Evaluate: $\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx$

Evaluate: ∫ (cos 5x+cos 4x)/(1-2cos 3x)dx

NCERT Exemplar · Q22

\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx

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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.

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The integrand simplifies to $-(\cos x+\cos 2x)$ , so the integral is $-\sin x-\dfrac12\sin 2x+C$ .

Idea. A fraction with $\cos 5x+\cos 4x$ on top and $1-2\cos 3x$ on the bottom looks hard, but it hides a clean identity: the whole quotient equals $-(\cos x+\cos 2x)$ . Once we confirm that, the integral is immediate.

1. Establish the identity

We claim

\frac{\cos 5x+\cos 4x}{1-2\cos 3x}=-(\cos x+\cos 2x).

Multiply the right side by the denominator and expand, using $2\cos A\cos B=\cos(A+B)+\cos(A-B)$ :

-(\cos x+\cos 2x)(1-2\cos 3x)=-\cos x-\cos 2x+2\cos 3x\cos x+2\cos 3x\cos 2x.

Now

$2\cos 3x\cos x=\cos 4x+\cos 2x,\qquad 2\cos 3x\cos 2x=\cos 5x+\cos x.$ …

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

KEAM 2024Set eng-2024-06054 marksMCQ
$\cos 18^\circ\cos 42^\circ\cos 78^\circ=$
›Reveal solutionSolution

Pairing $\cos42^\circ\cos78^\circ$ and combining reduces the product to $\frac14\sin36^\circ$ .

$\cos42^\circ\cos78^\circ=\tfrac12(\cos120^\circ+\cos36^\circ)=\tfrac12(-\tfrac12+\cos36^\circ)$ .

Multiplying by $\cos18^\circ$ and evaluating numerically: …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\dfrac{\sin 7x+\sin 5x}{\cos 7x+\cos 5x}=$
›Reveal solutionSolution

Apply sum-to-product to numerator and denominator; the $\cos x$ cancels leaving $\tan6x$ .

$\sin7x+\sin5x=2\sin6x\cos x,\qquad \cos7x+\cos5x=2\cos6x\cos x.$ …
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KEAM 2025Set eng-2025-04294 marksMCQ
$\sin\frac{5\pi}{12} \sin\frac{\pi}{12}$
›Reveal solutionSolution

$\sin\frac{5\pi}{12}\sin\frac{\pi}{12}=\sin75^\circ\sin15^\circ=\frac12[\cos60^\circ-\cos90^\circ]=\frac12\cdot\frac12=\frac14$ . …

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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$\sin \dfrac{3\pi}{14} \cos \dfrac{3\pi}{14} = k \cos \dfrac{\pi}{14}$
›Reveal solutionSolution
sin(3pi/14) cos(3pi/14) = (1/2) sin(3pi/7) = (1/2) cos(pi/14), so k = 1/2 and 4k = 2.

Concept and Intuition

The double-angle identity sin A cos A = (1/2) sin 2A, plus the co-function relation sin(pi/2 - x) = cos x, collapses the expression onto cos(pi/14).

Step-by-Step Solution

sin(3pi/14) cos(3pi/14) = (1/2) sin(6pi/14) = (1/2) sin(3pi/7).

3pi/7 = pi/2 - pi/14, so sin(3pi/7) = cos(pi/14). …
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KEAM 2025Set eng-2025-04234 marksMCQ
$f(x) = \cos x$
›Reveal solutionSolution
The expression simplifies identically to 0.

Concept and Intuition

Using the sum-to-product identity cosP+cosQ, the bracket collapses to 2cos x cos y; halving it exactly cancels the product term cos x cos y.

Step-by-Step Solution

$\cos(x+y) = \cos x\cos y - \sin x\sin y$ .

$\cos(y-x) = \cos y\cos x + \sin y\sin x$ .

Sum $= 2\cos x\cos y$ , so $\tfrac{1}{2}[\cdots] = \cos x\cos y$ . …
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KEAM 2026Set eng-2026-04224 marksMCQ
$4\cos 36^{\circ}\cos 72^{\circ}$
›Reveal solutionSolution
[!TLDR]

With $\cos36^\circ\cos72^\circ=\tfrac14$ , the expression $4\cos36^\circ\cos72^\circ=1$ .

Concept

The standard exact values are $\cos36^\circ=\dfrac{\sqrt5+1}{4}$ and $\cos72^\circ=\dfrac{\sqrt5-1}{4}$ (they can also be derived from multiple-angle identities in the NCERT/CBSE trigonometry syllabus).

Solution …
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KEAM 2025Set eng-2025-04234 marksMCQ
$\sin x + \sin y = a$
›Reveal solutionSolution
a/b = tan((x+y)/2) = tan(pi/3) = sqrt3.

Concept and Intuition

Sum-to-product turns sin x + sin y and cos x + cos y into forms sharing the common factor 2 cos((x-y)/2), so their ratio is simply tan of the half-sum (x+y)/2.

Step-by-Step Solution

a = sin x + sin y = 2 sin((x+y)/2) cos((x-y)/2).

b = cos x + cos y = 2 cos((x+y)/2) cos((x-y)/2).

a/b = tan((x+y)/2) = tan(pi/3) = sqrt3. …
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Q.Evaluate: $\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx$

Concept understanding — Product To Sum Identity

Why turn a product into a sum?

The core idea in one sentence

The four identities

Where they come from (the derivation)

Worked examples

1. Establish the identity

Method: Trigonometric simplification before integrating

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept

Solution …

Concept and Intuition

Step-by-Step Solution

Q.Evaluate: ∫cos⁡5x+cos⁡4x1−2cos⁡3x dx\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx∫1−2cos3xcos5x+cos4x​dx

1. Establish the identity

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Q.Evaluate: $\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx$

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