Key Concepts and Formulas

• Trigonometric Identities: We will use the half-angle identities for sine and cosine:
  • $\sin x = 2 \sin(x/2) \cos(x/2)$
  • $1 + \cos x = 2 \cos^2(x/2)$
• Integral of $\sec^2(u)$: The integral of $\sec^2 u$ with respect to $u$ is $\tan u + C$.
• Integral of $\csc u$: The integral of $\csc u$ with respect to $u$ is $-\log_e |\csc u + \cot u| + C$, or equivalently $\log_e |\csc u - \cot u| + C$.
• Integral of $\cot u$: The integral of $\cot u$ with respect to $u$ is $\log_e |\sin u| + C$.
• Substitution Rule: If $u = g(x)$, then $du = g'(x) dx$. This allows us to transform integrals into simpler forms.
• Definite Integration: The definite integral $\int_a^b f(x) dx$ is evaluated by finding an antiderivative $F(x)$ of $f(x)$ and computing $F(b) - F(a)$.

Step-by-Step Solution

The integral we need to evaluate is: I = \int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x

Step 1: Rewrite the integrand using half-angle formulas. To simplify the denominator, we will use the half-angle identities: $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1 + \cos x = 2 \cos^2(x/2)$. The integrand becomes: $\frac{(2+3 \sin x)}{\sin x(1+\cos x)} = \frac{2+3(2 \sin(x/2) \cos(x/2))}{(2 \sin(x/2) \cos(x/2))(2 \cos^2(x/2))}$ $= \frac{2+6 \sin(x/2) \cos(x/2)}{4 \sin(x/2) \cos^3(x/2)}$ Now, we can split this into two terms: $= \frac{2}{4 \sin(x/2) \cos^3(x/2)} + \frac{6 \sin(x/2) \cos(x/2)}{4 \sin(x/2) \cos^3(x/2)}$ $= \frac{1}{2 \sin(x/2) \cos^3(x/2)} + \frac{3}{2 \cos^2(x/2)}$ We can further simplify these terms: $= \frac{1}{2} \left(\frac{1}{\sin(x/2)\cos(x/2)}\right) \left(\frac{1}{\cos^2(x/2)}\right) + \frac{3}{2} \sec^2(x/2)$ Using $\sin x = 2 \sin(x/2) \cos(x/2)$, we have $\frac{1}{\sin(x/2)\cos(x/2)} = \frac{2}{\sin x} = 2 \csc x$. So the first term becomes: $\frac{1}{2} (2 \csc x) \sec^2(x/2) = \csc x \sec^2(x/2)$ This approach seems to be making the expression more complicated. Let's try another simplification of the original integrand.

Step 1 (Revised): Split the integrand into simpler fractions. We can rewrite the integrand as: $\frac{(2+3 \sin x)}{\sin x(1+\cos x)} = \frac{2}{\sin x(1+\cos x)} + \frac{3 \sin x}{\sin x(1+\cos x)}$ $= \frac{2}{\sin x(1+\cos x)} + \frac{3}{1+\cos x}$ Now we have two separate integrals to evaluate.

Step 2: Simplify the first term. Consider the first term: $\frac{2}{\sin x(1+\cos x)}$. We can use the half-angle identities here as well: $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1 + \cos x = 2 \cos^2(x/2)$. $\frac{2}{(2 \sin(x/2) \cos(x/2))(2 \cos^2(x/2))} = \frac{2}{4 \sin(x/2) \cos^3(x/2)} = \frac{1}{2} \frac{1}{\sin(x/2)\cos^3(x/2)}$ To make this easier to integrate, we can multiply the numerator and denominator by $\cos(x/2)$: $\frac{1}{2} \frac{\cos(x/2)}{\sin(x/2)\cos^4(x/2)} = \frac{1}{2} \frac{\cos(x/2)}{\sin(x/2)} \sec^4(x/2) = \frac{1}{2} \cot(x/2) \sec^4(x/2)$ This is still not straightforward. Let's try a different approach for the first term.

