Integrate the following function: $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$

Integrate the following function: (cos 2x + 2sin² x)/(cos² x)

Exercise 7.3 · Q18

\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}

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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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Using $\cos 2x = 1 - 2\sin^2 x$ , the numerator collapses to $1$ , so the integrand is $\sec^2 x$ and the integral is $\tan x + C$ .

Simplify the numerator. With the identity $\cos 2x = 1 - 2\sin^2 x$ ,

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

Rewrite the integrand. Dividing by $\cos^2 x$ , …

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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.Integrate the following function: $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$

Concept understanding — Trigonometric Simplification

Trigonometric Simplification

Your core toolkit

How the process feels

Strategies that usually work

Method: Collapse the numerator with a double-angle identity before dividing

Steps

Common Mistakes

Q.Integrate the following function: cos⁡2x+2sin⁡2xcos⁡2x\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}cos2xcos2x+2sin2x​

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Q.Integrate the following function: $\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}$

Unlock everything free for 14 days