First slide

Evaluation of definite integrals

Question

$\int_{\frac{π}{4}}^{\frac{3 π}{4}} \frac{x d x}{1 + \sin x} =$

Difficult

A

$π$
B

$\frac{π}{2} (\sqrt{2} - 1)$
C

$\frac{π}{2} (\sqrt{2} + 1)$
D

$π (\sqrt{2} - 1)$

Solution

$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x \to$
$I = \int_{\frac{π}{4}}^{\frac{3 π}{4}} \frac{x d x}{1 + \sin x}, I = \int_{\frac{π}{4}}^{\frac{3 π}{4}} \frac{(π - x)}{1 + \sin x} d x$

$2 I = π \int_{\frac{π}{4}}^{\frac{3 π}{4}} \frac{d x}{1 + \sin x}, t = x - \frac{π}{2}$

     $= π \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{d t}{1 + \cos t} = 2 π \int_{0}^{\frac{π}{4}} \frac{d t}{1 + \cos t}$

      $= π \int_{0}^{\frac{π}{4}} \sec^{2} \frac{t}{2} d t = 2 {π \tan \frac{t}{2} |}_{0}^{\frac{π}{4}}$           (Using $1 + \cos t = 2 \cos^{2} \frac{t}{2})$

      $= 2 π \tan \frac{π}{8} = 2 π (\sqrt{2} - 1) \to I = π (\sqrt{2} - 1)$

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