Q235 / 270 · NDA 2015 (Phase 1) · Mathematics
The integral $\int \frac{d x}{a \cos x+b \sin x}$ is of the form $\frac{1}{r} \operatorname{In}\left[\tan \left(\frac{x+\alpha}{2}\right)\right]$.
What is $r$ equal to?
- A$a^2+b^2$
- Correct answerB$\sqrt{a^2+b^2}$
- C$a+b$
- D$\sqrt{a^2-b^2}$
Solution
Given that, $\int \frac{\mathrm{dx}}{\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}}=\frac{1}{\mathrm{r}} \ln \left[\tan \left(\frac{\mathrm{x}+\alpha}{2}\right)\right]$
Let $\mathrm{a}=\mathrm{r} \sin \alpha, \mathrm{b}=\mathrm{r} \cos \alpha$
$\begin{aligned}
& \int \frac{d x}{r \sin \alpha \cos x+r \cos \alpha \sin x}=\frac{1}{r} \int \frac{1}{\sin (x+\alpha)} \\
& =\frac{1}{r} \int \operatorname{cosec}(x+\alpha) d x=\frac{1}{r} \ln \left[\tan \left(\frac{x+\alpha}{2}\right)\right]
\end{aligned}$
$\begin{aligned}
& a=r \sin \alpha \Rightarrow a^2=r^2 \sin ^2 \alpha ...(i)\\
& b=r \cos \alpha \Rightarrow b^2=r^2 \cos ^2 \alpha...(ii)
\end{aligned}$
Adding (i) and (ii), we get
$\Rightarrow \quad \begin{aligned}
\mathrm{r}^2 & =\mathrm{a}^2+\mathrm{b}^2 \\
\Rightarrow \quad \mathrm{r} & =\sqrt{\mathrm{a}^2+\mathrm{b}^2} .
\end{aligned}$
Hence, the correct answer is option: 'B'.