Question

The value of $\int \frac{dx}{a\sin x+b\cos x}=A[\ln (\tan \frac{x}{2}-\frac{a-\sqrt{a^2+b^2}}{b})-\ln (\tan \frac{x}{2}-\frac{a+\sqrt{a^2+b^2}}{b})]$ where $A$ is

• A. $\frac{1}{\sqrt{a^2+b^2}}$
• B. $\frac{1}{\sqrt{ab}}$
• C. $\sqrt{a+b}$
• D. $Noneofthese$

Answer 1

The correct option is A $\frac{1}{\sqrt{a^2+b^2}}$

$\int \frac{dx}{a\sin x+b\cos x}$

$=\int \frac{dx}{a\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}+b\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}$

$=\int \frac{(1+{\tan }^2\frac{x}{2})dx}{2a\tan \frac{x}{2}+b-b{\tan }^2\frac{x}{2}}$

$=-\int \frac{\left({\sec }^2\frac{x}{2}\right)dx}{b{\tan }^2\frac{x}{2}-2a\tan \frac{x}{2}-b}$

Let $t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx$

$\Rightarrow 2dt={\sec }^2\frac{x}{2}dx$

$=-2\int \frac{dt}{b{t}^2-2at-b}$

$=-2\int \frac{dt}{b(t^2-2\frac{a}{b}t-1)}$

$=-\frac{2}{b}\int \frac{dt}{(t^2-2\frac{a}{b}t+\frac{a^2}{b^2}-\frac{a^2}{b^2}-1)}$

$=-\frac{2}{b}\int \frac{dt}{{(t-\frac{a}{b})}^2-{\left(\frac{\sqrt{a^2+b^2}}{b}\right)}^2}$

We know that $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \frac{x-a}{x+a}+c$

We know that $\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\ln \frac{x-a}{x+a}+c$

$=-\frac{2}{b}\times \frac{1}{2\times \frac{\sqrt{a^2+b^2}}{b}}\left[\ln \frac{t-\frac{a}{b}-\frac{\sqrt{a^2+b^2}}{b}}{t-\frac{a}{b}+\frac{\sqrt{a^2+b^2}}{b}}\right]$

$=-\frac{1}{\sqrt{a^2+b^2}}[\ln (\tan \frac{x}{2}-\frac{a+\sqrt{a^2+b^2}}{b})-\ln (\tan \frac{x}{2}-\frac{a-\sqrt{a^2+b^2}}{b})]$

$=\frac{1}{\sqrt{a^2+b^2}}[\ln (\tan \frac{x}{2}-\frac{a-\sqrt{a^2+b^2}}{b})-\ln (\tan \frac{x}{2}-\frac{a+\sqrt{a^2+b^2}}{b})]$