Prove that ∫ dfrac1√(a²-x²)dx = sin⁻¹((x)/(a))+C.

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\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\left(\dfrac{x}{a}\right)+C

Manipur CohsemCOHSEM Manipur Higher Secondary Board 2019Subjective· 3mImportance★★★★★

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substitute x = a sinθ

Let $I=\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}$ . Put $x=a\sin\theta$ , $dx=a\cos\theta\,d\theta$ .

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 30 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. \int (from 0 to 1) dx/\sqrt(1-x²) is equal to:$
(a) π/2
(b) π/4
(c) π/3
(d) π/6
›Reveal solutionSolution

$\frac{1}{\sqrt{1-x^2}}$ is the derivative of $\sin^{-1}x$ , so the definite integral is just $\sin^{-1}(1) - \sin^{-1}(0)$ .

We know the standard result $\displaystyle\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$ .

So: …

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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int_{\sqrt2}^{x} \dfrac{dt}{t\sqrt{t^2-1}} = \dfrac{\pi}{12}$
(a) 2
(b) 1
(c) 0
(d) -1
›Reveal solutionSolution

The antiderivative of $\dfrac{1}{t\sqrt{t^2-1}}$ is $\sec^{-1}t$ ; apply the limits and solve for $x$ .

We know the standard result:

$\int\dfrac{dt}{t\sqrt{t^2-1}} = \sec^{-1}|t| + C$

So

$\int_{\sqrt2}^{x}\dfrac{dt}{t\sqrt{t^2-1}} = \sec^{-1}x - \sec^{-1}\sqrt2$

Since $\sec\dfrac{\pi}{4} = \sqrt2$ , we have $\sec^{-1}\sqrt2 = \dfrac{\pi}{4}$ .

Given this equals $\dfrac{\pi}{12}$ :

$\sec^{-1}x - \dfrac{\pi}{4} = \dfrac{\pi}{12}$ …

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CBSE 2024Set A1 mark
$\int \sqrt{x^2-a^2}\,dx$
(a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$
(b) $\sin^{-1}\dfrac{x}{a}+c$
(c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$
(d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$
(e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$
(f) $\sec^{-1}x$
(g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$
(h) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+c$
(i) $\tan^{-1}x$ . Which entry of Column B matches this Column A item?
›Reveal solutionSolution

$\displaystyle\int\sqrt{x^2-a^2}\,dx$ matches entry (e), the standard NCERT formula for this integral.

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CBSE 2024Set A1 mark
$\int \sqrt{a^2-x^2}\,dx$
(a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$
(b) $\sin^{-1}\dfrac{x}{a}+c$
(c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$
(d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$
(e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$
(f) $\sec^{-1}x$
(g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$
(h) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+c$
(i) $\tan^{-1}x$ . Which entry of Column B matches this Column A item?
›Reveal solutionSolution

$\displaystyle\int\sqrt{a^2-x^2}\,dx$ matches entry (c), the standard NCERT formula for this integral.

…

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CBSE 2024Set A1 mark
$\int \dfrac{dx}{\sqrt{x^2-a^2}}$
(a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$
(b) $\sin^{-1}\dfrac{x}{a}+c$
(c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$
(d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$
(e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$
(f) $\sec^{-1}x$
(g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$
(h) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+c$
(i) $\tan^{-1}x$ . Which entry of Column B matches this Column A item?
›Reveal solutionSolution

$\displaystyle\int\dfrac{dx}{\sqrt{x^2-a^2}}$ matches entry (d).

…

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CBSE 2024Set A1 mark
$\int \dfrac{dx}{\sqrt{a^2-x^2}}$
(a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$
(b) $\sin^{-1}\dfrac{x}{a}+c$
(c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$
(d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$
(e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$
(f) $\sec^{-1}x$
(g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$
(h) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+c$
(i) $\tan^{-1}x$ . Which entry of Column B matches this Column A item?
›Reveal solutionSolution

$\displaystyle\int\dfrac{dx}{\sqrt{a^2-x^2}}$ matches entry (b).