Step 1 (Further Revised): Decompose the integrand using partial fractions or by splitting the numerator differently. Let's split the original integrand as follows: $\frac{(2+3 \sin x)}{\sin x(1+\cos x)} = \frac{2}{\sin x(1+\cos x)} + \frac{3}{1+\cos x}$ We know that $\int \frac{3}{1+\cos x} dx = \int \frac{3}{2 \cos^2(x/2)} dx = \int \frac{3}{2} \sec^2(x/2) dx$. Let $u = x/2$, so $du = \frac{1}{2} dx$, which means $dx = 2 du$. $\int \frac{3}{2} \sec^2(x/2) (2 du) = \int 3 \sec^2 u du = 3 \tan u + C = 3 \tan(x/2) + C$ Now consider the first term: $\frac{2}{\sin x(1+\cos x)}$. We can rewrite $\sin x = \frac{2 \tan(x/2)}{1+\tan^2(x/2)}$ and $1+\cos x = \frac{2}{1+\tan^2(x/2)}$. $\frac{2}{\frac{2 \tan(x/2)}{1+\tan^2(x/2)} \cdot \frac{2}{1+\tan^2(x/2)}} = \frac{2(1+\tan^2(x/2))^2}{4 \tan(x/2)} = \frac{(1+\tan^2(x/2))^2}{2 \tan(x/2)}$ This is also complicated.

Let's go back to the initial split: $\frac{(2+3 \sin x)}{\sin x(1+\cos x)} = \frac{2}{\sin x(1+\cos x)} + \frac{3}{1+\cos x}$ Consider the term $\frac{2}{\sin x(1+\cos x)}$. We can write $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1+\cos x = 2 \cos^2(x/2)$. $\frac{2}{2 \sin(x/2) \cos(x/2) \cdot 2 \cos^2(x/2)} = \frac{1}{2 \sin(x/2) \cos^3(x/2)}$ This is not simplifying well. Let's try a direct substitution for the entire integrand.

Step 1 (Alternative Approach): Use substitution $t = \tan(x/2)$. If $t = \tan(x/2)$, then $dx = \frac{2 dt}{1+t^2}$. Also, $\sin x = \frac{2t}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$. The integrand becomes: $\frac{(2+3 \sin x)}{\sin x(1+\cos x)} = \frac{2+3 \left(\frac{2t}{1+t^2}\right)}{\left(\frac{2t}{1+t^2}\right) \left(1+\frac{1-t^2}{1+t^2}\right)}$ $= \frac{\frac{2(1+t^2)+6t}{1+t^2}}{\left(\frac{2t}{1+t^2}\right) \left(\frac{1+t^2+1-t^2}{1+t^2}\right)} = \frac{2+2t^2+6t}{1+t^2} \cdot \frac{(1+t^2)^2}{2t(2)}$ $= \frac{2(1+t^2+3t)}{1+t^2} \cdot \frac{(1+t^2)^2}{4t} = \frac{(1+t^2+3t)(1+t^2)}{2t}$ $= \frac{1+t^2+3t+t^2+t^4+3t^3}{2t} = \frac{t^4+3t^3+2t^2+3t+1}{2t}$ $= \frac{1}{2} \left( t^3 + 3t^2 + 2t + 3 + \frac{1}{t} \right)$ Now we need to integrate this with respect to $t$. The limits of integration change: When $x = \frac{\pi}{3}$, $t = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. When $x = \frac{\pi}{2}$, $t = \tan(\frac{\pi}{4}) = 1$. So the integral becomes: $I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{1}{2} \left( t^3 + 3t^2 + 2t + 3 + \frac{1}{t} \right) \frac{2 dt}{1+t^2}$ $I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t^3 + 3t^2 + 2t + 3 + \frac{1}{t}}{1+t^2} dt = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t^4+3t^3+2t^2+3t+1}{t(1+t^2)} dt$ This is still complicated. Let's go back to the split form.

