Question

$$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\csc x \cdot \cot x}{1 + \csc^2 x} \, dx =$$

• A. $\frac{\pi}{4} - \tan^{-1} 2$
• B. $\tan^{-1} 1$
• C. $\tan^{-1} 2$
• D. $\tan^{-1}\left(\frac{1}{3}\right)$

Hint:

When computing differences like $\tan^{-1}(2) - \tan^{-1}(1)$, you can quickly view it geometrically or use the identity to avoid missing steps. Remember that $\tan^{-1}(1) = \frac{\pi}{4}$, so the answer can also be written as $\tan^{-1}(2) - \frac{\pi}{4}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem requires us to calculate a definite integral containing trigonometric functions $\csc x$ and $\cot x$ within the integration interval boundaries $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$.

Step 2: Key Formula or Approach:
Notice that the derivative of $\csc x$ is $-\csc x \cdot \cot x$, which is exactly matching the functional structure of our numerator. Therefore, we will solve this definite integral using the method of integration by substitution: $$\text{Let } u = \csc x \implies du = -\csc x \cdot \cot x \, dx$$ We will transform the integrand function and shift the upper and lower limits accordingly.

Step 3: Detailed Explanation:
Let the given definite integral be: $$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \frac{\csc x \cdot \cot x}{1 + \csc^2 x} \, dx$$ Apply the substitution $u = \csc x$, which yields $du = -\csc x \cdot \cot x \, dx \implies \csc x \cdot \cot x \, dx = -du$. Now let's determine our new boundary integration limits in terms of variable $u$: 1. Lower limit: When $x = \frac{\pi}{6}$, $u = \csc\left(\frac{\pi}{6}\right) = 2$. 2. Upper limit: When $x = \frac{\pi}{2}$, $u = \csc\left(\frac{\pi}{2}\right) = 1$. Substitute these expressions into the integral $I$: $$I = \int_{2}^{1} \frac{-du}{1 + u^2}$$ We can eliminate the negative sign by reversing our upper and lower boundary integration limits: $$I = \int_{1}^{2} \frac{du}{1 + u^2}$$ The standard anti-derivative for this form is the inverse tangent function: $$I = \left[ \tan^{-1} u \right]_{1}^{2} = \tan^{-1}(2) - \tan^{-1}(1)$$ Using the trigonometric identity for the difference of two inverse tangent functions $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$: $$I = \tan^{-1}\left(\frac{2 - 1}{1 + (2)(1)}\right) = \tan^{-1}\left(\frac{1}{1 + 2}\right) = \tan^{-1}\left(\frac{1}{3}\right)$$

Step 4: Final Answer:
The value of the definite integral is $\tan^{-1}\left(\frac{1}{3}\right)$, which corresponds to option (D).