Find: $\displaystyle\int x^2 \cdot \sin^{-1}(x^{3/2})\, dx$

Find: ∫ x² · sin⁻¹(x3/2) dx — CBSE Class 12 Mathematics 2024 PYQ

Question

\displaystyle\int x^2 \cdot \sin^{-1}(x^{3/2})\, dx

CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The key idea is to simplify the integral using a substitution, then apply integration by parts, and finally resolve the resulting integral using trigonometric substitution.

Substitution: Let $t = x^{3/2}$ . Then $dt = \frac{3}{2} x^{1/2} dx$ . From $t = x^{3/2}$ , we have $x = t^{2/3}$ , so $x^2 = t^{4/3}$ . Also, $dx = \frac{2}{3} x^{-1/2} dt = \frac{2}{3} (t^{2/3})^{-1/2} dt = \frac{2}{3} t^{-1/3} dt$ . The integral becomes:

\int t^{4/3} \sin^{-1}(t) \cdot \frac{2}{3} t^{-1/3} dt = \frac{2}{3} \int t \sin^{-1}(t) dt

Integration by Parts: For $\int t \sin^{-1}(t) dt$ , let $U = \sin^{-1}(t)$ and $dV = t \, dt$ . Then $dU = \frac{1}{\sqrt{1-t^2}} dt$ and $V = \frac{t^2}{2}$ . Using $\int U \, dV = UV - \int V \, dU$ :

\int t \sin^{-1}(t) dt = \frac{t^2}{2} \sin^{-1}(t) - \int \frac{t^2}{2\sqrt{1-t^2}} dt

Trigonometric Substitution: To evaluate $\int \frac{t^2}{\sqrt{1-t^2}} dt$ , let $t = \sin\theta$ . Then $dt = \cos\theta \, d\theta$ .

\int \frac{\sin^2\theta}{\sqrt{1-\sin^2\theta}} \cos\theta \, d\theta = \int \frac{\sin^2\theta}{\cos\theta} \cos\theta \, d\theta = \int \sin^2\theta \, d\theta

Using the identity $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$:

\int \frac{1-\cos(2\theta)}{2} d\theta = \frac{1}{2} \left( \theta - \frac{\sin(2\theta)}{2} \right) = \frac{1}{2} (\theta - \sin\theta\cos\theta)

Substituting back $t = \sin\theta$, so $\theta = \sin^{-1}(t)$ and $\cos\theta = \sqrt{1-t^2}$:

\frac{1}{2} (\sin^{-1}(t) - t\sqrt{1-t^2})

Combine and Back-substitute: Substitute this result back into the integration by parts expression, and then multiply by $\frac{2}{3}$ : …

Start your 14-day free trial to unlock the full solution →

The integral is solved by first using a substitution $u = x^{3/2}$ to simplify the argument of the inverse sine function, followed by integration by parts, and finally a trigonometric substitution to evaluate the remaining integral, yielding $\frac{1}{6} (2x^3-1)\sin^{-1}(x^{3/2}) + \frac{1}{6}x^{3/2}\sqrt{1-x^3} + C$ .

The integral $\displaystyle\int x^2 \cdot \sin^{-1}(x^{3/2})\, dx$ involves a product of an algebraic term ( $x^2$ ) and an inverse trigonometric term ( $\sin^{-1}(x^{3/2})$ ). When dealing with inverse trigonometric functions, especially those with composite arguments, a common and effective strategy is to first simplify the argument using a substitution. This often transforms the integral into a more manageable form, typically one that can then be tackled with integration by parts.

In this specific problem, the argument of $\sin^{-1}$ is $x^{3/2}$ . If we let $u = x^{3/2}$ , the $\sin^{-1}(x^{3/2})$ term becomes $\sin^{-1}(u)$ , which is much simpler. The challenge then is to express $x^2\, dx$ in terms of $u$ and $du$ . By carefully performing this substitution, we convert the original integral into a product of an algebraic function and $\sin^{-1}(u)$ , which is a classic scenario for integration by parts. The LIATE rule helps us choose which function to differentiate and which to integrate in the integration by parts step.

Initial Substitution to Simplify the Argument: We begin by simplifying the argument of the inverse sine function. Let $u = x^{3/2}$ . To find $du$ , we differentiate $u$ with respect to $x$ : $du = \frac{3}{2}x^{1/2}\, dx$ . We need to express $x^2\, dx$ in terms of $u$ and $du$ . From $u = x^{3/2}$ , we can write $x = u^{2/3}$ . Then $x^2 = (u^{2/3})^2 = u^{4/3}$ . Also, $dx = \frac{2}{3}u^{-1/3}\, du$ . Substituting these into the original integral:

\int x^2 \cdot \sin^{-1}(x^{3/2})\, dx = \int u^{4/3} \cdot \sin^{-1}(u) \cdot \frac{2}{3}u^{-1/3}\, du

= \frac{2}{3} \int u^{(4/3 - 1/3)} \sin^{-1}(u)\, du

= \frac{2}{3} \int u \sin^{-1}(u)\, du

This substitution has successfully transformed the integral into a simpler form, which is now ready for integration by parts.