Step 2 (Back to Split Form): Integrate the second term. We found that $\int \frac{3}{1+\cos x} dx = 3 \tan(x/2)$. Let's evaluate this from $\frac{\pi}{3}$ to $\frac{\pi}{2}$: $\left[ 3 \tan(x/2) \right]_{\frac{\pi}{3}}^{\frac{\pi}{2}} = 3 \tan(\frac{\pi}{4}) - 3 \tan(\frac{\pi}{6}) = 3(1) - 3\left(\frac{1}{\sqrt{3}}\right) = 3 - \sqrt{3}$

Step 3: Integrate the first term. Consider the term $\frac{2}{\sin x(1+\cos x)}$. We can use the identity $\sin x = \frac{2 \tan(x/2)}{1+\tan^2(x/2)}$ and $1+\cos x = \frac{2}{1+\tan^2(x/2)}$. $\frac{2}{\sin x(1+\cos x)} = \frac{2}{\frac{2 \tan(x/2)}{1+\tan^2(x/2)} \cdot \frac{2}{1+\tan^2(x/2)}} = \frac{2(1+\tan^2(x/2))^2}{4 \tan(x/2)} = \frac{(1+\tan^2(x/2))^2}{2 \tan(x/2)}$ This is not yielding a simple integral.

Let's try to rewrite $\frac{2}{\sin x(1+\cos x)}$ in a different way. $\frac{2}{\sin x(1+\cos x)} = \frac{2}{2 \sin(x/2) \cos(x/2) \cdot 2 \cos^2(x/2)} = \frac{1}{2 \sin(x/2) \cos^3(x/2)}$ Divide numerator and denominator by $\cos^4(x/2)$: $= \frac{1}{2} \frac{\sec^4(x/2)}{\tan(x/2)}$ This is also not easy to integrate directly.

Let's try to manipulate the expression $\frac{2}{\sin x(1+\cos x)}$ to get terms involving $\csc x$ and $\cot x$. $\frac{2}{\sin x(1+\cos x)} = \frac{2}{\sin x + \sin x \cos x}$ This doesn't seem to simplify.

Let's go back to the split: $\frac{(2+3 \sin x)}{\sin x(1+\cos x)} = \frac{2}{\sin x(1+\cos x)} + \frac{3}{1+\cos x}$ We integrated the second part to get $3 \tan(x/2)$.

Consider the first part: $\frac{2}{\sin x(1+\cos x)}$. We can write $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1+\cos x = 2 \cos^2(x/2)$. $\frac{2}{2 \sin(x/2) \cos(x/2) \cdot 2 \cos^2(x/2)} = \frac{1}{2 \sin(x/2) \cos^3(x/2)}$ Let's divide the numerator and denominator by $\sin^4(x/2)$: $\frac{1}{2} \frac{\csc^4(x/2)}{\cot(x/2)}$ This is still complex.

Let's try to decompose the first term using $\sin x$ and $\cos x$: $\frac{2}{\sin x(1+\cos x)} = \frac{A}{\sin x} + \frac{B}{1+\cos x}$ $2 = A(1+\cos x) + B \sin x$ If $\cos x = -1$, then $x = \pi$. $\sin \pi = 0$. So $2 = A(0) + B(0)$, which is impossible. This means this partial fraction decomposition is not directly applicable in this form.

Let's try rewriting the first term using $\csc x$ and $\cot x$: $\frac{2}{\sin x(1+\cos x)} = \frac{2}{\sin x + \sin x \cos x}$ Multiply numerator and denominator by $(1-\cos x)$: $\frac{2(1-\cos x)}{\sin x(1+\cos x)(1-\cos x)} = \frac{2(1-\cos x)}{\sin x(1-\cos^2 x)} = \frac{2(1-\cos x)}{\sin x \sin^2 x} = \frac{2(1-\cos x)}{\sin^3 x}$ $= 2 \left( \frac{1}{\sin^3 x} - \frac{\cos x}{\sin^3 x} \right) = 2 (\csc^3 x - \cot x \csc^2 x)$ This is still hard to integrate.

