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UPSC NDA I 2015 Mathematics Solved Paper

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For the next two (2) items that follow :

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]

[NDA I 2015]

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Question : 70 of 120

Marks: +1, -0

What is r equal to ?

$a^2+b^2$
$\sqrt{a^2+b^2}$
$a+b$
$\sqrt{a^2-b^2}$

Solution:

The above integral can be solved in the following way

\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}

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

Now, the constant angle can written as

\sin \alpha = \frac{a}{\sqrt{a^2+b^2}}

\cos \alpha = \frac{b}{\sqrt{a^2+b^2}}

\tan \alpha = \frac{a}{b}

\alpha = \tan^{-1}\left(\frac{a}{b}\right)

Therefore

\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 \csc(\alpha+x) dx

\frac{-1}{\sqrt{a^2+b^2}} \ln \left| \csc(\alpha+x) + \cot(\alpha+x) \right|

\frac{1}{\sqrt{a^2+b^2}} \ln \left| \frac{1}{\csc(\alpha+x) + \cot(\alpha+x)} \right|

\frac{1}{\sqrt{a^2+b^2}} \ln \left| \frac{\sin(\alpha+x)}{1+\cos(\alpha+x)} \right|

= \; \frac{1}{\sqrt{a^2+b^2}} \ln \left| \; \frac{2 \sin\left(\frac{\alpha+x}{2}\right) \cos\left(\frac{\alpha+x}{2}\right)}{2 \cos^2\left(\frac{\alpha+x}{2}\right)} \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}

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