4 x}{\sin^4 x + \cos^4 x} \, dx$ is equal to (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{8}$ (C) $\dfrac{\pi}{16}$ (D) $\dfrac{\pi}{2}$ (E) $1$

   ›Reveal solutionSolution

   $\displaystyle\int_{\pi/8}^{3\pi/8}\dfrac{\sin^4 x}{\sin^4 x+\cos^4 x}\,dx=\dfrac{\pi}{8}$.

   Concept and Intuition

   The king property $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$ with $a+b=\pi/2$ pairs the integrand with its cosine-counterpart, and the two add to $1$.

   Step-by-Step Solution

   1. Let $I=\int_{\pi/8}^{3\pi/8}\dfrac{\sin^4 x}{\sin^4 x+\cos^4 x}dx$.
   2. Replace $x\to\tfrac\pi2-x$: $I=\int_{\pi/8}^{3\pi/8}\dfrac{\cos^4 x}{\cos^4 x+\sin^4 x}dx$. …

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8. KEAM 2025Set eng-2025-04274 marksMCQ
   Q.$\int_{-\pi/2}^{\pi/2} (x^5 + x^3 + x)\cos x\,dx =$ (A) $\dfrac{\pi}{4}$ (B) $\pi$ (C) $\dfrac{2\pi}{3}$ (D) $\dfrac{\pi}{2}$ (E) $0$

   ›Reveal solutionSolution

   The integrand is an odd function on a symmetric interval, so the integral is $0$.

   Let $g(x)=(x^5+x^3+x)\cos x$. Here $x^5+x^3+x$ is odd and $\cos x$ is even, so

   $g(-x)=(-x^5-x^3-x)\cos x=-g(x),$

   i.e. $g$ is odd. …

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9. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.$\int_0^2\dfrac{x^4}{x^4+(2-x)^4}\,dx=$ (A) $1$ (B) $2$ (C) $4$ (D) $8$ (E) $0$

   ›Reveal solutionSolution

   Apply the king property $x\to 2-x$; adding the two forms integrates to $2$, giving $I=1$.

   Let $I=\displaystyle\int_0^2\dfrac{x^{4}}{x^{4}+(2-x)^{4}}\,dx$. Replacing $x\to 2-x$,

   $I=\int_0^2\dfrac{(2-x)^{4}}{(2-x)^{4}+x^{4}}\,dx.$

   Adding the two expressions, …

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10. KEAM 2024Set eng-2024-06094 marksMCQ
    Q.$\int_{-\pi/2}^{\pi/2} \sin^9 x \cos^2 x\, dx =$ (A) $\dfrac{2}{3}$ (B) $1$ (C) $\dfrac{1}{11}$ (D) $\dfrac{7\pi}{6}$ (E) $0$

    ›Reveal solutionSolution

    $\sin^9x$ is odd and $\cos^2x$ is even, so the product is odd; integrating an odd function over $[-\frac{\pi}{2},\frac{\pi}{2}]$ gives $0$.

    Let $g(x)=\sin^9x\cos^2x$. Then $g(-x)=\sin^9(-x)\cos^2(-x)=-\sin^9x\cos^2x=-g(x)$, so $g$ is odd. …

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11. KEAM 2025Set eng-2025-04264 marksMCQ
    Q.Given that $\displaystyle\int_0^1 \tan^{-1}(t)\, dt = \dfrac{\pi}{4} - \dfrac{1}{2}\log 2$. Then $\displaystyle\int_0^1 \tan^{-1}(1-t)\, dt =$ (A) $\dfrac{\pi}{2} - \dfrac{1}{2}\log 2$ (B) $\dfrac{\pi}{4} - \dfrac{1}{2}\log 3$ (C) $\dfrac{\pi}{4} + \dfrac{1}{2}\log 2$ (D) $\dfrac{\pi}{4} + \dfrac{1}{2}\log 2$ (E) $\dfrac{\pi}{4} - \dfrac{1}{2}\log 2$

    ›Reveal solutionSolution

    [!TLDR]

    The substitution $u=1-t$ turns the required integral into the given one, so both equal $\frac{\pi}{4}-\frac{1}{2}\log 2$ — option (E).

    Concept

    The definite-integral property $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$ (a standard NCERT/CBSE result) lets us replace the variable by its reflection about the midpoint of the interval.

    Solution

    Let $u=1-t$, so $du=-dt$. The limits swap: $t=0\Rightarrow u=1$ and $t=1\Rightarrow u=0$. Then

    $\int_0^1 \tan^{-1}(1-t)\,dt=\int_1^0 \tan^{-1}(u)\,(-du)=\int_0^1 \tan^{-1}(u)\,du.$ …

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12. KEAM 2026Set eng-2026-04194 marksMCQ
    Q.If $I = \int_{-1}^{1} \left(\frac{x^4}{1 - x^4}\right)\cos^{-1}\left(\frac{2x}{1 + x^2}\right) dx$, then $2I$ is equal to (A) $\pi\int_{-1}^{1} \frac{x^4}{1 - x^4} dx$ (B) $2\pi\int_{-1}^{1} \frac{x^4}{1 - x^4} dx$ (C) $\int_{-1}^{1} \frac{x^4}{1 - x^4} dx$ (D) $\pi\int_{-1}^{1} \frac{x^4}{1 + x^4} dx$ (E) $-\pi\int_{-1}^{1} \frac{x^4}{1 - x^4} dx$

    ›Reveal solutionSolution

    The weight $x^4/(1-x^4)$ is even, and $\cos^{-1}\tfrac{2x}{1+x^2}$ satisfies $g(x)+g(-x)=\pi$; combining $I$ with its $x\to -x$ copy gives $2I=\pi\int_{-1}^{1}\tfrac{x^4}{1-x^4}dx$.

    Let $f(x)=\dfrac{x^4}{1-x^4}$, which is even, and $g(x)=\cos^{-1}\!\left(\dfrac{2x}{1+x^2}\right)$.

    Since $\dfrac{2(-x)}{1+x^2}=-\dfrac{2x}{1+x^2}$ and $\cos^{-1}(-a)=\pi-\cos^{-1}(a)$,

    $g(-x)=\pi-g(x)\;\Rightarrow\;g(x)+g(-x)=\pi.$ …

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