#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 58 Maths Textbook Solution.

$\left|=\frac{b}{a^{2}+b^{2}} \ln \right| \cos x+b \sin x \mid+\frac{a}{a^{2}+b^{2}} x+c$

Given:

$\int \frac{1}{a+\operatorname{btan} x} d x$

Hint:

To solve this equation, we have to use standard method form.

Solution:

$I=\int \frac{1}{a+b \tan x} d x$

$I=\int \frac{1}{a+b \frac{\sin x}{\cos x}} d x=\int \frac{\cos x}{a \cos x+\operatorname{ssin} x} d x$

$\text { Let } \cos x=A \frac{d}{d x}(a \cos x+b \sin x)+B(a \cos x+b \sin x)$

$=A(-\operatorname{asin} x+b \cos x)+B(\operatorname{acos} x+b \sin x)$

$=(A b+B a) \cos x+(B b-A a) \sin x$

$=A b+B a=1 \ldots \ldots . .(1) \quad B b-A a=0 . \ldots(2)$

On solving (1) and (2),

$B=\frac{a}{a^{2}+b^{2}}$

$A=\frac{b}{a^{2}+b^{2}}$

$I=\int \frac{A(-a \sin x+b \cos x)+B(\operatorname{acos} x+b \sin x)}{a \cos x+b \sin x} d x$

$I=\int A \frac{(-a \sin x+b \cos x)}{a \cos x+\operatorname{bsin} x}+B d x$

$I=\int A \frac{(-a \sin x+b \cos x)}{\operatorname{acos} x+\operatorname{bsin} x} d x+\int B d x$

$I=A \int \frac{(-a \sin x+b \cos x)}{a \cos x+b \sin x} d x+B \int d x \quad[\because d t=-a \sin x+b \cos x]$

$I=A \int \frac{d t}{t}+B x$

$I=\frac{b}{a^{2}+b^{2}} \ln |t|+\frac{a}{a^{2}+b^{2}} x c$

$I=\frac{b}{a^{2}+b^{2}} \ln |a \cos x+b \sin x|+\frac{a}{a^{2}+b^{2}} x+c$