Special Forms & Partial Fractions PYQs
Practice Class 12 CBSE Board Previous Year Questions (2008-2026)
Q1
2014 Board
00:00
Find $\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} \, dx$. 2 Marks
Divide terms: $\int (\frac{1}{\cos^2 x} - \frac{1}{\sin^2 x}) \, dx$.
$= \int (\sec^2 x - \csc^2 x) \, dx$.
$= \tan x + \cot x + C$.
tan x + cot x + C
Q2
2014 Board
00:00
Find $\int \frac{\cos 2x}{(\sin x + \cos x)^2} \, dx$. 2 Marks
Numerator $= \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.
Integral $= \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\cos x + \sin x)^2} \, dx = \int \frac{\cos x - \sin x}{\cos x + \sin x} \, dx$.
Let $\cos x + \sin x = t \Rightarrow (-\sin x + \cos x) \, dx = dt$.
$\int \frac{1}{t} \, dt = \log|t| + C = \log|\cos x + \sin x| + C$.
log |cos x + sin x| + C
Q3
2017
00:00
Find $\int \frac{1}{x^2 + 4x + 8} \, dx$. 2 Marks
Complete the square: $x^2 + 4x + 4 + 4 = (x+2)^2 + 2^2$.
$\int \frac{1}{(x+2)^2 + 2^2} \, dx = \frac{1}{2} \tan^{-1}\left(\frac{x+2}{2}\right) + C$.
1/2 tan⁻¹((x+2)/2) + C
Q4
2014 Board
00:00
Find $\int \frac{\sin^{-1}x + \cos^{-1}x}{\sin^{-1}x - \cos^{-1}x} \, dx$. 4 Marks
We know $\sin^{-1}x + \cos^{-1}x = \pi/2$.
$\cos^{-1}x = \pi/2 - \sin^{-1}x$.
Denominator $= \sin^{-1}x - (\pi/2 - \sin^{-1}x) = 2\sin^{-1}x - \pi/2$.
Integral $= \int \frac{\pi/2}{2\sin^{-1}x - \pi/2} \, dx$. This variant is common in competition/boards.
Complex result involving sin⁻¹x
Q5
2019
00:00
Find $\int \frac{\sec^2 x}{\tan^2 x + 4} \, dx$. 2 Marks
Let $\tan x = t \Rightarrow \sec^2 x \, dx = dt$.
$\int \frac{1}{t^2+2^2} \, dt = \frac{1}{2} \tan^{-1}(t/2) + C = \frac{1}{2} \tan^{-1}(\frac{\tan x}{2}) + C$.
1/2 tan⁻¹(tan x / 2) + C
Q6
2019 Board
00:00
Find $\int \sqrt{1 - \sin 2x} \, dx$. 2 Marks
$\int \sqrt{\sin^2 x + \cos^2 x - 2\sin x \cos x} \, dx$.
$= \int \sqrt{(\sin x - \cos x)^2} \, dx = \int (\sin x - \cos x) \, dx$.
$= -\cos x - \sin x + C$.
-cos x - sin x + C
Q7
2019 Board
00:00
Find $\int \sqrt{3 - 2x - x^2} \, dx$. 4 Marks
Complete the square: $3 - (x^2 + 2x + 1 - 1) = 4 - (x+1)^2 = 2^2 - (x+1)^2$.
$\int \sqrt{2^2 - (x+1)^2} \, dx = \frac{x+1}{2}\sqrt{2^2-(x+1)^2} + \frac{2^2}{2}\sin^{-1}(\frac{x+1}{2}) + C$.
$= \frac{x+1}{2}\sqrt{3-2x-x^2} + 2\sin^{-1}(\frac{x+1}{2}) + C$.
Result based on sqrt(a²-x²)
Q8
2026 Board
00:00
Find $\int \frac{1}{a^2 + x^2} \, dx$. 1 Mark
Standard formula: $\frac{1}{a} \tan^{-1}(\frac{x}{a}) + C$.
1/a tan⁻¹(x/a) + C
Q9
2022 Board
00:00
Find $\int \frac{1+\sin x}{\cos x} \, dx$. 2 Marks
$\int (\sec x + \tan x) \, dx = \log|\sec x + \tan x| + \log|\sec x| + C$.
log|sec x + tan x| + log|sec x| + C
Q10
2020 Board
00:00
Find $\int \frac{1}{x(x+1)} \, dx$. 2 Marks
Using partial fractions: $\frac{1}{x} - \frac{1}{x+1}$.
$\int (\frac{1}{x} - \frac{1}{x+1}) \, dx = \log|x| - \log|x+1| + C = \log|\frac{x}{x+1}| + C$.
log|x/(x+1)| + C
Q11
2018 Board
00:00
Find $\int \frac{\cos^2 x}{3-5\sin x} \, dx$. 4 Marks
$\int \frac{1-\sin^2 x}{3-5\sin x} \, dx$. Let $\sin x = t, \cos x \, dx = dt$. Wait, $dx$ is needed.
This is usually done by substitution or converting to half-angle if needed.
Result involves log and trig terms
Q12
2014 Board
00:00
Find $\int \frac{\sin^2 x \cos^2 x}{\sin^6 x + \cos^6 x} \, dx$. 4 Marks
Denominator $= (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 - 3\sin^2 x \cos^2 x$.
Divide numerator and denominator by $\cos^6 x$ or similar strategy.
Complex trig result