Let's reconsider the substitution $t = \tan(x/2)$. The integrand was $\frac{1}{2} \left( t^3 + 3t^2 + 2t + 3 + \frac{1}{t} \right)$ and $dx = \frac{2 dt}{1+t^2}$. The original integrand is $\frac{(2+3 \sin x)}{\sin x(1+\cos x)}$. Let's rewrite the numerator: $2+3 \sin x$. Let's rewrite the denominator: $\sin x (1+\cos x) = \sin x + \sin x \cos x$. Let's try to express the integrand in terms of $\tan(x/2)$. We had: $\frac{(2+3 \sin x)}{\sin x(1+\cos x)} = \frac{2+3 \left(\frac{2t}{1+t^2}\right)}{\left(\frac{2t}{1+t^2}\right) \left(1+\frac{1-t^2}{1+t^2}\right)} = \frac{2(1+t^2)+6t}{1+t^2} \cdot \frac{(1+t^2)^2}{2t \cdot 2}$ $= \frac{2(1+3t+t^2)}{1+t^2} \cdot \frac{(1+t^2)^2}{4t} = \frac{(1+3t+t^2)(1+t^2)}{2t}$ $= \frac{1+t^2+3t+3t^3+t^4+t^2}{2t} = \frac{t^4+3t^3+2t^2+3t+1}{2t}$ $= \frac{1}{2} \left( t^3 + 3t^2 + 2t + 3 + \frac{1}{t} \right)$ And $dx = \frac{2 dt}{1+t^2}$. So the integral is: $I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{1}{2} \left( t^3 + 3t^2 + 2t + 3 + \frac{1}{t} \right) \frac{2 dt}{1+t^2}$ $I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t^3 + 3t^2 + 2t + 3 + \frac{1}{t}}{1+t^2} dt$ Let's perform polynomial division or algebraic manipulation on the numerator: $t^3 + 3t^2 + 2t + 3 + \frac{1}{t} = \frac{t^4 + 3t^3 + 2t^2 + 3t + 1}{t}$ So the integral is: $I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t^4 + 3t^3 + 2t^2 + 3t + 1}{t(1+t^2)} dt$ This is still a complex rational function.

Let's go back to the split form and try to integrate the first term $\frac{2}{\sin x(1+\cos x)}$ using a different approach. $\frac{2}{\sin x(1+\cos x)} = \frac{2}{2 \sin(x/2) \cos(x/2) \cdot 2 \cos^2(x/2)} = \frac{1}{2 \sin(x/2) \cos^3(x/2)}$ Multiply numerator and denominator by $\sec^4(x/2)$: $\frac{1}{2} \frac{\sec^4(x/2)}{\tan(x/2)}$ Let $u = \tan(x/2)$, then $du = \frac{1}{2} \sec^2(x/2) dx$, so $dx = \frac{2 du}{\sec^2(x/2)} = \frac{2 du}{1+\tan^2(x/2)} = \frac{2 du}{1+u^2}$. The expression becomes: $\frac{1}{2} \frac{(\sec^2(x/2))^2}{\tan(x/2)} = \frac{1}{2} \frac{(1+u^2)^2}{u}$ The integral of this with respect to $x$ is: $\int \frac{1}{2} \frac{(1+u^2)^2}{u} \frac{2 du}{1+u^2} = \int \frac{(1+u^2)^2}{u(1+u^2)} du = \int \frac{1+u^2}{u} du$ $= \int \left( \frac{1}{u} + u \right) du = \log_e|u| + \frac{u^2}{2} + C$ Substituting back $u = \tan(x/2)$: $\log_e|\tan(x/2)| + \frac{\tan^2(x/2)}{2} + C$

So, the integral of $\frac{2}{\sin x(1+\cos x)}$ is $\log_e|\tan(x/2)| + \frac{\tan^2(x/2)}{2}$. And the integral of $\frac{3}{1+\cos x}$ is $3 \tan(x/2)$.

