pi-pi-cosxln-1-x-1-x-dx-

Question Number 95199 by Fikret last updated on 23/May/20

\int_{- π}^{π} c o s x l n \frac{1 + x}{1 - x} d x = ?

Answered by mr W last updated on 23/May/20

f (x) = \cos x \ln \frac{1 + x}{1 - x}

f (- x) = \cos (- x) \ln \frac{1 + (- x)}{1 - (- x)}

= \cos (- x) \ln \frac{1 - x}{1 + x}

= \cos x \ln {(\frac{1 + x}{1 - x})}^{- 1}

= - \cos x \ln \frac{1 + x}{1 - x}

= - f (x)

d o m a i n o f f u n c t i o n f (x) i s (- 1, + 1)

t h e r e f o r e \int_{- π}^{π} f (x) d x i s n o t d e f i n e d .

t h e r e f o r e i c h a n g e d i t t o

\int_{- π / 4}^{π / 4} f (x) d x = ? .

\int_{- π / 4}^{π / 4} f (x) d x = \int_{- π / 4}^{0} f (x) d x + \int_{0}^{π / 4} f (x) d x

= \int_{- π / 4}^{0} f (- x) d (- x) + \int_{0}^{π / 4} f (x) d x

= \int_{π / 4}^{0} f (x) d x + \int_{0}^{π / 4} f (x) d x

= - \int_{0}^{π / 4} f (x) d x + \int_{0}^{π / 4} f (x) d x

= 0

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