$\int \sin^{-1}\sqrt{\dfrac{x}{a+x}}\,dx$ (Hint: Put $x=a\tan^2\theta$)

> [!TLDR] > The key idea is to simplify the integrand using the given substitution $x = a \tan^2 \theta$, which turns the inverse sine into a simple angle. After substitution and integration by parts, the final result is $\boxed{(a+x) \tan^{-1}\sqrt{\frac{x}{a}} - \sqrt{ax} + C}$. The problem asks for the indefinite integral of $\sin^{-1}\sqrt{\frac{x}{a+x}}$. The expression inside the inverse sine looks messy, but the hint suggests a clever substitution: $x = a \tan^2 \theta$. Why does this work? Because $\sqrt{\frac{x}{a+x}}$ becomes something like $\sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}} = \sqrt{\frac{\tan^2 \theta}{1 + \tan^2 \theta}} = \sqrt{\sin^2 \theta} = \sin \theta$ (assuming $\theta$ in a suitable range). Then $\sin^{-1}(\sin \theta) = \theta$, which is much simpler to integrate. Let’s walk through it step by step. 1. **Substitute $x = a \tan^2 \theta$.** We need $dx$ in terms of $d\theta$. Differentiate: $$dx = a \cdot 2 \tan \theta \cdot \sec^2 \theta \, d\theta = 2a \tan \theta \sec^2 \theta \, d\theta.$$ Also, note that $a + x = a + a \tan^2 \theta = a(1 + \tan^2 \theta) = a \sec^2 \theta$. 2. **Simplify the integrand.** Compute $\sqrt{\frac{x}{a+x}}$: $$\sqrt{\frac{a \tan^2 \theta}{a \sec^2 \theta}} = \sqrt{\frac{\tan^2 \theta}{\sec^2 \theta}} = \sqrt{\sin^2 \theta} = \sin \theta,$$ taking $\theta \in [0, \pi/2)$ so $\sin \theta \ge 0$. Hence $\sin^{-1}\sqrt{\frac{x}{a+x}} = \sin^{-1}(\sin \theta) = \theta$. 3. **Rewrite the integral.** The integral becomes: $$I = \int \theta \cdot (2a \tan \theta \sec^2 \theta) \, d\theta = 2a \int \theta \tan \theta \sec^2 \theta \, d\theta.$$ 4. **Simplify the trigonometric part.** Notice $\tan \theta \sec^2 \theta = \frac{\sin \theta}{\cos^3 \theta}$. But a better approach: let $t = \tan \theta$, then $dt = \sec^2 \theta \, d\theta$, so $\tan \theta \sec^2 \theta \, d\theta = t \, dt$. However, we still have $\theta$ in terms of $t$: $\theta = \ta

∫ sin⁻¹√((x)/(a+x))dx (Hint: Put x=atan²θ)

NCERT Exemplar · Q40

\int \sin^{-1}\sqrt{\dfrac{x}{a+x}}\,dx

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The key idea is to simplify the integrand using the given substitution $x = a \tan^2 \theta$ , which turns the inverse sine into a simple angle. After substitution and integration by parts, the final result is $\boxed{(a+x) \tan^{-1}\sqrt{\frac{x}{a}} - \sqrt{ax} + C}$ .

The problem asks for the indefinite integral of $\sin^{-1}\sqrt{\frac{x}{a+x}}$ . The expression inside the inverse sine looks messy, but the hint suggests a clever substitution: $x = a \tan^2 \theta$ . Why does this work? Because $\sqrt{\frac{x}{a+x}}$ becomes something like $\sqrt{\frac{a \tan^2 \theta}{a + a \tan^2 \theta}} = \sqrt{\frac{\tan^2 \theta}{1 + \tan^2 \theta}} = \sqrt{\sin^2 \theta} = \sin \theta$ (assuming $\theta$ in a suitable range). Then $\sin^{-1}(\sin \theta) = \theta$ , which is much simpler to integrate.

Let’s walk through it step by step.

Substitute $x = a \tan^2 \theta$ . We need $dx$ in terms of $d\theta$ . Differentiate:

dx = a \cdot 2 \tan \theta \cdot \sec^2 \theta \, d\theta = 2a \tan \theta \sec^2 \theta \, d\theta.

Also, note that $a + x = a + a \tan^2 \theta = a(1 + \tan^2 \theta) = a \sec^2 \theta$ .

Simplify the integrand. Compute $\sqrt{\frac{x}{a+x}}$ :

\sqrt{\frac{a \tan^2 \theta}{a \sec^2 \theta}} = \sqrt{\frac{\tan^2 \theta}{\sec^2 \theta}} = \sqrt{\sin^2 \theta} = \sin \theta,

taking $\theta \in [0, \pi/2)$ so $\sin \theta \ge 0$ .