Therefore, the integrand can be integrated as: $\int \frac{(2+3 \sin x)}{\sin x(1+\cos x)} dx = \log_e|\tan(x/2)| + \frac{\tan^2(x/2)}{2} + 3 \tan(x/2) + C$

Now we need to evaluate this from $x = \frac{\pi}{3}$ to $x = \frac{\pi}{2}$. Let $F(x) = \log_e|\tan(x/2)| + \frac{\tan^2(x/2)}{2} + 3 \tan(x/2)$.

Evaluate at the upper limit $x = \frac{\pi}{2}$: $x/2 = \frac{\pi}{4}$. $\tan(\frac{\pi}{4}) = 1$. $F(\frac{\pi}{2}) = \log_e|1| + \frac{1^2}{2} + 3(1) = 0 + \frac{1}{2} + 3 = \frac{7}{2}$

Evaluate at the lower limit $x = \frac{\pi}{3}$: $x/2 = \frac{\pi}{6}$. $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. $F(\frac{\pi}{3}) = \log_e\left|\frac{1}{\sqrt{3}}\right| + \frac{\left(\frac{1}{\sqrt{3}}\right)^2}{2} + 3\left(\frac{1}{\sqrt{3}}\right)$ $= \log_e(3^{-1/2}) + \frac{\frac{1}{3}}{2} + \frac{3}{\sqrt{3}} = -\frac{1}{2} \log_e 3 + \frac{1}{6} + \sqrt{3}$ Note that $\log_e \sqrt{3} = \log_e 3^{1/2} = \frac{1}{2} \log_e 3$. So $-\frac{1}{2} \log_e 3 = -\log_e \sqrt{3}$. $F(\frac{\pi}{3}) = -\log_e \sqrt{3} + \frac{1}{6} + \sqrt{3}$

Now, subtract $F(\frac{\pi}{3})$ from $F(\frac{\pi}{2})$: $I = F(\frac{\pi}{2}) - F(\frac{\pi}{3}) = \frac{7}{2} - \left( -\log_e \sqrt{3} + \frac{1}{6} + \sqrt{3} \right)$ $I = \frac{7}{2} + \log_e \sqrt{3} - \frac{1}{6} - \sqrt{3}$ Combine the constant terms: $\frac{7}{2} - \frac{1}{6} = \frac{21}{6} - \frac{1}{6} = \frac{20}{6} = \frac{10}{3}$ So, the integral is: $I = \frac{10}{3} - \sqrt{3} + \log_e \sqrt{3}$

This matches option (A).

Step 4: Verification of the integration of the first term. Let's re-check the integration of $\frac{2}{\sin x(1+\cos x)}$. We used the substitution $u = \tan(x/2)$. The expression was $\frac{1}{2 \sin(x/2) \cos^3(x/2)}$. With $u = \tan(x/2)$, we have $\sin(x/2) = \frac{u}{\sqrt{1+u^2}}$ and $\cos(x/2) = \frac{1}{\sqrt{1+u^2}}$. $\frac{1}{2 \left(\frac{u}{\sqrt{1+u^2}}\right) \left(\frac{1}{\sqrt{1+u^2}}\right)^3} = \frac{1}{2} \frac{(1+u^2)^{3/2}}{u} \frac{1}{(1+u^2)^{3/2}} = \frac{1}{2u}$ This is incorrect.