…

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CBSE 2024Set A1 mark
$\int \dfrac{dx}{x^2-a^2}$
(a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$
(b) $\sin^{-1}\dfrac{x}{a}+c$
(c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$
(d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$
(e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$
(f) $\sec^{-1}x$
(g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$
(h) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+c$
(i) $\tan^{-1}x$ . Which entry of Column B matches this Column A item?
›Reveal solutionSolution

$\displaystyle\int\dfrac{dx}{x^2-a^2}$ matches entry (a).

…

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CBSE 2024Set A1 mark
$\int \dfrac{dx}{a^2-x^2}$
(a) $\dfrac{1}{2a}\log\left|\dfrac{x-a}{x+a}\right|+c$
(b) $\sin^{-1}\dfrac{x}{a}+c$
(c) $\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+c$
(d) $\log\left|x+\sqrt{x^2-a^2}\right|+c$
(e) $\dfrac{x}{2}\sqrt{x^2-a^2}-\dfrac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+c$
(f) $\sec^{-1}x$
(g) $\dfrac{1}{2a}\log\left|\dfrac{a+x}{a-x}\right|+c$
(h) $\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}+c$
(i) $\tan^{-1}x$ . Which entry of Column B matches this Column A item?
›Reveal solutionSolution

$\displaystyle\int\dfrac{dx}{a^2-x^2}$ matches entry (g).

…

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CBSE 2023Set A1 mark
$\int \tan x\,dx$
(a) $\log|\sin x| + c$
(b) $\cos 2x$
(c) $\sec^2 x$
(d) $2\cos 2x$
(e) $\log|\text{cosec}\,x - \cot x| + c$
(f) $-\log|\cos x| + c$
(g) $\log|\sec x + \tan x| + c$
›Reveal solutionSolution

$\displaystyle\int\tan x\,dx=-\log|\cos x|+c$ , matching option (f).

Write $\tan x=\dfrac{\sin x}{\cos x}$ and substitute $u=\cos x$ , $du=-\sin x\,dx$ : …

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CBSE 2023Set A1 mark
$\int \cot x\,dx$
(a) $\log|\sin x| + c$
(b) $\cos 2x$
(c) $\sec^2 x$
(d) $2\cos 2x$
(e) $\log|\text{cosec}\,x - \cot x| + c$
(f) $-\log|\cos x| + c$
(g) $\log|\sec x + \tan x| + c$
›Reveal solutionSolution

$\displaystyle\int\cot x\,dx=\log|\sin x|+c$ , matching option (a).

Write $\cot x=\dfrac{\cos x}{\sin x}$ and substitute $u=\sin x$ , $du=\cos x\,dx$ : …

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CBSE 2023Set A1 mark
$\int \sec x\,dx$
(a) $\log|\sin x| + c$
(b) $\cos 2x$
(c) $\sec^2 x$
(d) $2\cos 2x$
(e) $\log|\text{cosec}\,x - \cot x| + c$
(f) $-\log|\cos x| + c$
(g) $\log|\sec x + \tan x| + c$
›Reveal solutionSolution

$\displaystyle\int\sec x\,dx=\log|\sec x+\tan x|+c$ , matching option (g). This is a standard result — a memorised integral, verified by differentiating the RHS.

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CBSE 2023Set A1 mark
$\int \text{cosec}\,x\,dx$
(a) $\log|\sin x| + c$
(b) $\cos 2x$
(c) $\sec^2 x$
(d) $2\cos 2x$
(e) $\log|\text{cosec}\,x - \cot x| + c$
(f) $-\log|\cos x| + c$
(g) $\log|\sec x + \tan x| + c$
›Reveal solutionSolution

$\displaystyle\int \text{cosec}\,x\,dx=\log|\text{cosec}\,x-\cot x|+c$ , matching option (e). A standard integral, verified by differentiating the RHS.

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←Previous chapterCh 6Application of Derivatives

Q.Prove that $\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\left(\dfrac{x}{a}\right)+C$ .

Concept understanding — Standard Integral Forms

Q.Prove that ∫1a2−x2 dx=sin⁡−1(xa)+C\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\left(\dfrac{x}{a}\right)+C∫a2−x2​1​dx=sin−1(ax​)+C.

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Q.Prove that $\displaystyle\int \dfrac{1}{\sqrt{a^2-x^2}}\,dx = \sin^{-1}\left(\dfrac{x}{a}\right)+C$ .

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