Hence $\sin^{-1}\sqrt{\frac{x}{a+x}} = \sin^{-1}(\sin \theta) = \theta$ .

Rewrite the integral. The integral becomes:

I = \int \theta \cdot (2a \tan \theta \sec^2 \theta) \, d\theta = 2a \int \theta \tan \theta \sec^2 \theta \, d\theta.

Simplify the trigonometric part. Notice $\tan \theta \sec^2 \theta = \frac{\sin \theta}{\cos^3 \theta}$ . But a better approach: let $t = \tan \theta$ , then $dt = \sec^2 \theta \, d\theta$ , so $\tan \theta \sec^2 \theta \, d\theta = t \, dt$ . However, we still have $\theta$ in terms of $t$ : $\theta = \tan^{-1} t$ . So:

I = 2a \int \theta \cdot t \, dt = 2a \int (\tan^{-1} t) \cdot t \, dt.

Integrate by parts. Let $u = \tan^{-1} t$ and $dv = t \, dt$ . Then $du = \frac{1}{1+t^2} dt$ and $v = \frac{t^2}{2}$ . Integration by parts gives:

\int u \, dv = uv - \int v \, du,

so:

I = 2a \left[ \frac{t^2}{2} \tan^{-1} t - \int \frac{t^2}{2} \cdot \frac{1}{1+t^2} dt \right] = a \left[ t^2 \tan^{-1} t - \int \frac{t^2}{1+t^2} dt \right].

Simplify the remaining integral. Write $\frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2}$ . Then:

\int \frac{t^2}{1+t^2} dt = \int \left(1 - \frac{1}{1+t^2}\right) dt = t - \tan^{-1} t + C_1.

Substitute back: …

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

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CBSE 2026Set A1 markMCQ
$\int\log x\,dx =$
(a) $\frac{1}{x} + k$
(b) $x\log x + k$
(c) $x\log x - x + k$
(d) $x\log x + x + k$
›Reveal solutionSolution

Integrate by parts: $\int\log x\,dx=x\log x-x+k$ .

Take $u=\log x$ and $dv=dx$ , so $du=\tfrac{1}{x}\,dx$ and $v=x$ :

…

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CBSE 2026Set A1 markMCQ
$\int\cos\sqrt{x}\,dx =$
(a) $\sin\sqrt{x} + \cos\sqrt{x} + k$
(b) $\frac{1}{2}(\sqrt{x}\sin\sqrt{x} - \cos\sqrt{x}) + k$
(c) $2(\sqrt{x}\sin\sqrt{x} + \cos\sqrt{x}) + k$
(d) $\sin\sqrt{x} + k$
›Reveal solutionSolution

Put $t=\sqrt{x}$ ; the integral becomes $2\int t\cos t\,dt=2(t\sin t+\cos t)+k$ .

Let $t=\sqrt{x}$ , so $x=t^2$ and $dx=2t\,dt$ . Then

$\int\cos\sqrt{x}\,dx=\int\cos t\,(2t\,dt)=2\int t\cos t\,dt$ .

Integrate $\int t\cos t\,dt$ by parts ( $u=t,\ dv=\cos t\,dt$ ): $=t\sin t-\int\sin t\,dt=t\sin t+\cos t$ .

…

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CBSE 2026Set A1 markMCQ
$\int e^x\left(\tan^{-1}x + \frac{1}{1+x^2}\right)dx =$
(a) $e^x\tan^{-1}x + k$
(b) $e^x \cdot \frac{1}{1+x^2} + k$
(c) $e^x + k$
(d) $\tan^{-1}x + k$
›Reveal solutionSolution

Recognise $\int e^x[f(x)+f'(x)]\,dx=e^x f(x)+k$ ; here $f(x)=\tan^{-1}x$ .

Note $\dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^2}$ . So the integrand is $e^x\big[\tan^{-1}x+(\tan^{-1}x)'\big]$ , which matches the standard pattern

…

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CBSE 2026Set A1 markMCQ
$\int_0^1 x e^x\,dx =$
(a) $1$
(b) $0$
(c) $2$
(d) $-1$
›Reveal solutionSolution

By parts: $\int_0^1 xe^x\,dx=[(x-1)e^x]_0^1=1$ .

Take $u=x,\ dv=e^x\,dx$ , so $du=dx,\ v=e^x$ :

$\int x e^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x=(x-1)e^x$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int e^x(\sin x-\cos x)\,dx$
(a) $-e^x\cos x+c$
(b) $e^x\sin x+c$
(c) $-e^x\sec x+c$
(d) $e^x\operatorname{cosec} x+c$
›Reveal solutionSolution

Recognising the standard form $\int e^x[f(x)+f'(x)]\,dx=e^xf(x)+c$ gives $-e^x\cos x+c$ .

