(a) Evaluate: ∫₀π/2 sin 2x cos 3x dx OR (b) If (d)/(dx) — CBSE Class 12 Mathematics 2024 PYQ

Question

\displaystyle\int_{0}^{\pi/2} \sin 2x \cos 3x\, dx

(OR)

(b) If

\dfrac{d}{dx} F(x) = \dfrac{1}{\sqrt{2x - x^2}}

and

F(1) = 0

, find

F(x)

CBSECBSE Class XII Board 2024Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Product To Sum Identity

Why turn a product into a sum?

Products of trig functions are messy — they don't integrate nicely and are hard to simplify. Sums are clean: you can separate and integrate them term-by-term. The Product-to-Sum identities convert a product of sines and cosines into a sum (or difference), turning something like $\sin 3x \cos 5x$ into $\frac12[\sin 8x + \sin(-2x)]$ .

The core idea in one sentence

Important

Any product of two sines and/or cosines can be rewritten as half the sum (or difference) of two sine/cosine functions whose arguments are the sum and difference of the original angles.

The four identities

For any two angles $A$ and $B$ :

\begin{aligned} \sin A \cos B &= \frac12 \bigl[ \sin(A+B) + \sin(A-B) \bigr] \\[4pt] \cos A \sin B &= \frac12 \bigl[ \sin(A+B) - \sin(A-B) \bigr] \\[4pt] \cos A \cos B &= \frac12 \bigl[ \cos(A+B) + \cos(A-B) \bigr] \\[4pt] \sin A \sin B &= \frac12 \bigl[ \cos(A-B) - \cos(A+B) \bigr] \end{aligned}

The pattern:

Same functions ( $\cos\cos$ or $\sin\sin$ ) → result uses cosines.
Different functions ( $\sin\cos$ or $\cos\sin$ ) → result uses sines.
The $\sin\sin$ case has a minus before $\cos(A+B)$ — the one that trips people up.

Where they come from (the derivation)

They follow directly from the sum and difference formulas:

\begin{aligned} \sin(A+B) &= \sin A \cos B + \cos A \sin B \\ \sin(A-B) &= \sin A \cos B - \cos A \sin B \\ \cos(A+B) &= \cos A \cos B - \sin A \sin B \\ \cos(A-B) &= \cos A \cos B + \sin A \sin B \end{aligned}

Add the first two: $\sin(A+B) + \sin(A-B) = 2\sin A \cos B$ . Divide by 2 → the first identity. Subtract them → the second. Add/subtract the cosine formulas → the other two.

Tip

If you forget an identity, derive it in 10 seconds from the sum/difference formulas.

Worked examples

Simplify $\sin 5x \cos 2x$ . With $A = 5x$ , $B = 2x$ (first identity):

\sin 5x \cos 2x = \frac12 \bigl[ \sin 7x + \sin 3x \bigr]

Simplify $\sin 4x \sin x$ . With $A = 4x$ , $B = x$ (fourth identity):

\sin 4x \sin x = \frac12 \bigl[ \cos 3x - \cos 5x \bigr]

Watch out

The $\sin\sin$ identity has a minus between the two cosines, with $\cos(A-B)$ first. Writing $\frac12[\cos(A+B) - \cos(A-B)]$ is wrong.

--- …

Part (b)Concept understanding — Integration By Completing Square

Integration by Completing the Square

You can integrate $\dfrac{1}{x^2+1}$ or $\dfrac{1}{x^2-a^2}$ on sight. But a quadratic denominator such as $x^2+4x+5$ fits neither standard form directly. The fix is to rewrite the quadratic as a perfect square plus (or minus) a constant, turning it into a form you already know.

The move

For $x^2+bx+c$ , add and subtract $\left(\dfrac{b}{2}\right)^2$ :

x^2+bx+c=\left(x+\frac{b}{2}\right)^2+\left(c-\frac{b^2}{4}\right).

After substituting $u=x+\dfrac{b}{2}$ the integral collapses to $\displaystyle\int\frac{du}{u^2+k^2}$ or $\displaystyle\int\frac{du}{u^2-k^2}$ .

