Class 12 Integrals PYQs

Q1 2024

00:00

Find $\int \tan x \, dx$. 1 Mark

$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx$.

Let $\cos x = t \Rightarrow -\sin x \, dx = dt$.

$-\int \frac{1}{t} \, dt = -\log |t| + C = -\log |\cos x| + C$.

$= \log |\sec x| + C$.

log |sec x| + C

Q2 2025 Board

00:00

Find $\int \frac{x^2+1}{x} \, dx$. 1 Mark

$\int (x + \frac{1}{x}) \, dx$.

$= \frac{x^2}{2} + \log |x| + C$.

x²/2 + log |x| + C

Q3 2018

00:00

Find $\int \frac{\cos^2 x}{\cos 2x + 2\sin^2 x} \, dx$. 2 Marks

We know $\cos 2x = \cos^2 x - \sin^2 x$.

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

Integral $= \int \cos^2 x \, dx = \int \frac{1 + \cos 2x}{2} \, dx$.

$= \frac{1}{2}x + \frac{\sin 2x}{4} + C$.

x/2 + (sin 2x)/4 + C

Q4 2019

00:00

Find $\int \sin x \log(\cos x) \, dx$. 2 Marks

Let $\cos x = t \Rightarrow -\sin x \, dx = dt$.

$-\int \log t \, dt$.

Using integration by parts: $-(t \log t - t) + C = t(1 - \log t) + C$.

$= \cos x(1 - \log(\cos x)) + C$.

cos x(1 - log(cos x)) + C

Q5 2011

00:00

Find $\int \frac{\cos^2 x}{1-\sin x} \, dx$. 1 Mark

$\int \frac{1-\sin^2 x}{1-\sin x} \, dx = \int \frac{(1-\sin x)(1+\sin x)}{1-\sin x} \, dx$.

$= \int (1+\sin x) \, dx = x - \cos x + C$.

x - cos x + C

Q6 2014 Board

00:00

Find $\int x e^{x^2} \, dx$. 2 Marks

Let $x^2 = t \Rightarrow 2x \, dx = dt \Rightarrow x \, dx = dt/2$.

$\int \frac{1}{2} e^t \, dt = \frac{1}{2} e^t + C = \frac{1}{2} e^{x^2} + C$.

1/2 e^{x^2} + C

Q7 2011 Board

00:00

Find $\int \frac{\sin 2x}{\sin x + \cos x} \, dx$. 2 Marks

$\int \frac{2\sin x \cos x}{\sin x + \cos x} \, dx$. Add and subtract 1 in numerator?

Alternative: $(\sin x + \cos x)^2 = 1 + \sin 2x \Rightarrow \sin 2x = (\sin x + \cos x)^2 - 1$.

$\int \frac{(\sin x + \cos x)^2 - 1}{\sin x + \cos x} \, dx = \int (\sin x + \cos x - \frac{1}{\sin x + \cos x}) \, dx$.

$= -\cos x + \sin x - \frac{1}{\sqrt{2}} \log|\tan(x/2 + \pi/8)| + C$.

-cos x + sin x - (1/√2) log|tan(x/2 + π/8)| + C

Q8 2008 Board