Find : $\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx$ OR Find : $\int \frac{x - 3}{(x - 1)^3} e^x dx$

> [!TLDR] > **Part (a):** splitting the fraction gives $\sec x\tan x+\csc x\cot x$, so the integral is $\sec x-\csc x+C$. **Part (b):** using the $e^x\big(f+f'\big)$ pattern gives $\dfrac{e^x}{(x-1)^2}+C$. ## Part (a) The denominator is a product of squares, so we split the single fraction into two simpler ones and simplify each. 1. **Split the fraction.** $$\frac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}=\frac{\sin^3 x}{\sin^2 x\cos^2 x}+\frac{\cos^3 x}{\sin^2 x\cos^2 x}.$$ 2. **Simplify each term.** $$\frac{\sin^3 x}{\sin^2 x\cos^2 x}=\frac{\sin x}{\cos^2 x}=\sec x\tan x,\qquad \frac{\cos^3 x}{\sin^2 x\cos^2 x}=\frac{\cos x}{\sin^2 x}=\csc x\cot x.$$ 3. **Integrate term by term.** Recall $\dfrac{d}{dx}(\sec x)=\sec x\tan x$ and $\dfrac{d}{dx}(\csc x)=-\csc x\cot x$, so $$\int\sec x\tan x\,dx=\sec x,\qquad \int\csc x\cot x\,dx=-\csc x.$$ 4. **Combine.** $$\int \frac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}\,dx=\sec x-\csc x+C.$$ …

Find : ∫ (sin³ x + cos³ x)/(sin² x cos² x) dx OR Find : — CBSE Class 12 Mathematics 2019 PYQ

Question

\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx

(OR)

\int \frac{x - 3}{(x - 1)^3} e^x dx

CBSECBSE Class XII Board 2019Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Integration By Expansion

Integration by Expansion

The idea

Some integrands look forbidding only because they are written as a product or a power. If you first multiply them out — expand them — into a plain sum of standard terms, you can then integrate the sum term by term using the basic formulas you already know. Expansion is not a new rule of integration; it is a rewriting step that turns the integrand into something the sum rule and the power rule can finish.

Note

This works because integration is linear: $\int(f\pm g)\,dx = \int f\,dx \pm \int g\,dx$ . Once the integrand is a sum, each piece is handled separately.

Algebraic expansion

Products and powers of polynomials are expanded first:

\int (x+2)^2\,dx = \int (x^2+4x+4)\,dx = \frac{x^3}{3}+2x^2+4x+C.

Similarly, split a fraction into separate terms before integrating:

\int \frac{x^2+3x-1}{x}\,dx = \int\Big(x+3-\frac{1}{x}\Big)\,dx = \frac{x^2}{2}+3x-\log|x|+C.

Trigonometric expansion

Many trigonometric integrands have no direct formula in the form given, but do have one after a standard identity is used to expand them into a sum:

\sin^2 x = \frac{1-\cos 2x}{2}, \qquad \cos^2 x = \frac{1+\cos 2x}{2}

So, for example,

\int \sin^2 x \, dx = \int \frac{1-\cos 2x}{2}\,dx = \frac{x}{2}-\frac{\sin 2x}{4}+C.

Product-to-sum identities do the same job for products such as $\sin 3x\cos 5x$ , and $\sin^3 x$ , $\cos^3 x$ can be expanded using their triple-angle forms.

How to use it

Look at the integrand — is it a product, a power, or a single fraction over $x$ ?
Expand it (multiply out, or apply a trig identity) into a sum of standard terms. …

Part (b)Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Some integrands are a product of two very different functions — $x e^x$ , $x\cos x$ , $\log x$ , $x\sin^{-1}x$ — where substitution gets you nowhere. Integration by parts is the tool for these. It comes straight from reversing the product rule for differentiation.

Starting from $\dfrac{d}{dx}(uv)=u\,v' + u'\,v$ and integrating both sides gives the working formula:

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx.

In words: integral of (first $\times$ derivative-of-second) $=$ first $\times$ integral-of-second $-$ integral of (derivative-of-first $\times$ integral-of-second).

Choosing $u$ : the ILATE rule

The whole game is picking which factor is $u$ (to differentiate) and which is $dv$ (to integrate). Pick $u$ by ILATE — the first type that appears:

Inverse trig ( $\sin^{-1}x$ ), Logarithmic ( $\log x$ ), Algebraic ( $x^2$ ), Trigonometric ( $\sin x$ ), Exponential ( $e^x$ ).

Whatever comes first in ILATE becomes $u$ ; the rest is $dv$ . This makes the new integral $\int v\,du$ simpler than the one you started with.

Worked idea

For $\int x e^x\,dx$ : algebraic before exponential, so $u=x$ , $dv=e^x dx$ . Then $du=dx$ , $v=e^x$ :

$\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1)+C.$ …

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Part (a): splitting the fraction gives $\sec x\tan x+\csc x\cot x$ , so the integral is $\sec x-\csc x+C$ . Part (b): using the $e^x\big(f+f'\big)$ pattern gives $\dfrac{e^x}{(x-1)^2}+C$ .

Part (a)

The denominator is a product of squares, so we split the single fraction into two simpler ones and simplify each.

Split the fraction.

\frac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}=\frac{\sin^3 x}{\sin^2 x\cos^2 x}+\frac{\cos^3 x}{\sin^2 x\cos^2 x}.

Simplify each term.

\frac{\sin^3 x}{\sin^2 x\cos^2 x}=\frac{\sin x}{\cos^2 x}=\sec x\tan x,\qquad \frac{\cos^3 x}{\sin^2 x\cos^2 x}=\frac{\cos x}{\sin^2 x}=\csc x\cot x.

Integrate term by term. Recall $\dfrac{d}{dx}(\sec x)=\sec x\tan x$ and $\dfrac{d}{dx}(\csc x)=-\csc x\cot x$ , so

\int\sec x\tan x\,dx=\sec x,\qquad \int\csc x\cot x\,dx=-\csc x.

Combine. $\int \frac{\sin^3 x+\cos^3 x}{\sin^2 x\cos^2 x}\,dx=\sec x-\csc x+C.$ …

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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 48 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 $v=x$ :

…

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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 2026Set ANNUAL1 mark
$Q.Evaluate \int \dfrac{2 - 3\sin x}{\cos^2 x},dx.$
›Reveal solutionSolution

Split the integrand into two standard integrals, $\sec^2x$ and $\sec x\tan x$ .

Working: Split the fraction:

$\int \frac{2-3\sin x}{\cos^2 x}\,dx = \int \frac{2}{\cos^2 x}\,dx - \int\frac{3\sin x}{\cos^2 x}\,dx = 2\int\sec^2x\,dx - 3\int \sec x\tan x\,dx$

…

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CBSE 2026Set ANNUAL1 mark
$Q.Find the integral: \int \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^2 dx$
›Reveal solutionSolution

$\displaystyle\int\left(\sqrt x-\tfrac{1}{\sqrt x}\right)^2 dx=\dfrac{x^2}{2}-2x+\ln|x|+c$ .

Concept. Expand the integrand into simple power terms, then integrate term by term using $\int x^n dx=\dfrac{x^{n+1}}{n+1}$ and $\int \dfrac1x dx=\ln|x|$ .

Steps. …

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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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Q.Find : ∫sin⁡3x+cos⁡3xsin⁡2xcos⁡2xdx\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx∫sin2xcos2xsin3x+cos3x​dx

Part (a)

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Q.Find : $\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx$

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