First slide

Evaluation of definite integrals

Question

$\int_{0}^{\frac{π}{4}} {(\frac{x}{x \sin x + \cos x})}^{2} d x =$

Difficult

A

$\frac{5 - π}{5 + π}$
B

$\frac{2}{4 + π}$
C

$\frac{4 - π}{4 + π}$
D

$\frac{4 + π}{4 - π}$

Solution

$\int_{0}^{\frac{π}{4}} \frac{x^{2} d x}{{(x \sin x + \cos x)}^{2}} = \int_{0}^{\frac{π}{4}} x \sec x . \frac{x \cos x d x}{{(x \sin x + \cos x)}^{2}}$
                                    $= \int_{0}^{\frac{π}{4}} x \sec x \frac{d}{d x} (\frac{- 1}{(x \sin x + \cos x)}) d x$
                                    ${= \frac{- x s e c x}{x \sin x + \cos x} |}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} \frac{\sec x + x \sec x \tan x}{x \sin x + \cos x} d x$
                                     $= - \frac{2 π}{π + 4} + \int_{0}^{\frac{π}{4}} \sec^{2} x d x = \frac{- 2 π}{π + 4} + 1$
                                    $= \frac{4 - π}{4 + π}$

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