Test Index

NDA Math Indefinite Integrals

Show Para Hide Para

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]

© examsnet.com

Question : 31 of 38

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}

© examsnet.com

Go to Question:

Prev Question

Next Question