UPSC NDA I 2015 Mathematics Paper

Question Numbers: 70-71

For the next two (2) items that follow:

The integral

∫\frac{dx}{acosx+bsinx}

is of the form

\frac{1}{r}ln[tan(\frac{x+a}{2})]

Question : 70 of 120

Marks: +1, -0

What is r equal to?

Solution:

Concept:

Given

a = r \sin \alpha

and

b = r \cos \alpha

, we find

r

by squaring and adding these equations.

Explanation:

We start from the substitution used in the integral:

a = r \sin \alpha

and

b = r \cos \alpha

Here

r

is the magnitude and

\alpha

is an angle.

Square both equations:

a^2 = r^2 \sin^2 \alpha

and

b^2 = r^2 \cos^2 \alpha

Add them:

a^2 + b^2 = r^2 (\sin^2 \alpha + \cos^2 \alpha)

Since

\sin^2 \alpha + \cos^2 \alpha = 1

, we get

a^2 + b^2 = r^2

Taking the positive square root (as

r

is a magnitude):

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

Answer:

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

, which corresponds to option B.

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