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: Concept
Use the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$.

Step 2: Analysis
Let $I = \int_{0}^{1} \tan^{-1}(1-t) dt$. Applying the property where $a=1$: $I = \int_{0}^{1} \tan^{-1}(1 - (1-t)) dt$.

Step 3: Conclusion
$I = \int_{0}^{1} \tan^{-1}(t) dt$. The value is identical to the given integral. Final Answer: (E)