∫ sin² x dx = (a) (x)/(2)-(sin 2x)/(4)+c (b) (sin – Class 12 Mathematics | ConceptFirst

Question of 373

\int \sin^2 x \,dx =

(a)

\frac{x}{2}-\frac{\sin 2x}{4}+c

(b)

\frac{\sin 2x}{4}-\frac{x}{2}+c

(c)

\frac{\sin 2x}{2}+\frac{x}{2}+c

(d) none of these

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Concept understanding — Sine Double Angle Integration

Sine Double Angle Integration — From Intuition to Formula

Suppose you want the area under $\sin(2x)$ from $x = 0$ to $\pi/2$ . This graph oscillates twice as fast as a regular sine wave, completing a cycle in $\pi$ units instead of $2\pi$ — the "double angle" inside compresses the wave horizontally.

The catch: you can't integrate $\sin(2x)$ the same way as $\sin x$ . Differentiating $\cos(2x)$ gives $-2\sin(2x)$ — not $-\sin(2x)$ — so the antiderivative needs a factor to compensate for that extra $2$ .

The Precise Statement

\int \sin(ax) \, dx = -\frac{1}{a} \cos(ax) + C

For $a = 2$ :

\int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + C

This is the sine double angle integration formula — a direct application of the reverse chain rule.

Why It Works

Differentiate the right-hand side:

\frac{d}{dx} \left[ -\frac{1}{2} \cos(2x) + C \right] = -\frac{1}{2} \cdot (-\sin(2x)) \cdot 2 = \sin(2x)

The $-\frac{1}{2}$ cancels the $-2$ from the chain rule, leaving exactly $\sin(2x)$ .

Watch out

A common mistake is writing $\int \sin(2x)\,dx = -\cos(2x) + C$ . Differentiating $-\cos(2x)$ gives $2\sin(2x)$ , not $\sin(2x)$ . Always check by differentiating your answer.

Definite Integrals

\int_a^b \sin(2x) \, dx = \left[ -\frac{1}{2} \cos(2x) \right]_a^b = -\frac{1}{2} \left[ \cos(2b) - \cos(2a) \right]

Example: $\int_0^{\pi/2} \sin(2x) \, dx = -\frac{1}{2} [\cos(\pi) - \cos(0)] = -\frac{1}{2} [(-1) - 1] = 1$ .

The General Pattern

\int \sin(kx) \, dx = -\frac{1}{k} \cos(kx) + C

The $k$ in the denominator is the "compensation factor" for the chain rule, working for any constant $k \neq 0$ . …

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Rewrite sin^2 x using the double-angle identity, then integrate term by term.

Using $\sin^2x=\dfrac{1-\cos2x}{2}$ : …

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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 17 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$\displaystyle\int_0^{\pi/2} \sqrt{1+\cos 2x}\, dx =$
(a) 0
(b) 1
(c) $\dfrac{1}{\sqrt{2}}$
(d) None of these
›Reveal solutionSolution

Use $1+\cos 2x = 2\cos^2 x$ to simplify the square root, then integrate $\cos x$ .

Since $1+\cos 2x = 2\cos^2 x$ , we have $\sqrt{1+\cos 2x} = \sqrt{2}\,|\cos x|$ . On $[0,\pi/2]$ , $\cos x \ge 0$ , so this is $\sqrt{2}\cos x$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int \cos^2 x\, dx$
(a) $\dfrac14 \sin 2x + c$
(b) $\dfrac{x}{2} + \dfrac14 \sin 2x + c$
(c) $\sin^2 x + c$
(d) $\dfrac{x}{2} - \dfrac14 \cos 2x + c$
›Reveal solutionSolution

Use the identity $\cos^2 x = \dfrac{1+\cos 2x}{2}$ .

…

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CBSE 2025Set E1 markMCQ
$\int_{0}^{\pi/2} \sin x \cdot \cos x\,dx =$
(a) $1$
(b) $\frac{1}{2}$
(c) $-1$
(d) $\frac{1}{4}$
›Reveal solutionSolution

With $u=\sin x$ , $\int_0^{\pi/2}\sin x\cos x\,dx=\big[\tfrac{\sin^2x}{2}\big]_0^{\pi/2}=\tfrac12$ .

Let $u=\sin x$ , so $du=\cos x\,dx$ . Then

…

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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int_{0}^{\pi/4} \sin 2x\,dx$
(a) 0
(b) 1
(c) 2
(d) $\dfrac{1}{2}$
›Reveal solutionSolution

Antiderivative of $\sin2x$ is $-\cos2x/2$ ; evaluating from $0$ to $\pi/4$ gives $1/2$ .

$\int_{0}^{\pi/4}\sin2x\,dx=\left[-\dfrac{\cos2x}{2}\right]_{0}^{\pi/4}$

…

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CBSE 20241 markMCQ
$\int \frac{\cos 2x}{\sin^2 x \cos^2 x} dx$
›Reveal solutionSolution
The integral simplifies by rewriting $\cos 2x$ as $\cos^2 x - \sin^2 x$ , splitting into two simple integrals, and integrating term by term. The result is $-\cot x - \tan x + c$ , which matches option (D).

