First slide

Evaluation of definite integrals

Question

$\int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{(π + 4 x^{3}) d x}{2 - \cos (| x | + \frac{π}{3})} =$

Difficult

A

$\frac{4 π}{\sqrt{3}}$
B

$\frac{4 π}{3} \tan (\frac{1}{2})$
C

$\frac{4 π}{3} \tan^{- 1} (\frac{1}{3})$
D

$\frac{4 π}{\sqrt{3}} \tan^{- 1} (\frac{1}{2})$

Solution

$I_{1} = \int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{π d x}{2 - \cos (|x| + \frac{π}{3})},$ dropping the odd term
          $= 2 π \int_{0}^{\frac{π}{3}} \frac{d x}{2 - \cos (x + \frac{π}{3})}, x + \frac{π}{3} = t$
          $= 2 π \int_{\frac{π}{3}}^{\frac{2 π}{3}} \frac{d t}{2 - \cos t}, u = \tan \frac{t}{2}$
        $= 4 π \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{d u}{2 (1 + u^{2}) - (1 - u^{2})}$
           $= 4 {π \int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} \frac{d u}{1 + 3 u^{2}} = \frac{4 π}{\sqrt{3}} \tan^{- 1} \sqrt{3} u |}_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}$
             $= \frac{4 π}{\sqrt{3}} (\tan^{- 1} 3 - \tan^{- 1} 1) = \frac{4 π}{\sqrt{3}} \tan^{- 1} (\frac{1}{2})$

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