First slide

Evaluation of definite integrals

Question

If $I_{1} = \int_{1 / e}^{\tan x} \frac{t}{1 + t^{2}} d t$ and $I_{2} = \int_{1 / e}^{\cot x} \frac{d t}{t (1 + t^{2})}$ then the value of $I_{1} + I_{2}$ is

Moderate

A

1/2
B

1
C

e/2
D

$(1 / 2) (e + 1 / e)$

Solution

Putting $t = 1 / u$ in $I_{2}$ we have
$I_{2} = - \int_{e}^{\tan x} \frac{u d u}{1 + u^{2}} = - \int_{1 / e}^{\tan x} \frac{u d u}{1 + u^{2}} + \int_{1 / e}^{e} \frac{u d u}{1 + u^{2}}$
$= - I_{1} + \frac{1}{2} \int_{1 / e}^{e} \frac{2 u d u}{1 + u^{2}}$
so ${I_{1} + I_{2} = \frac{1}{2} \log (u^{2} + 1)]}_{1 / e}^{e}$
$\begin{array}{l} = \frac{1}{2} [\log (e^{2} + 1) - \log (\frac{e^{2} + 1}{e^{2}})] \\ = \frac{1}{2} \times 2 = 1 \end{array}$

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