First slide

Evaluation of definite integrals

Question

$\int_{\frac{π}{6}}^{\frac{π}{3}} \frac{1}{1 + \tan x} d x =$

Easy

A

$\frac{π}{4}$
B

$\frac{π}{8}$
C

$\frac{π}{12}$
D

$\frac{π}{16}$

Solution

$\int_{π / 6}^{π / 3} \frac{1}{1 + T a n x} d x = \int_{π / 6}^{π / 3} \frac{C o s x}{C o s x + S i n x} d x (\sin c e \int_{a}^{b} \frac{f (x)}{f (x) + f (a + b - x)} d x = \frac{b - a}{2})$
                             $= \frac{b - a}{2}$
                               $= \frac{\frac{π}{3} - \frac{π}{6}}{2}$
                               $= \frac{π}{12}$

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