Definite Integration Question 11
Question 11 - 29 January - Shift 2
The value of the integral $\int _{1 / 2}^{2} \frac{\tan ^{-1} x}{x} d x$ is equal to
(1) $\pi \log _e 2$
(2) $\frac{1}{2} \log _e 2$
(3) $\frac{\pi}{4} \log _e 2$
(4) $\frac{\pi}{2} \log _e 2$
Show Answer
Answer: (4)
Solution:
Formula: Properties of Inverse Trigonometric Function, Standard formulas for Indefinite Integration, Integration by substitution
$ I=\int _{1 / 2}^{2} \frac{\tan ^{-1} x}{x} d x \ldots (i) $
Substitute $\quad x=\frac{1}{t} \quad d x=-\frac{1}{t^{2}} d t$
$ \begin{aligned} & I=-\int_2^{1 / 2} \frac{\tan ^{-1} \frac{1}{t}}{\frac{1}{t}} \cdot \frac{1}{t^{2}} dt=-\int_2^{1 / 2} \frac{\tan ^{-1} \frac{1}{t}}{t} dt \\ & I=\int _{1 / 2}^{2} \frac{\cot ^{-1} t}{t} dt=\int _{1 / 2}^{2} \frac{\cot ^{-1} x}{x} dx \ldots (ii) \end{aligned} $
On adding both equations, we get
$ \begin{aligned} 2 I & =\int _{1 / 2}^{2} \frac{\tan ^{-1} x+\cot ^{-1} x}{x} d x=\frac{\pi}{2} \int _{1 / 2}^{2} \frac{d x}{x}=\frac{\pi}{2}(\ln 2) _{1 / 2}^{2} \\ & =\frac{\pi}{2}(\ln 2-\ln \frac{1}{2})=\pi \ell n 2 \\ & I=\frac{\pi}{2} \ln 2 \end{aligned} $
