Q236 / 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 $\alpha$ equal to?

  • Correct answer
    A
    $\tan ^{-1}\left(\frac{a}{b}\right)$
  • B
    $\tan ^{-1}\left(\frac{b}{a}\right)$
  • C
    $\tan ^{-1}\left(\frac{a+b}{a-b}\right)$
  • D
    $\tan ^{-1}\left(\frac{a-b}{a+b}\right)$

Solution

$\begin{aligned} & a=r \sin \alpha \\ & b=r \cos \alpha \end{aligned}$ Dividing (i) from (ii), $\begin{aligned} & \frac{\mathrm{a}}{\mathrm{~b}}=\tan \alpha \\ & \alpha=\tan ^{-1}\left(\frac{\mathrm{a}}{\mathrm{~b}}\right) . \end{aligned}$

Hence, the correct answer is option: 'A'.

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