Question

If \(\int_{0}^{1} \frac{dx}{e^x + e^{-x}} = \tan^{-1} e + k\), then the value of \(k\) is :

• A. \(e\)
• B. \(\frac{\pi}{4}\)
• C. \(0\)
• D. \(-\frac{\pi}{4}\)

Hint:

Whenever you see \(e^x\) and \(e^{-x}\) in a denominator, multiplying by \(e^x\) is almost always the correct first step.
Don't forget to change the limits of integration when using substitution in definite integrals.
Familiarize yourself with the values of inverse trigonometric functions for standard values like 1, 0, and \(\sqrt{3}\).

Answer 1

The Correct Option is D

Concept:
• Definite integral evaluation involves finding the antiderivative and applying limits.
• Simplify expressions with \(e^x\) and \(e^{-x}\) by multiplying the numerator and denominator by \(e^x\).
• Substitution method: Let \(e^x = t\).

Step 1: Simplify the integrand
Multiply the numerator and denominator by \(e^x\):
\[ I = \int_{0}^{1} \frac{e^x}{e^x(e^x + e^{-x})} dx = \int_{0}^{1} \frac{e^x}{e^{2x} + 1} dx \]

Step 2: Apply substitution and change limits
Let \(e^x = t \implies e^x dx = dt\).
Changing the limits of integration:
When \(x = 0\), \(t = e^0 = 1\).
When \(x = 1\), \(t = e^1 = e\).
The integral becomes:
\[ I = \int_{1}^{e} \frac{dt}{t^2 + 1} \]

Step 3: Integrate and evaluate the definite integral
We know that \(\int \frac{dt}{1 + t^2} = \tan^{-1} t\).
\[ I = [\tan^{-1} t]_{1}^{e} \] \[ I = \tan^{-1} e - \tan^{-1} 1 \] We know \(\tan^{-1} 1 = \frac{\pi}{4}\):
\[ I = \tan^{-1} e - \frac{\pi}{4} \]

Step 4: Compare with the given expression to find \(k\)
The question states \(I = \tan^{-1} e + k\).
Comparing our result with this:
\[ \tan^{-1} e - \frac{\pi}{4} = \tan^{-1} e + k \] Subtracting \(\tan^{-1} e\) from both sides:
\[ k = -\frac{\pi}{4} \]