Types of Integrals

1 Mark Questions

1. Find <math display="block">\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx</math>.

Delhi 2014C

Let <math>I = \int \frac{\sin^2 x}{\sin^2 x \cos^2 x} dx - \int \frac{\cos^2 x}{\sin^2 x \cos^2 x} dx</math> <math display="block">= \int \sec^2 dx - \int \csc^2 x \, dx</math> <math>= \tan x + \cot x + C</math> (1)

2. Find <math display="block">\int \frac{\sin^6 x}{\cos^8 x} dx.</math>

All India 2014C

Let <math>I = \int \frac{\sin^6 x}{\cos^8 x} dx = \int \tan^6 x \sec^2 x dx</math> Put <math>\tan x = t</math> <math>\sec^2 x \, dx = dt</math> <math display="block">\therefore I = \int t^6 dt = \frac{t^7}{7} + C = \frac{\tan^7 x}{7} + C </math> (1)

1. Evaluate <math display="block">\int \frac{dx}{\sin^2 x \cos^2 x}</math>.

Delhi 2014C; Foreign 2014

Firstly, divide numerator and denominator by <math>\cos^4 x</math> and use <math>\sec^2 x = 1 + \tan^2 x</math>, then put <math>tan x = t</math> and integrate.

Let <math display="block">I = \int \frac{dx}{\sin^2 \cos^2 x}</math>

On dividing the numerator and denominator by <math>\cos^4 x</math>, we get

<math display="block">I = \int \frac{\sec^2 x \cdot \sec^2 x}{\tan^2 x} dx</math>

<math display="block">I = \int \frac{(1 + \tan^2 x) \cdot \sec^2 x}{\tan^2 x} dx</math>

Put <math>\tan x = t \implies \sec^2 x \, dx = dt</math>

:. <math display="block">I = \int \frac{1+t^2}{t^2} dt = \int 1 dt + \int \frac{1}{t^2} dt</math> <math>\Rightarrow I = t - \frac{1}{t} + C</math> <math>I = \tan x - \cot x + C</math> (1) 4. Evaluate <math>\int \cos^{-1}(\sin x)dx</math>.

Delhi 2014C