Question:

Step 1: Simplify the integrand using trigonometric identities. We need to simplify $\dfrac{2+\sin 2x}{1+\cos 2x}$. Using the double angle formulas: - $\sin 2x = 2\sin x\cos x$ - $\cos 2x = 2\cos^2 x - 1$, which gives $1 + \cos 2x = 2\cos^2 x$ $$\frac{2+\sin 2x}{1+\cos 2x} = \frac{2 + 2\sin x\cos x}{2\cos^2 x} = \frac{2(1 + \sin x\cos x)}{2\cos^2 x} = \frac{1}{\cos^2 x} + \frac{\sin x}{\cos x}$$ Therefore: $$\frac{2+\sin 2x}{1+\cos 2x} = \sec^2 x + \tan x$$ Step 2: Rewrite the integral with the simplified integrand. $$I = \int_{\pi/4}^{\pi/3} e^x(\sec^2 x + \tan x)\, dx$$ Step 3: Apply the standard integration formula for products of exponential and derivatives. We recognize that this integral has the form $\int e^x(f(x) + f'(x))\, dx$ where $f(x) = \tan x$ and $f'(x) = \sec^2 x$. Using the formula $\int e^x(f(x) + f'(x))\, dx = e^x f(x) + C$: $$I = [e^x \tan x]_{\pi/4}^{\pi/3}$$ Step 4: Evaluate the definite integral at the bounds. $$I = e^{\pi/3}\tan\left(\frac{\pi}{3}\right) - e^{\pi/4}\tan\left(\frac{\pi}{4}\right)$$ Substituting the values $\tan(\pi/3) = \sqrt{3}$ and $\tan(\pi/4) = 1$: $$I = \sqrt{3}\, e^{\pi/3} - e^{\pi/4}$$ Step 5: Factor out the common exponential term. Factor out $e^{\pi/4}$: $$I = e^{\pi/4}\left(\sqrt{3}\, e^{\pi/3 - \pi/4} - 1\right)$$ Computing the exponent: $\dfrac{\pi}{3} - \dfrac{\pi}{4} = \dfrac{4\pi - 3\pi}{12} = \dfrac{\pi}{12}$ $$I = e^{\pi/4}\left(\sqrt{3}\, e^{\pi/12} - 1\right)$$ Step 6: Compare with the given form and identify parameters. The integral is expressed as $e^{\pi/a}\left(b\, e^{\pi/c} - 1\right)$. Comparing: - $a = 4$ - $b = \sqrt{3}$ - $c = 12$ Step 7: Calculate the final answer. $$\frac{b^2 c}{a} = \frac{(\sqrt{3})^2 \cdot 12}{4} = \frac{3 \cdot 12}{4} = \frac{36}{4} = 9$$ The value of $\dfrac{b^2 c}{a}$ is $\boxed{9}$, which corresponds to **Option 3**.