Key Concepts and Formulas

• Half-Angle Tangent Substitution (Weierstrass Substitution): For integrals involving rational functions of $\sin x$ and $\cos x$, the substitution $t = \tan(x/2)$ is highly effective. It transforms trigonometric functions as follows:
  • $\sin x = \frac{2t}{1+t^2}$
  • $\cos x = \frac{1-t^2}{1+t^2}$
  • $dx = \frac{2 dt}{1+t^2}$
• Definite Integral Limit Transformation: When performing a substitution in a definite integral, the limits of integration must be changed according to the substitution variable.
• Integration of Rational Functions: Integrals of the form $\int \frac{P(t)}{Q(t)} dt$, where $P(t)$ and $Q(t)$ are polynomials, can often be solved by partial fraction decomposition or by completing the square in the denominator, leading to standard integral forms like $\int \frac{1}{a^2+t^2} dt = \frac{1}{a} \tan^{-1}(\frac{t}{a}) + C$.

Step-by-Step Solution

Step 1: Identify the Integral Type and Apply the Half-Angle Substitution

The given integral is $I = \int\limits_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} \mathrm{~d} x$. The integrand is a rational function of $\sin x$ and $\cos x$. This suggests using the half-angle tangent substitution, $t = \tan(x/2)$.

Step 2: Transform the Trigonometric Functions and Differential Element

We apply the standard substitutions:

• $\sin x = \frac{2t}{1+t^2}$
• $\cos x = \frac{1-t^2}{1+t^2}$
• $dx = \frac{2 dt}{1+t^2}$

Substitute these into the integrand: $\frac{1}{3+2 \sin x+\cos x} = \frac{1}{3 + 2\left(\frac{2t}{1+t^2}\right) + \left(\frac{1-t^2}{1+t^2}\right)}$ Combine the terms in the denominator: $= \frac{1}{\frac{3(1+t^2) + 4t + (1-t^2)}{1+t^2}} = \frac{1+t^2}{3+3t^2+4t+1-t^2} = \frac{1+t^2}{2t^2+4t+4}$ $= \frac{1+t^2}{2(t^2+2t+2)}$ Now, substitute $dx$: $dx = \frac{2 dt}{1+t^2}$ So, the integral becomes: $I = \int \frac{1+t^2}{2(t^2+2t+2)} \cdot \frac{2 dt}{1+t^2}$ The $(1+t^2)$ terms cancel out: $I = \int \frac{1}{t^2+2t+2} dt$

Step 3: Transform the Limits of Integration

We need to change the limits of integration from $x$ to $t$.

• When $x = 0$, $t = \tan(0/2) = \tan(0) = 0$.
• When $x = \frac{\pi}{2}$, $t = \tan((\pi/2)/2) = \tan(\pi/4) = 1$.

So, the definite integral in terms of $t$ is: $I = \int_{0}^{1} \frac{1}{t^2+2t+2} dt$

Step 4: Integrate the Rational Function

The denominator $t^2+2t+2$ is a quadratic. We can complete the square to simplify the integration. $t^2+2t+2 = (t^2+2t+1) + 1 = (t+1)^2 + 1$ The integral becomes: $I = \int_{0}^{1} \frac{1}{(t+1)^2 + 1} dt$ This integral is in the form $\int \frac{1}{u^2+a^2} du = \frac{1}{a} \tan^{-1}(\frac{u}{a}) + C$, where $u = t+1$ and $a = 1$.

Let $u = t+1$. Then $du = dt$. When $t=0$, $u = 0+1 = 1$. When $t=1$, $u = 1+1 = 2$. The integral in terms of $u$ is: $I = \int_{1}^{2} \frac{1}{u^2 + 1^2} du$

Step 5: Evaluate the Definite Integral

Now, apply the integration formula: $I = \left[ \frac{1}{1} \tan^{-1}\left(\frac{u}{1}\right) \right]_{1}^{2} = \left[ \tan^{-1}(u) \right]_{1}^{2}$ Evaluate at the limits: $I = \tan^{-1}(2) - \tan^{-1}(1)$ We know that $\tan^{-1}(1) = \frac{\pi}{4}$. $I = \tan^{-1}(2) - \frac{\pi}{4}$

Common Mistakes & Tips

• Incorrect Limit Transformation: Forgetting to change the limits of integration after applying a substitution is a common error. Always re-evaluate the limits based on the new variable.
• Algebraic Errors: Simplifying the integrand after substitution can lead to algebraic mistakes. Double-check all arithmetic and algebraic manipulations carefully.
• Forgetting the $dx$ term: Ensure that the $dx$ is also substituted correctly as $\frac{2 dt}{1+t^2}$ when using the half-angle tangent substitution.

Summary

The integral was solved using the half-angle tangent substitution ($t = \tan(x/2)$), which transformed the trigonometric integral into an integral of a rational function. After applying the substitution, the limits of integration were updated from $0$ to $\pi/2$ for $x$, to $0$ to $1$ for $t$. The resulting rational function was integrated by completing the square in the denominator, leading to a standard arctangent form. Evaluating the definite integral at the transformed limits yielded the final answer.

The final answer is $\boxed{\tan ^{-1}(2)-\frac{\pi}{4}}$ which corresponds to option (B).