Q.$\int_{0}^{\pi} \sin x \, dx =$ ___.

Showing the 12 most recent of 27 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.The antiderivative of $\frac{1}{x\sqrt{x^2-1}}$, $x>1$ with respect to $x$

   (a) $\sin^{-1}x+c$

   (b) $\cos^{-1}x+c$

   (c) $-\operatorname{cosec}^{-1}x+c$

   (d) $\cot^{-1}x+c$

   ›Reveal solutionSolution

   The standard antiderivative is $\sec^{-1}x$, which equals $-\operatorname{cosec}^{-1}x+c$; answer (c).

   The standard result is

   $\int\frac{dx}{x\sqrt{x^2-1}}=\sec^{-1}|x|+c,\qquad x>1.$

   Since $\sec^{-1}x+\operatorname{cosec}^{-1}x=\dfrac{\pi}{2}$ (a constant), we have $\sec^{-1}x=\dfrac{\pi}{2}-\operatorname{cosec}^{-1}x$, so the antiderivative can equally be written …

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2. CBSE 2025Set A1 mark
   Q.Match the correct pair. Column A (i): $\displaystyle\int \dfrac{dx}{\sqrt{x^2+a^2}}$. Column B options:

   (a) $\log|\sin x| + c$

   (b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$

   (c) $\csc^{-1}x$

   (d) $\log\left|x+\sqrt{x^2-9}\right| + c$

   (e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (g) $\log|\sec x| + c$. Write the letter of the option that matches Column A (i).

   ›Reveal solutionSolution

   This is the standard NCERT integral formula $\displaystyle\int\dfrac{dx}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right|+c$.

   This is one of the standard results proved (in the NCERT text) using the substitution $x = a\tan\theta$: …

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3. CBSE 2025Set A1 mark
   Q.Match the correct pair. Column A (ii): $\displaystyle\int \sqrt{x^2+a^2}\,dx$. Column B options:

   (a) $\log|\sin x| + c$

   (b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$

   (c) $\csc^{-1}x$

   (d) $\log\left|x+\sqrt{x^2-9}\right| + c$

   (e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (g) $\log|\sec x| + c$. Write the letter of the option that matches Column A (ii).

   ›Reveal solutionSolution

   This is the standard NCERT formula for $\int\sqrt{x^2+a^2}\,dx$.

   This is a standard result (derived in the NCERT text via integration by parts): …

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4. CBSE 2025Set A1 mark
   Q.Match the correct pair. Column A (iii): $\displaystyle\int \tan x\,dx$. Column B options:

   (a) $\log|\sin x| + c$

   (b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$

   (c) $\csc^{-1}x$

   (d) $\log\left|x+\sqrt{x^2-9}\right| + c$

   (e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (g) $\log|\sec x| + c$. Write the letter of the option that matches Column A (iii).

   ›Reveal solutionSolution

   Write $\tan x = \sin x/\cos x$ and integrate by substitution $u=\cos x$.

   $\int \tan x\,dx = \int \dfrac{\sin x}{\cos x}\,dx$

   Let $u = \cos x$, $du = -\sin x\,dx$: …

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5. CBSE 2025Set A1 mark
   Q.Match the correct pair. Column A (iv): $\displaystyle\int \cot x\,dx$. Column B options:

   (a) $\log|\sin x| + c$

   (b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$

   (c) $\csc^{-1}x$

   (d) $\log\left|x+\sqrt{x^2-9}\right| + c$

   (e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (g) $\log|\sec x| + c$. Write the letter of the option that matches Column A (iv).

   ›Reveal solutionSolution

   Write $\cot x = \cos x/\sin x$ and integrate by substitution $u=\sin x$.

   $\int \cot x\,dx = \int \dfrac{\cos x}{\sin x}\,dx$

   Let $u = \sin x$, $du = \cos x\,dx$: …

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6. CBSE 2025Set A1 mark
   Q.Match the correct pair. Column A (v): $\displaystyle\int \sqrt{x^2-9}\,dx$. Column B options:

   (a) $\log|\sin x| + c$

   (b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$

   (c) $\csc^{-1}x$

   (d) $\log\left|x+\sqrt{x^2-9}\right| + c$

   (e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (g) $\log|\sec x| + c$. Write the letter of the option that matches Column A (v).

   ›Reveal solutionSolution

   This is the standard NCERT formula $\int\sqrt{x^2-a^2}\,dx$ with $a=3$.

   The standard result (by parts, same family as $\int\sqrt{x^2+a^2}\,dx$ but with a sign change) is:

   $\int \sqrt{x^2-a^2}\,dx = \dfrac{x}{2}\sqrt{x^2-a^2} - \dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right| + c$

   With $a^2 = 9$ (i.e. $a=3$): …

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7. CBSE 2025Set A1 mark
   Q.Match the correct pair. Column A (vi): $\displaystyle\int \dfrac{dx}{\sqrt{x^2-9}}$. Column B options:

   (a) $\log|\sin x| + c$

   (b) $\dfrac{x}{2}\sqrt{x^2-9} - \dfrac{9}{2}\log\left|x+\sqrt{x^2-9}\right| + c$

   (c) $\csc^{-1}x$

   (d) $\log\left|x+\sqrt{x^2-9}\right| + c$

   (e) $\dfrac{x}{2}\sqrt{x^2+a^2} + \dfrac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (f) $\log\left|x+\sqrt{x^2+a^2}\right| + c$

   (g) $\log|\sec x| + c$. Write the letter of the option that matches Column A (vi).