Case A — leads to inverse tangent

\int\frac{dx}{x^2+4x+5}.

Complete the square: $x^2+4x+5=(x+2)^2+1$ . With $u=x+2$ ,

\int\frac{du}{u^2+1}=\tan^{-1}u+C=\tan^{-1}(x+2)+C.

Tip

$\displaystyle\int\frac{du}{u^2+a^2}=\frac1a\tan^{-1}\frac{u}{a}+C$ is the workhorse when the constant left over is positive.

Case B — leads to a logarithm

\int\frac{dx}{x^2-6x+5}.

Here $x^2-6x+5=(x-3)^2-4$ . With $u=x-3$ the denominator $u^2-4$ factors, so use partial fractions:

$\int\frac{du}{u^2-4}=\frac14\log\left|\frac{u-2}{u+2}\right|+C=\frac14\log\left|\frac{x-5}{x-1}\right|+C.$ …

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Part (a): a product-to-sum identity gives $\displaystyle\int_0^{\pi/2}\sin2x\cos3x\,dx=-\dfrac25$ . Part (b): completing the square inside the root gives $F(x)=\sin^{-1}(x-1)$ , and $F(1)=0$ fixes $C=0$ .

Part (a)

1. Convert the product to a sum. With $\sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)]$ and $A=2x$ , $B=3x$ :

\sin2x\cos3x=\tfrac12[\sin5x+\sin(-x)]=\tfrac12[\sin5x-\sin x].

2. Integrate term by term (using $\int\sin ax\,dx=-\tfrac1a\cos ax$ ):

\int_0^{\pi/2}\tfrac12(\sin5x-\sin x)\,dx=\tfrac12\Big[-\tfrac15\cos5x+\cos x\Big]_0^{\pi/2}.

3. Apply the limits.

At $x=\tfrac{\pi}{2}$ : $\cos\tfrac{5\pi}{2}=\cos\tfrac{\pi}{2}=0$ and $\cos\tfrac{\pi}{2}=0$ , giving $0$ .
At $x=0$ : $-\tfrac15\cos0+\cos0=-\tfrac15+1=\tfrac45$ . …

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

CBSE 2026Set ANNUAL1 mark
$\int \frac{dx}{\sqrt{2x - x^2}}$
›Reveal solutionSolution

Complete the square under the root, then use the standard integral $\int dx/\sqrt{a^2-x^2}=\sin^{-1}(x/a)+c$ .

$2x-x^2 = 1-(x-1)^2$

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. \int dx/(x² - 2x + 2) is ................. .$
(a) tan⁻¹(x−1) + c
(b) tan⁻¹(x+1) + c
(c) tan⁻¹(x+2) + c
(d) tan⁻¹(x−2) + c
›Reveal solutionSolution

Complete the square in the denominator, then use the standard $\int \frac{dt}{t^2+a^2}$ form.

$x^2-2x+2 = (x-1)^2+1$

…

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CBSE 2025Set ANNUAL1 mark
$\displaystyle\int\sqrt{x^2+3x+\dfrac{9}{4}}\,dx=$
›Reveal solutionSolution

The expression under the root is a perfect square, so the square root simplifies to a linear expression.

Note that $x^2+3x+\dfrac{9}{4}=\left(x+\dfrac{3}{2}\right)^2$ .

So $\sqrt{x^2+3x+\dfrac{9}{4}}=\left|x+\dfrac{3}{2}\right|$ , which (taking the positive branch) is $x+\dfrac{3}{2}$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$\displaystyle\int \sqrt{1+x^2}\, dx$
(a) $\dfrac{x}{2}\sqrt{1+x^2} + \dfrac{1}{2}\log\left|x+\sqrt{1+x^2}\right| + c$
(b) $\dfrac{2}{3}(1+x^2)^{3/2} + c$
(c) $\dfrac{2}{3}x(1+x^2)^{3/2} + c$
(d) $\dfrac{x^2}{2}\sqrt{1+x^2} + \dfrac{1}{2}x^2\log\left|x+\sqrt{1+x^2}\right| + c$
›Reveal solutionSolution

This is a standard integral of the form $\int\sqrt{x^2+a^2}\,dx$ .

