Evaluate: $\int \dfrac{\cos x-\cos 2x}{1-\cos x}\,dx$

Evaluate: ∫ (cos x-cos 2x)/(1-cos x)dx

NCERT Exemplar · Q25

\int \dfrac{\cos x-\cos 2x}{1-\cos x}\,dx

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Concept understanding — Trigonometric Simplification

Trigonometric Simplification

You know $\sin^2 x + \cos^2 x = 1$ — but the skill of turning a messy trig expression into that kind of clean form is trigonometric simplification. Because sines, cosines and their relatives are all tied together by identities from the unit circle, a tangled combination can almost always be rewritten as something shorter: a single term, a constant, or an easier combination.

Your core toolkit

\sin^2\theta+\cos^2\theta=1,\quad 1+\tan^2\theta=\sec^2\theta,\quad 1+\cot^2\theta=\csc^2\theta

How the process feels

Simplify $\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}$ . Over a common denominator the numerator is $\sin^2 x+(1+\cos x)^2 = \sin^2 x + 1 + 2\cos x + \cos^2 x$ . The Pythagorean identity turns $\sin^2 x+\cos^2 x$ into $1$ , giving $2+2\cos x = 2(1+\cos x)$ , so

\frac{2(1+\cos x)}{\sin x\,(1+\cos x)} = \frac{2}{\sin x} = 2\csc x.

A two-term sum collapses to one term.

Strategies that usually work

Convert everything to sines and cosines — cancellations then appear.
Spot Pythagorean pairs and replace them with $1$ (or $\sec^2$ , $\csc^2$ ).
Factor and cancel as you would with ordinary algebra.
Multiply by a conjugate — e.g. multiply $\frac{1}{1+\sin x}$ by $\frac{1-\sin x}{1-\sin x}$ to unlock a Pythagorean identity. …

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Rewrite $\cos 2x$ with the double-angle identity, factor the numerator, and cancel $1-\cos x$ . The integrand reduces to $2\cos x+1$ , giving $\int = 2\sin x + x + C$ .

1. Rewrite $\cos 2x$ . Using $\cos 2x = 2\cos^2 x - 1$ ,

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

2. Factor the numerator. Since $2\cos^2 x - \cos x - 1 = (2\cos x + 1)(\cos x - 1)$ ,

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

3. Cancel with the denominator. For $\cos x \neq 1$ , …

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

Higher Secondary (+2 Stage) Examination 2025Set ANNUAL3 marks
$\tan^{-1}\left(\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\dfrac{\pi}{4}-\dfrac{1}{2}\cos^{-1}x$
›Reveal solutionSolution

The substitution $x=\cos2\theta$ turns $\sqrt{1\pm x}$ into a clean $\sqrt2\cos\theta$ / $\sqrt2\sin\theta$ pair, collapsing the whole expression to $\tan(\pi/4-\theta)$ .

Let $x=\cos2\theta$ , so $\theta=\dfrac12\cos^{-1}x$ . Using $1+\cos2\theta=2\cos^2\theta$ and $1-\cos2\theta=2\sin^2\theta$ :

$\sqrt{1+x}=\sqrt{2\cos^2\theta}=\sqrt2\,\cos\theta,\qquad \sqrt{1-x}=\sqrt{2\sin^2\theta}=\sqrt2\,\sin\theta$

(taking $\theta$ in a range where $\cos\theta,\sin\theta\ge0$ , consistent with the given domain $-\tfrac{1}{\sqrt2}\le x\le1$ ).

Substitute into the given expression:

$\dfrac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}=\dfrac{\sqrt2\cos\theta-\sqrt2\sin\theta}{\sqrt2\cos\theta+\sqrt2\sin\theta}=\dfrac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}.$

Divide numerator and denominator by $\cos\theta$ :

$=\dfrac{1-\tan\theta}{1+\tan\theta}=\tan\left(\dfrac{\pi}{4}-\theta\right),$ …

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

Concept understanding — Trigonometric Simplification

Trigonometric Simplification

Your core toolkit

How the process feels

Strategies that usually work

Method: Simplify a trig quotient by factoring before integrating

Steps

Common Mistakes

Q.Evaluate: ∫cos⁡x−cos⁡2x1−cos⁡x dx\int \dfrac{\cos x-\cos 2x}{1-\cos x}\,dx∫1−cosxcosx−cos2x​dx

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

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