Let's use the substitution $t = \tan(x/2)$ directly on the entire integrand. We found the integrand in terms of $t$ to be $\frac{1}{2} \left( t^3 + 3t^2 + 2t + 3 + \frac{1}{t} \right)$. And $dx = \frac{2 dt}{1+t^2}$. So the integral is: $I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{1}{2} \left( t^3 + 3t^2 + 2t + 3 + \frac{1}{t} \right) \frac{2 dt}{1+t^2}$ $I = \int_{\frac{1}{\sqrt{3}}}^{1} \frac{t^3 + 3t^2 + 2t + 3 + \frac{1}{t}}{1+t^2} dt$ Let's try to split the numerator in a way that relates to $1+t^2$. $\frac{t^3 + 3t^2 + 2t + 3 + \frac{1}{t}}{1+t^2} = \frac{t^4 + 3t^3 + 2t^2 + 3t + 1}{t(1+t^2)}$ We can perform partial fraction decomposition on $\frac{t^4 + 3t^3 + 2t^2 + 3t + 1}{t(1+t^2)}$. However, the degree of the numerator is higher than the denominator. This indicates an error in the transformation or the initial split.

Let's return to the split form: $\frac{2}{\sin x(1+\cos x)} + \frac{3}{1+\cos x}$. We integrated $\frac{3}{1+\cos x}$ correctly as $3 \tan(x/2)$.

Consider $\frac{2}{\sin x(1+\cos x)}$. We can write $\sin x = 2 \sin(x/2) \cos(x/2)$ and $1+\cos x = 2 \cos^2(x/2)$. $\frac{2}{2 \sin(x/2) \cos(x/2) \cdot 2 \cos^2(x/2)} = \frac{1}{2 \sin(x/2) \cos^3(x/2)}$ Let's divide numerator and denominator by $\cos^4(x/2)$: $\frac{1}{2} \frac{\sec^4(x/2)}{\tan(x/2)}$ Let $u = \tan(x/2)$. Then $du = \frac{1}{2} \sec^2(x/2) dx$. $\int \frac{1}{2} \frac{\sec^4(x/2)}{\tan(x/2)} dx = \int \frac{1}{2} \frac{(\sec^2(x/2))^2}{\tan(x/2)} dx$ $= \int \frac{1}{2} \frac{(1+\tan^2(x/2))^2}{\tan(x/2)} dx$ Let $t = \tan(x/2)$, so $dt = \frac{1}{2} \sec^2(x/2) dx$. This substitution is still problematic for the entire expression.

Let's try to express $\frac{1}{2 \sin(x/2) \cos^3(x/2)}$ in terms of $\csc x$ and $\cot x$. $\frac{1}{2 \sin(x/2) \cos^3(x/2)} = \frac{1}{\sin x \cos^2(x/2)}$ Using $\cos^2(x/2) = \frac{1+\cos x}{2}$: $\frac{1}{\sin x \left(\frac{1+\cos x}{2}\right)} = \frac{2}{\sin x (1+\cos x)}$ This brings us back to the start.

Let's use the identity: $\frac{1}{\sin x(1+\cos x)} = \frac{1}{2 \sin(x/2) \cos^3(x/2)}$ Consider integrating this term. Let $t = \tan(x/2)$. Then $\sin(x/2) = \frac{t}{\sqrt{1+t^2}}$, $\cos(x/2) = \frac{1}{\sqrt{1+t^2}}$. $\frac{1}{2 \frac{t}{\sqrt{1+t^2}} \left(\frac{1}{\sqrt{1+t^2}}\right)^3} = \frac{1}{2} \frac{(1+t^2)^{3/2}}{t} \frac{1}{(1+t^2)^{3/2}} = \frac{1}{2t}$ This is incorrect.

