(a) Find: ∫ (x + sin x)/(1 + cos x)dx OR (b) Evaluate: ∫₀π/4 fracdxcos³ x √(2sin 2x)

Question

\int \frac{x + \sin x}{1 + \cos x}\,dx

(OR)

(b) Evaluate:

\int_0^{\pi/4} \frac{dx}{\cos^3 x \sqrt{2\sin 2x}}

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★

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Part (a)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 $du=dx$ , $v=e^x$ :

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

Part (b)Concept understanding — Definite Substitution Method

Substitution in Definite Integrals

You already know substitution for indefinite integrals: set $u = g(x)$ , rewrite in terms of $u$ , integrate, then substitute back. For a definite integral there is a cleaner twist — instead of substituting back, you convert the limits of integration to the new variable and finish entirely in $u$ .

Why the limits must change

The limits $a$ and $b$ are $x$ -values. Once you switch to $u = g(x)$ , those numbers no longer describe the start and end of the integration — the corresponding $u$ -values do. Keeping the old numbers would integrate over the wrong interval, like reading a distance in kilometres off a scale marked in miles.

\int_{a}^{b} f\big(g(x)\big)\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du

The steps

Choose $u = g(x)$ , picking something whose derivative already appears in the integrand.
Differentiate: $du = g'(x)\,dx$ .
Convert the limits: the lower limit becomes $u = g(a)$ , the upper becomes $u = g(b)$ .
Integrate in $u$ — no substituting back needed.

Example. Evaluate $\int_0^2 2x\,(x^2+1)^3\,dx$ .

Let $u = x^2 + 1$ , so $du = 2x\,dx$ . When $x=0$ , $u=1$ ; when $x=2$ , $u=5$ . Then

\int_0^2 (x^2+1)^3\,(2x\,dx) = \int_1^5 u^3\,du = \left[\tfrac{u^4}{4}\right]_1^5 = \tfrac{625-1}{4} = 156.

We never returned to $x$ — the converted limits carried the work. …

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Part (a): $\displaystyle\int\frac{x+\sin x}{1+\cos x}\,dx=x\tan\tfrac{x}{2}+C$ . Part (b): $\displaystyle\int_0^{\pi/4}\frac{dx}{\cos^3 x\sqrt{2\sin 2x}}=\frac{6}{5}$ .

Part (a): $\displaystyle\int \frac{x + \sin x}{1 + \cos x}\,dx$

The integrand mixes an algebraic factor $x$ with trigonometric terms, so we first simplify using half-angle identities:

1+\cos x = 2\cos^2\tfrac{x}{2},\qquad \sin x = 2\sin\tfrac{x}{2}\cos\tfrac{x}{2}.

Hence

\frac{x}{1+\cos x}=\frac{x}{2\cos^2\tfrac{x}{2}}=\frac{x}{2}\sec^2\tfrac{x}{2},\qquad \frac{\sin x}{1+\cos x}=\frac{2\sin\tfrac{x}{2}\cos\tfrac{x}{2}}{2\cos^2\tfrac{x}{2}}=\tan\tfrac{x}{2}.

So the integral splits as

\int \frac{x}{2}\sec^2\tfrac{x}{2}\,dx+\int \tan\tfrac{x}{2}\,dx.

For the first integral integrate by parts with $u=x$ , $dv=\tfrac12\sec^2\tfrac{x}{2}\,dx$ , so $v=\tan\tfrac{x}{2}$ :

\int \frac{x}{2}\sec^2\tfrac{x}{2}\,dx = x\tan\tfrac{x}{2}-\int \tan\tfrac{x}{2}\,dx.

Adding the leftover $\int\tan\tfrac{x}{2}\,dx$ from the split, the two $\tan\tfrac{x}{2}$ integrals cancel exactly: …

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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 47 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
$\int_1^e\frac{(\log x)^2}{x}\,dx =$
(a) $\frac{1}{3}$
(b) $\frac{1}{3}e^3$
(c) $\frac{1}{3}(e^3 - 1)$
(d) $e^3$
›Reveal solutionSolution

Substitute $t=\log x$ : the integral becomes $\int_0^1 t^2\,dt=\tfrac13$ .

Let $t=\log x$ , so $dt=\tfrac{dx}{x}$ . Limits: $x=1\to t=0$ , $x=e\to t=1$ . Then

…

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CBSE 2026Set A1 markMCQ
$\int_0^{\pi/4}\frac{e^{\tan x}}{\cos^2 x}\,dx =$
(a) $e - 1$
(b) $e + 1$
(c) $\frac{1}{e} + 1$
(d) $\frac{1}{e} - 1$
›Reveal solutionSolution

Substitute $t=\tan x$ (so $dt=\sec^2 x\,dx$ ): the integral becomes $\int_0^1 e^t\,dt=e-1$ .

Let $t=\tan x$ . Then $dt=\sec^2 x\,dx=\dfrac{dx}{\cos^2 x}$ . Limits: $x=0\to t=0$ , $x=\tfrac{\pi}{4}\to t=1$ . So

…

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

Substitute $t=\sqrt{x}$ ; the integral becomes $2\int_0^1 e^t\,dt=2(e-1)$ .

Let $t=\sqrt{x}$ , so $dt=\dfrac{dx}{2\sqrt{x}}$ , i.e. $\dfrac{dx}{\sqrt{x}}=2\,dt$ . Limits: $x=0\to t=0$ , $x=1\to t=1$ . Then

…

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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 A1 markMCQ
$\int_0^a\sqrt{a^2 - x^2}\,dx =$
(a) $\frac{\pi}{4}$
(b) $\frac{a^2}{4}$
(c) $\frac{\pi a^2}{4}$
(d) $\pi$
›Reveal solutionSolution

$\int_0^a\sqrt{a^2-x^2}\,dx$ is a quarter-circle area $=\tfrac{\pi a^2}{4}$ .

Using the standard formula $\int\sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}$ , evaluate from $0$ to $a$ :

…

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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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Q.(a) Find: ∫x+sin⁡x1+cos⁡x dx\int \frac{x + \sin x}{1 + \cos x}\,dx∫1+cosxx+sinx​dx

Part (a): ∫x+sin⁡x1+cos⁡x dx\displaystyle\int \frac{x + \sin x}{1 + \cos x}\,dx∫1+cosxx+sinx​dx

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Q.(a) Find: $\int \frac{x + \sin x}{1 + \cos x}\,dx$

Part (a): $\displaystyle\int \frac{x + \sin x}{1 + \cos x}\,dx$

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