Question

Given that $\int_{0}^{1}\tan^{-1}(t)\,dt = \frac{\pi}{4} - \frac{1}{2}\log 2$, then $\int_{0}^{1}\tan^{-1}(1-t)\,dt =$

• A. $\frac{\pi}{2}-\frac{1}{2}\log 2$
• B. $\frac{\pi}{4}-\frac{1}{2}\log 3$
• C. $\frac{\pi}{4}+\frac{1}{2}\log 2$
• D. $\frac{\pi}{4}+\frac{1}{2}\log 2$
• E. $\frac{\pi}{4}-\frac{1}{2}\log 2$

Hint:

Property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ is the most frequently tested property of definite integrals.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
We are asked to evaluate a definite integral by using the value of a related integral. This problem hinges on a key property of definite integrals.
Step 2: Key Formula or Approach:
We will use the King's property of definite integrals, which states:
\[ \int_0^a f(x) dx = \int_0^a f(a-x) dx \] Let's apply this property to the integral we want to find.
Step 3: Detailed Explanation:
Let the integral we want to find be I.
\[ I = \int_0^1 \tan^{-1}(1-t) dt \] Here, the function is \( f(t) = \tan^{-1}(1-t) \) and the upper limit is \( a=1 \).
According to the property, this integral is equal to:
\[ I = \int_0^1 f(1-t) dt \] Let's find \( f(1-t) \). We replace 't' with '(1-t)' in the function \( f(t) \):
\[ f(1-t) = \tan^{-1}(1 - (1-t)) = \tan^{-1}(1 - 1 + t) = \tan^{-1}(t) \] So, the property tells us that:
\[ \int_0^1 \tan^{-1}(1-t) dt = \int_0^1 \tan^{-1}(t) dt \] We are given the value of the integral on the right-hand side.
\[ \int_0^1 \tan^{-1}(t) dt = \frac{\pi}{4} - \frac{1}{2}\log 2 \] Therefore, the integral we are looking for has the same value.
Step 4: Final Answer:
The value of \( \int_0^1 \tan^{-1}(1-t) dt \) is \( \frac{\pi}{4} - \frac{1}{2}\log 2 \).