First slide

Evaluation of definite integrals

Question

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

Moderate

A

$\frac{π}{12}$
B

$\frac{π}{4}$
C

$\frac{π}{3}$
D

$\frac{π}{6}$

Solution

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

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