<p>The value of definite integral \(\displaystyle\int_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{\cos^{-1}\!\left(\dfrac{2x}{1+x^2}\right) + \tan^{-1}\!\left(\dfrac{2x}{1-x^2}\right)}{e^x + 1}\,dx\) is equal to:</p>
<p>\(\dfrac{\pi}{2\sqrt{3}}\)</p>
<p>\(\dfrac{\pi}{\sqrt{3}}\)</p>
<p>\(\dfrac{\pi}{4\sqrt{3}}\)</p>
<p>\(\dfrac{\pi}{3\sqrt{3}}\)</p>
Step-by-Step Solution
Key Concept: Use the property that for an integral of the form ∫[−a to a] f(x)/(e^x+1) dx, we can apply the technique f(x)/(e^x+1) + f(−x)/(e^(−x)+1) = f(x), which simplifies the integrand by canceling exponential terms. Additionally, recognize the inverse trigonometric substitutions: cos⁻¹(2x/(1+x²)) = 2tan⁻¹(x) and tan⁻¹(2x/(1−x²)) = 2tan⁻¹(x) for appropriate domains.
Step 1: Simplify the inverse trigonometric functions
For $x \in \left[-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right]$, which is a subset of $(-1, 1)$, we can use the substitution $x = \tan\theta$.
Then $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$.
The first term is $\cos^{-1}\left(\frac{2x}{1+x^2}\right)$.
Substituting $x=\tan\theta$:
$$ \cos^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right) = \cos^{-1}(\sin(2\theta)) $$
Using the identity $\sin(2\theta) = \cos\left(\frac{\pi}{2}-2\theta\right)$:
$$ \cos^{-1}\left(\cos\left(\frac{\pi}{2}-2\theta\right)\right) $$
Since $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$, we have $2\theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$.
Therefore, $\frac{\pi}{2}-2\theta \in \left[\frac{\pi}{2}-\frac{\pi}{3}, \frac{\pi}{2}+\frac{\pi}{3}\right] = \left[\frac{\pi}{6}, \frac{5\pi}{6}\right]$.
In this interval, $\cos^{-1}(\cos(\phi)) = \phi$.
So, $\cos^{-1}\left(\frac{2x}{1+x^2}\right) = \frac{\pi}{2}-2\theta = \frac{\pi}{2}-2\tan^{-1}(x)$.
The second term is $\tan^{-1}\left(\frac{2x}{1-x^2}\right)$.
Substituting $x=\tan\theta$:
$$ \tan^{-1}\left(\frac{2\tan\theta}{1-\tan^2\theta}\right) = \tan^{-1}(\tan(2\theta)) $$
Since $\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right]$, we have $2\theta \in \left[-\frac{\pi}{3}, \frac{\pi}{3}\right]$.
In this interval, $\tan^{-1}(\tan(\phi)) = \phi$.
So, $\tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\theta = 2\tan^{-1}(x)$.
Step 2: Combine the numerator
Adding the simplified terms:
$$ \left(\frac{\pi}{2}-2\tan^{-1}(x)\right) + 2\tan^{-1}(x) = \frac{\pi}{2} $$
Step 3: Rewrite the integral
The integral becomes:
$$ \int_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{\frac{\pi}{2}}{e^x + 1}\,dx = \frac{\pi}{2} \int_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{1}{e^x + 1}\,dx $$
Step 4: Evaluate the definite integral $\int_{-a}^{a} \frac{1}{e^x+1} dx$
Let $J = \int_{-a}^{a} \frac{1}{e^x+1} dx$.
Using the property $\int_A^B f(x) dx = \int_A^B f(A+B-x) dx$, we have:
$$ J = \int_{-a}^{a} \frac{1}{e^{(-a+a-x)}+1} dx = \int_{-a}^{a} \frac{1}{e^{-x}+1} dx $$
Multiply the numerator and denominator by $e^x$:
$$ J = \int_{-a}^{a} \frac{e^x}{1+e^x} dx $$
Adding the two expressions for $J$:
$$ 2J = \int_{-a}^{a} \left( \frac{1}{e^x+1} + \frac{e^x}{e^x+1} \right) dx = \int_{-a}^{a} \frac{1+e^x}{e^x+1} dx = \int_{-a}^{a} 1\,dx $$
$$ 2J = [x]_{-a}^{a} = a - (-a) = 2a $$
Therefore, $J = a$.
Step 5: Apply with $a = \frac{1}{\sqrt{3}}$
Using the result from Step 4, with $a = \frac{1}{\sqrt{3}}$:
$$ \int_{\frac{-1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{1}{e^x + 1}\,dx = \frac{1}{\sqrt{3}} $$
Step 6: Calculate the final answer
Substitute this back into the expression from Step 3:
$$ \frac{\pi}{2} \times \frac{1}{\sqrt{3}} = \frac{\pi}{2\sqrt{3}} $$
Correct Answer: C