There is a standard integration result:

$\int e^x\big[f(x)+f'(x)\big]\,dx=e^x f(x)+c$

Compare the integrand $\sin x-\cos x$ with $f(x)+f'(x)$ . Try $f(x)=-\cos x$ ; then $f'(x)=\sin x$ , so

$f(x)+f'(x)=-\cos x+\sin x=\sin x-\cos x$ …

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CBSE 2026Set ANNUAL1 markMCQ
$\int e^x (\log \sec x + \tan x)\, dx =$
(a) $e^x + C$
(b) $e^x \tan x + C$
(c) $e^x(\log \sec x) + C$
(d) None of these
›Reveal solutionSolution

This is of the standard form $\int e^x[f(x)+f'(x)]\,dx = e^x f(x)+C$ .

Let $f(x) = \log\sec x$ . Then $f'(x) = \dfrac{\sec x\tan x}{\sec x} = \tan x$ .

…

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CBSE 2025Set X11 markMCQ
$\int e^x (\sin x - \cos x)\, dx$
(a) $-e^x \cos x$
(b) $e^x \cos x$
(c) $e^x \sin x$
(d) $e^x \sin^2 x$
›Reveal solutionSolution

Integral of the form $\int e^x(f+f')\,dx = e^x f$ — correct option (a). …

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int x \cdot e^x\,dx$
›Reveal solutionSolution

Apply integration by parts with $u=x$ , $dv=e^x dx$ .

…

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CBSE 2025Set E1 markMCQ
$\int \log x^2\,dx =$
(a) $\frac{1}{x^2} + k$
(b) $\frac{2}{x} + k$
(c) $x\log x - x + k$
(d) $2(x\log x - x) + k$
›Reveal solutionSolution

Bring down the power, then integrate $\log x$ by parts; result $2(x\log x-x)+k$ .

First $\log x^2=2\log x$ . Now integrate $\int\log x\,dx$ by parts with $u=\log x$ , $dv=dx$ : …

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int x\sin x\,dx$
›Reveal solutionSolution

Use integration by parts (ILATE): take $u=x$ (algebraic) and $dv=\sin x\,dx$ .

Let $u=x$ , $dv=\sin x\,dx$ , so $du=dx$ , $v=-\cos x$ .

…

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int x\log 2x\, dx$
›Reveal solutionSolution

Use integration by parts, taking the logarithm as the first function.

Let $I=\displaystyle\int x\log 2x\,dx$ . Apply $\int u\,dv=uv-\int v\,du$ with

$u=\log 2x,\qquad dv=x\,dx.$

Then

$du=\frac{1}{2x}\cdot 2\,dx=\frac1x\,dx,\qquad v=\frac{x^2}{2}.$

Therefore

$I=\log 2x\cdot\frac{x^2}{2}-\int\frac{x^2}{2}\cdot\frac1x\,dx=\frac{x^2}{2}\log 2x-\frac12\int x\,dx.$ …

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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int e^x\sec x(1+\tan x)\, dx$
(a) $e^x\cos x + c$
(b) $e^x\sec x + c$
(c) $e^x\sin x + c$
(d) $e^x\tan x + c$
›Reveal solutionSolution

Recognise the form $\int e^x\{f(x)+f'(x)\}\,dx=e^xf(x)+c$ .

Expand the integrand:

$e^x\sec x(1+\tan x)=e^x\big(\sec x+\sec x\tan x\big).$

Let $f(x)=\sec x$ . Then

$f'(x)=\sec x\tan x.$

So the integrand is exactly $e^x\{f(x)+f'(x)\}$ , and by the standard result

$\int e^x\{f(x)+f'(x)\}\,dx=e^xf(x)+c=e^x\sec x+c.$

Verification (differentiate the answer): …

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Q. $\int \sin^{-1}\sqrt{\dfrac{x}{a+x}}\,dx$ (Hint: Put $x=a\tan^2\theta$ )

Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Choosing $u$ : the ILATE rule

Worked idea

Method: Rationalising a nested radical by trig substitution

Steps

Common Mistakes

Q.∫sin⁡−1xa+x dx\int \sin^{-1}\sqrt{\dfrac{x}{a+x}}\,dx∫sin−1a+xx​​dx (Hint: Put x=atan⁡2θx=a\tan^2\thetax=atan2θ)

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Q. $\int \sin^{-1}\sqrt{\dfrac{x}{a+x}}\,dx$ (Hint: Put $x=a\tan^2\theta$ )

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