Question

$\int_0^1 \tan^{-1} x \, dx = $ ______.

• A. $\pi/4 - \log 2$
• B. $\pi/4 - \log \sqrt{2}$
• C. $\pi/4 + \log 2$
• D. $\pi/4 + \log \sqrt{2}$

Hint:

Whenever you have to integrate a lone logarithmic ($\ln x$) or inverse trig function ($\sin^{-1} x$, $\tan^{-1} x$), ALWAYS multiply it by 1 and use Integration by Parts!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We must evaluate the definite integral of an isolated inverse trigonometric function. This classically requires the method of Integration by Parts.

Step 2: Detailed Explanation:
We evaluate $I = \int_{0}^{1} \tan^{-1}(x) \cdot 1 \, dx$.
Apply Integration by Parts using the LIATE rule (Logarithmic, Inverse trig, Algebraic, Trig, Exponential).
Let $u = \tan^{-1} x \implies du = \frac{1}{1+x^2} dx$
Let $dv = 1 \, dx \implies v = x$
The integration by parts formula is $\int u \, dv = uv - \int v \, du$:
$I = \left[ x \tan^{-1} x \right]_0^1 - \int_{0}^{1} x \left( \frac{1}{1+x^2} \right) dx$
Evaluate the first part at the limits:
$\left[ x \tan^{-1} x \right]_0^1 = (1 \cdot \tan^{-1}(1)) - (0 \cdot \tan^{-1}(0))$
$= (1 \cdot \frac{\pi}{4}) - 0 = \frac{\pi}{4}$
Now, evaluate the remaining integral: $\int_{0}^{1} \frac{x}{1+x^2} dx$.
Use substitution: let $t = 1+x^2 \implies dt = 2x dx \implies x dx = \frac{dt}{2}$.
When $x = 0, t = 1$. When $x = 1, t = 2$.
$\int_{1}^{2} \frac{1}{t} \left(\frac{dt}{2}\right) = \frac{1}{2} [\log|t|]_1^2 = \frac{1}{2} (\log 2 - \log 1) = \frac{1}{2} \log 2$.
Combine both parts back together:
$I = \frac{\pi}{4} - \frac{1}{2} \log 2$
By the laws of logarithms ($a \log B = \log B^a$):
$I = \frac{\pi}{4} - \log(2^{1/2})$
$I = \frac{\pi}{4} - \log(\sqrt{2})$

Step 3: Final Answer:
The integral evaluates to $\pi/4 - \log \sqrt{2}$, matching option (b).