The standard formula (derivable by integration by parts, treating $\sqrt{1+x^2}=\sqrt{1+x^2}\cdot 1$ ) is:

$\int \sqrt{x^2+a^2}\,dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right| + c$

With $a=1$ : …

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CBSE 2024Set D1 markMCQ
$\int \sqrt{a^2 - x^2}\,dx =$
(a) $\frac{x}{2}\sqrt{a^2 - x^2}\,dx$
(b) $\frac{a^2}{2}\sin^{-1}\frac{x}{a} + c$
(c) $\frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + c$
(d) $\frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\sin^{-1}\frac{x}{a} + c$
›Reveal solutionSolution

Standard result: $\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+c$ .

This is a memorised standard form (derivable by the substitution $x=a\sin\theta$ ):

…

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CBSE 2024Set D1 markMCQ
$\int_0^{\pi/6} \cos x \cdot \cos 2x\,dx =$
(a) $5/6$
(b) $1/6$
(c) $5/12$
(d) $-5/12$
›Reveal solutionSolution

$\cos x\cos 2x=\tfrac12(\cos 3x+\cos x)$ ; integrating from $0$ to $\tfrac\pi6$ gives $\tfrac{5}{12}$ .

Use product-to-sum: $\cos x\cos 2x = \dfrac{1}{2}(\cos 3x + \cos x)$ .

$\displaystyle\int_0^{\pi/6}\cos x\cos 2x\,dx = \dfrac{1}{2}\left[\dfrac{\sin 3x}{3} + \sin x\right]_0^{\pi/6}$

…

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CBSE 2023Set E1 markMCQ
$\displaystyle\int\dfrac{3\,dx}{\sqrt{1-9x^{2}}} =$
(a) $\tan^{-1}3x+k$
(b) $\sec^{-1}3x+k$
(c) $\sin^{-1}3x+k$
(d) $\cos^{-1}3x+k$
›Reveal solutionSolution

With $u=3x$ , $du=3\,dx$ , the integral becomes $\int\dfrac{du}{\sqrt{1-u^{2}}}=\sin^{-1}3x+k$ .

Let $u=3x$ , so $du=3\,dx$ . The numerator $3\,dx=du$ .

…

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CBSE 2021Set ANNUAL1 markMCQ
$\displaystyle\int \dfrac{dx}{x^{2}+2x+2}$
(a) $x\tan^{-1}(x+1)+C$
(b) $\tan^{-1}(x+1)+C$
(c) $(x+1)\tan^{-1}x+C$
(d) $\tan^{-1}x+C$
›Reveal solutionSolution

Completing the square, $x^2+2x+2=(x+1)^2+1$ , giving $\tan^{-1}(x+1)+C$ .

$x^2+2x+2 = (x+1)^2+1$

…

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CBSE 2019Set ANNUAL1 mark
$\int \sin 2x \cos 3x \, dx$
›Reveal solutionSolution

Product-to-sum gives ½(sin5x − sinx); integrating gives −cos5x/10 + cosx/2 + C.

Step 1: Use sinA cosB = ½[sin(A+B) + sin(A−B)] with A = 2x, B = 3x:

sin2x cos3x = ½[sin5x + sin(−x)] = ½[sin5x − sinx].

Step 2: Integrate term by term:

∫½ sin5x dx = ½ · (−cos5x/5) = −cos5x/10. …

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Q.(a) Evaluate: ∫0π/2sin⁡2xcos⁡3x dx\displaystyle\int_{0}^{\pi/2} \sin 2x \cos 3x\, dx∫0π/2​sin2xcos3xdx

Part (a)

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Q.(a) Evaluate: $\displaystyle\int_{0}^{\pi/2} \sin 2x \cos 3x\, dx$

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