Question

The integral $\int \frac{dx}{a\cos x+b\sin x}$ is of the form $\frac{1}{r}\ln \left[\tan \left(\frac{x+\alpha }{2}\right)\right]$.

What is $r$ equal to?

• A. $a^2+b^2$
• B. $\sqrt{a^2+b^2}$
• C. $a+b$
• D. $\sqrt{a^2-b^2}$

Answer 1

The correct option is B $\sqrt{a^2+b^2}$

$\int \frac{dx}{a\cos x+b\sin x}=\frac{1}{\sqrt{a^2+b^2}}\int \frac{dx}{\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x}$

Let $\frac{a}{\sqrt{a^2+b^2}}=\sin \alpha ,\frac{b}{\sqrt{a^2+b^2}}=\cos \alpha$

$\Rightarrow \int \frac{dx}{a\cos x+b\sin x}=\frac{1}{\sqrt{a^2+b^2}}\int \frac{dx}{\sin \alpha \cos x+\cos \alpha \sin x}$

$=\frac{1}{\sqrt{a^2+b^2}}\int \frac{dx}{\sin (\alpha +x)}$

$=\frac{1}{\sqrt{a^2+b^2}}\int \frac{dx}{2\sin \left(\frac{\alpha +x}{2}\right)\cos \left(\frac{\alpha +x}{2}\right)}$

$=\frac{1}{\sqrt{a^2+b^2}}\ln \left(\tan \left(\frac{\alpha +x}{2}\right)\right)$

$\therefore r=\sqrt{a^2+b^2}$