Question

\(\displaystyle\int_0^{\frac{{\rm{\pi }}}{4}} \frac{1}{{{{\rm{e}}^{\tan {\rm{x}}}}\left( {1 - {{\sin }^2}{\rm{x}}} \right)}}{\rm{dx}}\) is equal to?
1. \(​-1 - \frac{1}{{\rm{e}}}\)
2. \(​1 + \frac{1}{{\rm{e}}}\)
3. \(​1 - \frac{1}{{\rm{e}}}\)
4. None of the above
5. e

Answer 1

Correct Answer - Option 3 : \(​1 - \frac{1}{{\rm{e}}}\)

Concept:

\(\smallint {{\rm{e}}^{ - {\rm{x}}}}{\rm{dx}} = - {{\rm{e}}^{ - {\rm{x}}}} + {\rm{c}}\)

Calculation:

I = \(\displaystyle\int_0^{\frac{{\rm{\pi }}}{4}} \frac{1}{{{{\rm{e}}^{\tan {\rm{x}}}}\left( {1 - {{\sin }^2}{\rm{x}}} \right)}}{\rm{dx}} \)

\(= {\rm{\;}}\displaystyle\int_0^{\frac{{\rm{\pi }}}{4}} \frac{1}{{{{\rm{e}}^{\tan {\rm{x}}}}{{ \ \cos }^2}{\rm{x}}}}{\rm{dx}} \)

\(= \displaystyle\int_0^{\frac{{\rm{\pi }}}{4}} \frac{{{{\sec }^2}{\rm{x}}}}{{{{\rm{e}}^{\tan {\rm{x}}}}}}{\rm{dx}}\)

Let tan x = t

⇒ sec² x dx = dt

x	0	π/4
t	1	1

 

\({\rm{I}} = \displaystyle\int_0^1 \frac{1}{{{{\rm{e}}^{\rm{t}}}}}{\rm{dt}} \)

\(= \displaystyle\int_0^1 {{\rm{e}}^{ - {\rm{t}}}}{\rm{dt}} \)

\(= \left[ { - {{\rm{e}}^{ - {\rm{t}}}}} \right]_0^1 \)

\(= - \left[ {{{\rm{e}}^{ - 1}} - {{\rm{e}}^0}} \right] \)

\(= 1 - \frac{1}{{\rm{e}}}\)