Integrate the following function: (cos 2x + 2sin² – Class 12 Mathematics | ConceptFirst

Exercise 7.3 · Q18

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

Gujarat GsebTextbookSubjective· 2mImportance★★★★★

23% · 86/373 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

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 $\cos^2 x$ , …

Start your 14-day free trial to unlock the full solution →

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

Showing the 12 most recent of 20 on this concept.

GUJCET 2021Set 151 markMCQ
$\int \tan\left(\frac{\pi}{4} - x\right) \cdot (2 + 2\sin 2x)\,dx = \underline{\hspace{2cm}} + C$
›Reveal solutionSolution

Simplify the integrand to $2\cos2x$ before integrating.

Concept: Using $\tan\!\left(\tfrac{\pi}{4}-x\right)=\dfrac{\cos x-\sin x}{\cos x+\sin x}$ and $2+2\sin2x=2(\sin x+\cos x)^2$ : …
Start your 14-day free trial to see the full solution →
GUJCET 2021Set 151 markMCQ
$\cos^2\left(\sin^{-1} x\right) + \sin^2\left(\cos^{-1} x\right) = \underline{\hspace{2cm}}$
›Reveal solutionSolution

Both terms equal $1-x^2$ .

Concept: If $\theta=\sin^{-1}x$ then $\cos^2\theta=1-x^2$ . If $\phi=\cos^{-1}x$ then $\sin^2\phi=1-x^2$ . Adding, …
Start your 14-day free trial to see the full solution →
GUJCET 2019Set 171 markMCQ
$\int \dfrac{\sin x}{\sin(x-\alpha)}\,dx = px - q\log|\sin(x-\alpha)| + C$
›Reveal solutionSolution

Split $\sin x = \sin((x-\alpha)+\alpha)$ ; integrate → $p=\cos\alpha,\ q=-\sin\alpha$ .

Concept: Expand the numerator relative to $(x-\alpha)$ :
$\sin x = \sin(x-\alpha)\cos\alpha + \cos(x-\alpha)\sin\alpha$ $\frac{\sin x}{\sin(x-\alpha)} = \cos\alpha + \sin\alpha\,\cot(x-\alpha)$
Integrating:

$\int = x\cos\alpha + \sin\alpha\,\log|\sin(x-\alpha)| + C$ …
Start your 14-day free trial to see the full solution →
GUJCET 2026Set x1 markMCQ
$\tan^{-1}\left[2\cos\left(2\sin^{-1}\frac{1}{2}\right)\right] =$
›Reveal solutionSolution

Evaluate the innermost inverse function first, then work outward.

$\sin^{-1}\frac{1}{2}=\frac{\pi}{6}$ , so $2\sin^{-1}\frac12=\frac{\pi}{3}$ . …
Start your 14-day free trial to see the full solution →
GUJCET 2019Set 171 markMCQ
$\cos\left(\cot^{-1}\left(\csc\left(\cos^{-1}a\right)\right)\right) = \underline{\hspace{2cm}}$
›Reveal solutionSolution
Let $\theta=\cos^{-1}a$ ; then $\csc\theta=\dfrac{1}{\sqrt{1-a^2}}$ , and $\cos(\cot^{-1}(\csc\theta))=\dfrac{1}{\sqrt{2-a^2}}$ .

Concept: Peel the composition from the inside.

$\theta=\cos^{-1}a\Rightarrow\cos\theta=a,\ \sin\theta=\sqrt{1-a^2}$ , so $\csc\theta=\dfrac{1}{\sqrt{1-a^2}}$ . …
Start your 14-day free trial to see the full solution →
GUJCET 2023Set 091 markMCQ
$\cos\left(\sec^{-1}2\right) + \tan\left(\cot^{-1}\sqrt{3}\right) + \sin\left(\operatorname{cosec}^{-1}\frac{2}{\sqrt{3}}\right) =$
›Reveal solutionSolution
Evaluate each inverse-trig term from a reference right triangle, then add.

