Question

The value of \( \int_1^{\sqrt{3}} \frac{dx}{1 + x^2} \) will be :

• A. \( \frac{\pi}{3} \)
• B. \( \frac{2\pi}{3} \)
• C. \( \frac{\pi}{6} \)
• D. \( \frac{\pi}{12} \)

Hint:

Memorizing standard integral forms, especially those involving trigonometric and inverse trigonometric functions, is crucial for speed and accuracy. For definite integrals, be careful with the arithmetic when finding a common denominator.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a definite integral problem. We need to find the antiderivative of the integrand and then evaluate it at the upper and lower limits of integration, according to the Fundamental Theorem of Calculus.
Step 2: Key Formula or Approach:
The standard integral formula required is:
\[ \int \frac{1}{1 + x^2} dx = \tan^{-1}(x) + C \] The Fundamental Theorem of Calculus states:
\[ \int_a^b f(x) dx = F(b) - F(a) \], where \( F'(x) = f(x) \).
Step 3: Detailed Explanation or Calculation:
First, find the antiderivative of the integrand:
\[ \int \frac{dx}{1 + x^2} = \tan^{-1}(x) \] Now, apply the limits of integration from 1 to \( \sqrt{3} \):
\[ \int_1^{\sqrt{3}} \frac{dx}{1 + x^2} = [\tan^{-1}(x)]_1^{\sqrt{3}} \] \[ = \tan^{-1}(\sqrt{3}) - \tan^{-1}(1) \] We need to find the principal values for these inverse trigonometric functions.
- The angle whose tangent is \( \sqrt{3} \) is \( \frac{\pi}{3} \). So, \( \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \).
- The angle whose tangent is 1 is \( \frac{\pi}{4} \). So, \( \tan^{-1}(1) = \frac{\pi}{4} \).
Now, perform the subtraction:
\[ \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \] Step 4: Final Answer:
The value of the definite integral is \( \frac{\pi}{12} \).