Let's rewrite the integrand as: $\frac{2}{\sin x(1+\cos x)} = \frac{2}{2 \sin(x/2) \cos(x/2) \cdot 2 \cos^2(x/2)} = \frac{1}{2 \sin(x/2) \cos^3(x/2)}$ Divide numerator and denominator by $\cos^4(x/2)$: $\frac{\sec^4(x/2)}{2 \tan(x/2)} = \frac{(1+\tan^2(x/2))^2}{2 \tan(x/2)}$ Let $t = \tan(x/2)$. The expression is $\frac{(1+t^2)^2}{2t}$. The integral of this with respect to $x$ is: $\int \frac{(1+t^2)^2}{2t} \frac{2 dt}{1+t^2} = \int \frac{1+t^2}{t} dt = \int \left( \frac{1}{t} + t \right) dt = \log_e|t| + \frac{t^2}{2} + C$ Substituting back $t = \tan(x/2)$: $\log_e|\tan(x/2)| + \frac{\tan^2(x/2)}{2} + C$ This is the integral of the first term $\frac{2}{\sin x(1+\cos x)}$.

So, the full integral is: $\int \frac{(2+3 \sin x)}{\sin x(1+\cos x)} dx = \left( \log_e|\tan(x/2)| + \frac{\tan^2(x/2)}{2} \right) + 3 \tan(x/2) + C$

Now, evaluate the definite integral: Limits are from $\frac{\pi}{3}$ to $\frac{\pi}{2}$. Upper limit $x = \frac{\pi}{2}$: $x/2 = \frac{\pi}{4}$. $\tan(\frac{\pi}{4}) = 1$. $\log_e(1) + \frac{1^2}{2} + 3(1) = 0 + \frac{1}{2} + 3 = \frac{7}{2}$ Lower limit $x = \frac{\pi}{3}$: $x/2 = \frac{\pi}{6}$. $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$. $\log_e\left(\frac{1}{\sqrt{3}}\right) + \frac{\left(\frac{1}{\sqrt{3}}\right)^2}{2} + 3\left(\frac{1}{\sqrt{3}}\right)$ $= -\frac{1}{2} \log_e 3 + \frac{1}{6} + \sqrt{3} = -\log_e \sqrt{3} + \frac{1}{6} + \sqrt{3}$ Subtracting the lower limit value from the upper limit value: $\frac{7}{2} - \left( -\log_e \sqrt{3} + \frac{1}{6} + \sqrt{3} \right) = \frac{7}{2} + \log_e \sqrt{3} - \frac{1}{6} - \sqrt{3}$ $= \left(\frac{7}{2} - \frac{1}{6}\right) - \sqrt{3} + \log_e \sqrt{3} = \left(\frac{21-1}{6}\right) - \sqrt{3} + \log_e \sqrt{3}$ $= \frac{20}{6} - \sqrt{3} + \log_e \sqrt{3} = \frac{10}{3} - \sqrt{3} + \log_e \sqrt{3}$ This result matches option (A).

Common Mistakes & Tips

• Incorrect Trigonometric Substitution: When using $t = \tan(x/2)$, ensure all trigonometric functions ($\sin x, \cos x$) and the differential $dx$ are correctly expressed in terms of $t$.
• Algebraic Errors: Simplifying complex rational functions can be prone to algebraic mistakes. Double-check each step, especially when expanding and combining terms.
• Logarithm Properties: Remember that $\log_e(a^b) = b \log_e a$ and $\log_e(1/a) = -\log_e a$. This is useful for simplifying terms like $\log_e(1/\sqrt{3})$.
• Limits of Integration: When performing a substitution, remember to change the limits of integration to be in terms of the new variable.

Summary

The problem requires evaluating a definite integral involving trigonometric functions. The strategy employed was to simplify the integrand by splitting it into simpler terms and using the substitution $t = \tan(x/2)$. This substitution transforms the trigonometric integral into an integral of a rational function, which can then be integrated. After finding the antiderivative, the definite integral was evaluated by applying the limits of integration, leading to the final answer.

The final answer is $\boxed{\frac{10}{3}-\sqrt{3}+\log _{e} \sqrt{3}}$.