Question

The value of integral \(\mathop \smallint \nolimits_0^1 \frac{{dx}}{{{e^x} + {e^{ - x}}}}\) is

1. cot-1 e
2. tan^-1 e
3. \( {\tan ^{ - 1}}{e^{ }}- \frac{\pi }{4}\)
4. \( {\cot ^{ - 1}}{e^{ }}- \frac{\pi }{4}\)

Answer 1

Correct Answer - Option 3 : \( {\tan ^{ - 1}}{e^{ }}- \frac{\pi }{4}\)

Concept:

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

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

Calculation:

\(I = \mathop \smallint \nolimits_0^1 \frac{{dx}}{{{e^x} + {e^{ - x}}}}\)

This can be written as:

\(I = \mathop \smallint \nolimits_0^1 \frac{{{e^x}\;dx}}{{{e^{2x}} + 1}}\)

Let e^x = t

Differentiating both the sides, we get:

e^x dx = 1. dt

\(I = \mathop \smallint \nolimits_1^e \frac{{dt}}{{{t^2} + 1}} \)

\(= \left[ {{{\tan }^{ - 1}}t} \right]_1^e = {\tan ^{ - 1}}{e^{ }}- \frac{\pi }{4}\)