The key insight here is that the denominator $\sin^2 x \cos^2 x$ is a product of squares, and the numerator $\cos 2x$ is begging to be expressed in terms of $\sin^2 x$ and $\cos^2 x$ . Once you do that, the fraction splits naturally into two separate terms, each of which is a standard integral.

Let’s walk through it.

Rewrite $\cos 2x$ using a double-angle identity.

The most useful form here is $\cos 2x = \cos^2 x - \sin^2 x$ . This directly matches the squares in the denominator.

Split the integrand into two fractions.

$\frac{\cos 2x}{\sin^2 x \cos^2 x} = \frac{\cos^2 x - \sin^2 x}{\sin^2 x \cos^2 x} = \frac{\cos^2 x}{\sin^2 x \cos^2 x} - \frac{\sin^2 x}{\sin^2 x \cos^2 x}$
Cancel common factors:
$= \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x}$

Recognise the standard forms.

$\frac{1}{\sin^2 x} = \csc^2 x, \quad \frac{1}{\cos^2 x} = \sec^2 x$
So the integral becomes:
$\int (\csc^2 x - \sec^2 x) \, dx$

Integrate term by term. Recall:

$\int \csc^2 x \, dx = -\cot x + c, \quad \int \sec^2 x \, dx = \tan x + c$
Therefore:
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CBSE 2024Set D1 markMCQ
$\int_0^{\pi/2} \cos 2x\,dx =$
(a) $0$
(b) $1$
(c) $-1$
(d) $2$
›Reveal solutionSolution

The antiderivative $\frac{\sin 2x}{2}$ evaluates to $0$ at both limits.

…

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CBSE 2024Set ANNUAL1 markMCQ
$\int \sin^2 x \,dx =$
(a) $\frac{x}{2}-\frac{\sin 2x}{4}+c$
(b) $\frac{\sin 2x}{4}-\frac{x}{2}+c$
(c) $\frac{\sin 2x}{2}+\frac{x}{2}+c$
(d) none of these
›Reveal solutionSolution

Rewrite sin^2 x using the double-angle identity, then integrate term by term.

Using $\sin^2x=\dfrac{1-\cos2x}{2}$ : …

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CBSE 2023Set AX1 markMCQ
$\int \cos^2 x\, dx$
(a) $\dfrac{x}{2}+\dfrac{1}{4}\sin 2x+c$
(b) $\dfrac{x}{4}-\dfrac{1}{2}\sin 2x+c$
(c) $\cos^2 x-\sin^2 x+c$
(d) $2\cos x\sin x+\dfrac{x}{2}+c$
›Reveal solutionSolution

Reduce the square with the double-angle identity, then integrate term by term: $\dfrac{x}{2}+\dfrac14\sin 2x+c$ , option (a).

Concept. A squared trig function is integrated by lowering the power with a double-angle identity, turning $\cos^2x$ into a simple linear combination of $1$ and $\cos 2x$ .

Solution. Using $\cos^2x=\dfrac{1+\cos 2x}{2}$ , …

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CBSE 2023Set ANNUAL1 markMCQ
$\int \cos^2 x\,dx$
(a) $\frac{x}{2}+\frac{1}{4}\sin 2x+C$
(b) $x^2+\frac{1}{4}\sin 2x+C$
(c) $\frac{x}{4}+\frac{1}{2}\sin x+C$
(d) $\frac{x^2}{2}+\frac{1}{2}\sin^2 x+C$
›Reveal solutionSolution

Use the identity $\cos^2 x=\frac{1+\cos 2x}{2}$ to make the integral straightforward.

$\int\cos^2x\,dx=\int\dfrac{1+\cos 2x}{2}dx=\dfrac{1}{2}\int dx+\dfrac{1}{2}\int\cos 2x\,dx$

…

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CBSE 2023Set ANNUAL1 mark
$\displaystyle\int_0^{\pi/2} \cos 2x\, dx$
›Reveal solutionSolution

Integrate $\cos2x$ to $\tfrac12\sin2x$ and evaluate the limits.

$\int_0^{\pi/2}\cos 2x\,dx=\left[\frac{\sin 2x}{2}\right]_0^{\pi/2}=\frac{\sin\pi}{2}-\frac{\sin 0}{2}=\frac{0}{2}-\frac{0}{2}=0.$

…

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CBSE 2022Set ANNUAL1 markMCQ
$\int_{\pi/2}^{\pi/2} \sin^2 x\,dx =$
(a) $0$
(b) $1$
(c) $5$
(d) $11$
›Reveal solutionSolution

The limits of integration are identical, so the integral is $0$ .

The printed integral is $\int_{\pi/2}^{\pi/2}\sin^2 x\,dx$ . By the property $\int_a^a f(x)\,dx = 0$ , any definite integral whose lower and upper …

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CBSE 2021Set I1 markMCQ
$\int_0^{\pi/2} \cos x\,dx =$
(a) $1$
(b) $-1$
(c) $0$
(d) $2$
›Reveal solutionSolution

Antiderivative of $\cos x$ is $\sin x$ ; evaluate at the limits.

…

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