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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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

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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