   ›Reveal solutionSolution

   This is the standard NCERT formula $\int \dfrac{dx}{\sqrt{x^2-a^2}}$ with $a=3$.

   The standard result (by substitution $x=a\sec\theta$) is:

   $\int \dfrac{dx}{\sqrt{x^2-a^2}} = \log\left|x+\sqrt{x^2-a^2}\right| + c$

   With $a=3$: …

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8. CBSE 2025Set ANNUAL1 mark
   Q.Find an anti-derivative of $(ax+b)^2$ by the method of inspection. OR Find the order and degree of the differential equation $\dfrac{d^2y}{dx^2} + 5x\left(\dfrac{dy}{dx}\right)^2 - 6y = \log x$

   ›Reveal solutionSolution

   Guess a cube and adjust the constant so its derivative reproduces $(ax+b)^2$.

   We seek $F(x)$ with $F'(x)=(ax+b)^2$. Trying $F(x)=(ax+b)^3$:

   $\frac{d}{dx}(ax+b)^3=3(ax+b)^2\cdot a=3a(ax+b)^2,$

   which is $3a$ times too large. Dividing by $3a$:

   $\frac{d}{dx}\left[\frac{(ax+b)^3}{3a}\right]=(ax+b)^2.$

   …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.$\int_0^1 \frac{1}{\sqrt{1-x^2}} \,dx =$

   (a) $\frac{\pi}{2}$

   (b) $\frac{\pi}{4}$

   (c) $\frac{2\pi}{3}$

   (d) $\frac{3\pi}{2}$

   ›Reveal solutionSolution

   Integrate 1/√(1-x^2) to get sin^{-1}x, then evaluate between 0 and 1.

   …

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10. CBSE 2024Set ANNUAL1 mark
    Q.Evaluate : $\displaystyle\int \dfrac{dx}{\sqrt{9-25x^2}}$ OR Integrate w.r.t. $x$ : $\dfrac{1}{x+x\log x}$

    ›Reveal solutionSolution

    Reduce to the standard form $\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac{x}{a}$.

    Rewrite the denominator:

    $\sqrt{9-25x^2} = \sqrt{25\left(\dfrac{9}{25}-x^2\right)} = 5\sqrt{\left(\dfrac{3}{5}\right)^2 - x^2}.$

    So

    $\int \dfrac{dx}{\sqrt{9-25x^2}} = \dfrac{1}{5}\int \dfrac{dx}{\sqrt{(3/5)^2 - x^2}} = \dfrac{1}{5}\sin^{-1}\!\dfrac{x}{3/5} + C = \dfrac{1}{5}\sin^{-1}\!\dfrac{5x}{3} + C.$

    …

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11. CBSE 2023Set 65/1/11 markMCQ
    Q.$\int_0^{\pi/6} \sec^2\left(x - \frac{\pi}{6}\right) dx$ is equal to:

    (a) $\frac{1}{\sqrt{3}}$

    (b) $-\frac{1}{\sqrt{3}}$

    (c) $\sqrt{3}$

    (d) $-\sqrt{3}$

    ›Reveal solutionSolution

    The integral simplifies by noticing the derivative of $\tan$ is $\sec^2$, so the antiderivative is $\tan\left(x - \frac{\pi}{6}\right)$. Evaluating from $0$ to $\frac{\pi}{6}$ gives $\frac{1}{\sqrt{3}}$, which is option (a).

    The key here is recognising that $\sec^2$ is the derivative of $\tan$. That means the integral is essentially a direct application of the fundamental theorem — no substitution or trickery needed. The only subtlety is handling the shift inside the argument: $x - \frac{\pi}{6}$ just shifts the tangent function horizontally, which doesn't change the fact that its derivative is $\sec^2$ of the same argument.

    1. Identify the antiderivative. We know $\frac{d}{dx} \tan u = \sec^2 u \cdot \frac{du}{dx}$. Here $u = x - \frac{\pi}{6}$, so $\frac{du}{dx} = 1$. Therefore,

    $\int \sec^2\left(x - \frac{\pi}{6}\right) dx = \tan\left(x - \frac{\pi}{6}\right) + C.$

    2. Apply the limits. The definite integral from $0$ to $\frac{\pi}{6}$ is:

    $\left[ \tan\left(x - \frac{\pi}{6}\right) \right]_{0}^{\pi/6} = \tan\left(\frac{\pi}{6} - \frac{\pi}{6}\right) - \tan\left(0 - \frac{\pi}{6}\right).$

    3. Simplify each term.
       • At the upper limit: $\tan(0) = 0$.
       • At the lower limit: $\tan\left(-\frac{\pi}{6}\right)$. Since $\tan$ is an odd function, $\tan(-\theta) = -\tan\theta$, so

    $\tan\left(-\frac{\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}.$

    4. Compute the difference. …

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12. CBSE 2023Set E1 markMCQ
    Q.$\displaystyle\int\sin\dfrac{3x}{4}\,dx =$

    (a) $k-\dfrac{3}{4}\cos\dfrac{3x}{4}$

    (b) $k+\dfrac{3}{4}\cos\dfrac{3x}{4}$

    (c) $k-\dfrac{4}{3}\cos\dfrac{3x}{4}$

    (d) $k+\dfrac{4}{3}\cos\dfrac{3x}{4}$

    ›Reveal solutionSolution

    $\displaystyle\int\sin\dfrac{3x}{4}\,dx=-\dfrac{4}{3}\cos\dfrac{3x}{4}+k$.

    Using $\int\sin(ax)\,dx=-\dfrac{1}{a}\cos(ax)$ with $a=\dfrac{3}{4}$:

    …

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