2. Integration by Parts:

We now need to evaluate $I' = \int u \sin^{-1}(u)\, du$ . This is a product of an algebraic function ( $u$ ) and an inverse trigonometric function ( $\sin^{-1}(u)$ ). According to the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential), we choose the inverse trigonometric function as $f(u)$ (the part to differentiate) and the algebraic function as $g'(u)$ (the part to integrate).

> [!FORMULA]
> The integration by parts formula is: $\int f(u)g'(u)\, du = f(u)g(u) - \int f'(u)g(u)\, du$.

Let:
$f(u) = \sin^{-1}(u) \implies f'(u) = \frac{1}{\sqrt{1-u^2}}$
$g'(u) = u \implies g(u) = \frac{u^2}{2}$

Applying the integration by parts formula:

\int u \sin^{-1}(u)\, du = \sin^{-1}(u) \cdot \frac{u^2}{2} - \int \frac{1}{\sqrt{1-u^2}} \cdot \frac{u^2}{2}\, du

= \frac{u^2}{2}\sin^{-1}(u) - \frac{1}{2} \int \frac{u^2}{\sqrt{1-u^2}}\, du

Let's denote the remaining integral as $J = \int \frac{u^2}{\sqrt{1-u^2}}\, du$.

3. Evaluating the Remaining Integral $J$ :

The integral $J = \int \frac{u^2}{\sqrt{1-u^2}}\, du$ is a standard form that can be solved using a trigonometric substitution.

Let $u = \sin\theta$ .

Then $du = \cos\theta\, d\theta$ .

And $\sqrt{1-u^2} = \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} = \cos\theta$ (assuming $\theta \in [-\pi/2, \pi/2]$ for the principal value of $\sin^{-1}(u)$ ).

Substituting these into $J$:

J = \int \frac{\sin^2\theta}{\cos\theta} \cdot \cos\theta\, d\theta = \int \sin^2\theta\, d\theta

We use the double-angle identity $\sin^2\theta = \frac{1-\cos(2\theta)}{2}$:

J = \int \frac{1-\cos(2\theta)}{2}\, d\theta = \frac{1}{2} \int (1-\cos(2\theta))\, d\theta

J = \frac{1}{2} \left( \theta - \frac{\sin(2\theta)}{2} \right) + C_1

Using the identity $\sin(2\theta) = 2\sin\theta\cos\theta$:

J = \frac{1}{2} \left( \theta - \frac{2\sin\theta\cos\theta}{2} \right) + C_1 = \frac{1}{2} (\theta - \sin\theta\cos\theta) + C_1

Now, we substitute back in terms of $u$:
Since $u = \sin\theta$, we have $\theta = \sin^{-1}(u)$.
And $\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-u^2}$.
So,
$$ J = \frac{1}{2} (\sin^{-1}(u) - u\sqrt{1-u^2}) + C_1 $$ …

Start your 14-day free trial to unlock the full solution →

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 40 on this concept.

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 $v=x$ :

…

Start your 14-day free trial to see the full solution →
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 ( $u=t,\ dv=\cos t\,dt$ ): $=t\sin t-\int\sin t\,dt=t\sin t+\cos t$ .

…

Start your 14-day free trial to see the full solution →
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

…

Start your 14-day free trial to see the full solution →
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 $du=dx,\ v=e^x$ :

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

…

Start your 14-day free trial to see the full solution →
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

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

Start your 14-day free trial to see the full solution →
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 $f'(x) = \dfrac{\sec x\tan x}{\sec x} = \tan x$ .

…

Start your 14-day free trial to see the full solution →
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). …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$\displaystyle\int x \cdot e^x\,dx$
›Reveal solutionSolution

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

…

Start your 14-day free trial to see the full solution →
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 $u=\log x$ , $dv=dx$ : …

Start your 14-day free trial to see the full solution →
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 $du=dx$ , $v=-\cos x$ .

…

Start your 14-day free trial to see the full solution →
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

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

Start your 14-day free trial to see the full solution →
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): …

Start your 14-day free trial to see the full solution →

Q.Find: $\displaystyle\int x^2 \cdot \sin^{-1}(x^{3/2})\, dx$

Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Choosing $u$ : the ILATE rule

Worked idea

Q.Find: ∫x2⋅sin⁡−1(x3/2) dx\displaystyle\int x^2 \cdot \sin^{-1}(x^{3/2})\, dx∫x2⋅sin−1(x3/2)dx

🎓Unlock everything free for 14 days

Q.Find: $\displaystyle\int x^2 \cdot \sin^{-1}(x^{3/2})\, dx$

Unlock everything free for 14 days