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UPSC NDA II 2018 Mathematics Solved Paper

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Question : 74 of 120

Marks: +1, -0

What is

\overset{1}{\underset{-1}{∫}}{\frac{d}{dx}(tan^{-1}\frac{1}{x})}dx

equal to?

0
$-\frac{\pi }{4}$
$-\frac{\pi }{2}$
$\frac{\pi }{2}$

Solution:

Concept:

The integrand

\frac{d}{dx}\left(\tan^{-1}\frac{1}{x}\right)

simplifies using the identity

\tan^{-1}\frac{1}{x} = \cot^{-1}x

.

The integral of an even function over symmetric limits

[-a,a]

equals

2\int_{0}^{a} f(x)\,dx

.

Explanation:

Let

I = \overset{1}{\underset{-1}{\int}} \frac{d}{dx}\left(\tan^{-1}\frac{1}{x}\right) dx

.

Use

\tan^{-1}\frac{1}{x} = \cot^{-1}x

. Then

\frac{d}{dx}(\cot^{-1}x) = \frac{-1}{1+x^{2}}

.

So

I = \overset{1}{\underset{-1}{\int}} \frac{-1}{1+x^{2}}\,dx

.

Define

f(x) = \frac{-1}{1+x^{2}}

. Check even/odd:

f(-x) = \frac{-1}{1+(-x)^{2}} = \frac{-1}{1+x^{2}} = f(x)

.

Since

f(x)

is even,

I = 2\overset{1}{\underset{0}{\int}} \frac{-1}{1+x^{2}}\,dx = -2\overset{1}{\underset{0}{\int}} \frac{1}{1+x^{2}}\,dx

.

Now

\int \frac{1}{1+x^{2}} dx = \tan^{-1}x + C

. Evaluate:

[-2\tan^{-1}x]_{0}^{1} = -2(\tan^{-1}1 - \tan^{-1}0) = -2\left(\frac{\pi}{4} - 0\right) = -\frac{\pi}{2}

.

Thus

I = -\frac{\pi}{2}

.

Answer:

-\frac{\pi}{2}

\quad

(Option C)

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