What is ^{-1}( x + x) \, dx equal to?

Correct Answer is : πx4+x24+cπ x/4 + x²/4 + c4πx+4x2+c

Correct Answer is : πx4+x24+cπ x/4 + x²/4 + c4πx+4x2+c

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UPSC NDA 2 2017 Mathematics Solved Paper

Question : 67 of 120

Marks: +1, -0

What is

\int \tan^{-1}(\sec x + \tan x) \, dx

equal to?

Solution:

Concept:

Simplify the integrand

\tan^{-1}(\sec x + \tan x)

into a linear form

\frac{\pi}{4} + \frac{x}{2}

using trigonometric identities, then integrate directly.

Explanation:

Let

I = \int \tan^{-1}(\sec x + \tan x) \, dx

Rewrite

\sec x + \tan x = \frac{1 + \sin x}{\cos x}

Use half‑angle identities:

1 + \sin x = \left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2

and

\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2} = \left(\cos\frac{x}{2} - \sin\frac{x}{2}\right)\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)

Thus

\frac{1+\sin x}{\cos x} = \frac{\sin\frac{x}{2} + \cos\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}

Divide numerator and denominator by

\cos\frac{x}{2}

\frac{\tan\frac{x}{2} + 1}{1 - \tan\frac{x}{2}}

This equals

\tan\left(\frac{\pi}{4} + \frac{x}{2}\right)

because

\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}

with

A = \frac{\pi}{4}

B = \frac{x}{2}

Hence

\tan^{-1}(\sec x + \tan x) = \frac{\pi}{4} + \frac{x}{2}

(since

\tan^{-1}(\tan\theta) = \theta

in the principal range).

Now integrate:

\int\left(\frac{\pi}{4} + \frac{x}{2}\right) dx = \frac{\pi}{4}x + \frac{1}{2} \cdot \frac{x^2}{2} + c = \frac{\pi x}{4} + \frac{x^2}{4} + c

Answer:

\frac{\pi x}{4} + \frac{x^2}{4} + c

, which corresponds to option A.

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