First slide

Introduction to integration

Question

$\int \frac{d t}{a^{2} \sin^{2} t + b^{2} \cos^{2} t} is equal to$

Moderate

A

$\frac{1}{a} \tan^{- 1} (\frac{a \tan t}{b}) + k$
B

$\frac{- 1}{a b} \tan^{- 1} (\frac{b}{a} \tan t) + c$
C

$\frac{1}{a b} \tan^{- 1} (\frac{b}{a} \tan t) + k$
D

$\frac{1}{a b} \tan^{- 1} (\frac{a}{b} \tan t) + k$

Solution

$\int \frac{1}{a^{2} \sin^{2} t + b^{2} \cos^{2} t} d t = \int \frac{\frac{1}{a^{2} \cos^{2} t} d t}{\tan^{2} t + \frac{b^{2}}{a^{2}}}$
$= \frac{1}{a^{2}} \int \frac{\sec^{2} t d t}{\tan^{2} t + {(\frac{b}{a})}^{2}}$
$Put \tan t = r$
$\sec^{2} t d t = d r$
$Use \int \frac{1}{a^{2} + x^{2}} d x = \frac{1}{a} \tan^{- 1} (\frac{x}{a}) + c$
$\begin{array}{l} = \frac{1}{a^{2}} \int \frac{d r}{r^{2} + {(\frac{b}{a})}^{2}} \\ = \frac{1}{a^{2}} \frac{1}{(\frac{b}{a})} \tan^{- 1} (\frac{r}{\frac{b}{a}}) + k \\ = \frac{1}{a b} \tan^{- 1} (\frac{a \tan t}{b}) + k \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App