Concept:

$\cos(\sec^{-1}2)=\cos\tfrac{\pi}{3}=\tfrac12.$

$\tan(\cot^{-1}\sqrt3)=\tan\tfrac{\pi}{6}=\tfrac{1}{\sqrt3}.$

$\sin\!\left(\operatorname{cosec}^{-1}\tfrac{2}{\sqrt3}\right)=\tfrac{\sqrt3}{2}.$ …
Start your 14-day free trial to see the full solution →
GUJCET 2019Set 171 markMCQ
$\sin^2\left(\sin^{-1}\dfrac{1}{2}\right) + \tan^2\left(\sec^{-1}2\right) + \cot^2\left(\csc^{-1}4\right) = \underline{\hspace{2cm}}$
›Reveal solutionSolution
Evaluate each inverse-trig term on a right triangle.

Steps.

$\sin^2(\sin^{-1}\tfrac12)=\left(\tfrac12\right)^2=\tfrac14$ .

$\sec^{-1}2 = 60^\circ$ , so $\tan^2 60^\circ = 3$ . …
Start your 14-day free trial to see the full solution →
GUJCET 2025Set 031 markMCQ
$\tan^{-1}\left[\frac{\sqrt{2}}{\sqrt{3}}\cos\left(5\sin^{-1}\frac{1}{\sqrt{2}}\right)\right] =$
›Reveal solutionSolution

Evaluate the inner inverse-sine, then the cosine, then the outer arctan.

$\sin^{-1}\tfrac{1}{\sqrt2}=\tfrac{\pi}{4}$ , so $5\sin^{-1}\tfrac1{\sqrt2}=\tfrac{5\pi}{4}$ and $\cos\tfrac{5\pi}{4}=-\tfrac{1}{\sqrt2}$ . …
Start your 14-day free trial to see the full solution →
GUJCET 2020Set 071 markMCQ
$\tan^2\left(\sec^{-1} 3\right) + \csc^2\left(\cot^{-1} 2\right) + \cos^2\left(\cos^{-1}\frac{2}{3} + \sin^{-1}\frac{2}{3}\right) =$
›Reveal solutionSolution

The three terms evaluate to $8+5+0=13$ .

Term 1. $\tan^2(\sec^{-1}3)=\sec^2-1=3^2-1=8.$

Term 2. $\csc^2(\cot^{-1}2)=1+\cot^2=1+2^2=5.$ …
Start your 14-day free trial to see the full solution →
GUJCET 2022Set 081 markMCQ
$\sin^{-1}a=\alpha+\beta$
›Reveal solutionSolution

Use sin(α+β)sin(α−β) = sin²α − sin²β together with cos²β = 1 − sin²β.

Concept. Given $a=\sin(\alpha+\beta)$ and $b=\sin(\alpha-\beta)$ , the identity $\sin(\alpha+\beta)\sin(\alpha-\beta)=\sin^2\alpha-\sin^2\beta$ gives $ab=\sin^2\alpha-\sin^2\beta$ .

Solution. …
Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$\sin\left[2\cot^{-1}\left(-\dfrac{5}{12}\right)\right] =$
(a) $24/169$
(b) $-120/169$
(c) $-24/169$
(d) $120/169$
›Reveal solutionSolution

Build a reference right triangle from $\cot\theta=-5/12$ , find $\sin\theta,\cos\theta$ , then use the double-angle formula.

Let $\theta=\cot^{-1}(-5/12)$ . Since the range of $\cot^{-1}$ is $(0,\pi)$ and $\cot\theta<0$ , $\theta$ lies in the second quadrant, so $\sin\theta>0,\cos\theta<0$ .

…

Start your 14-day free trial to see the full solution →
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$\sin^{-1}\left\{\cos\left(\sin^{-1}\dfrac{1}{2}\right)\right\}$
(a) $\dfrac{\pi}{3}$
(b) $\dfrac{\pi}{4}$
(c) $\dfrac{\pi}{6}$
(d) $-\dfrac{\pi}{3}$
›Reveal solutionSolution

Work from the inside out, evaluating each inverse-trig layer in turn.

…

Start your 14-day free trial to see the full solution →

←Previous question · 7.3.2Q17 — Integrate the following function: (sin³ x + cos³ x)/(sin² x cos² x)→Next question · 7.3.2Q19 — Integrate the following function: (1)/(sin x cos³ x)