Question-59647

Question Number 59647 by aliesam last updated on 12/May/19

Commented by Mr X pcx last updated on 13/May/19

I = \int_{0}^{π} \frac{c o s x e^{∣ s i n x ∣}}{1 + e^{t a n x}} d x + \int_{π}^{2 π} \frac{c o s x e^{∣ s i n x ∣}}{1 + e^{t a n x}} d x

b u t

\int_{π}^{2 π} \frac{c o s x e^{∣ s i n x ∣}}{1 + e^{t a n x}} d x =_{x = π + t} - \int_{0}^{π} \frac{c o s t e^{∣ s i n t ∣}}{1 + e^{t a n x}} d x

\Rightarrow I = 0 .

Answered by MJS last updated on 12/May/19

f (x) = \frac{e^{∣ \sin x ∣} \cos x}{1 + e^{\tan x}}

\underset{0}{\int^{π / 2}} f (x) d x = - \underset{π}{\int^{3 π / 2}} f (x) d x

\underset{π / 2}{\int^{π}} f (x) d x = - \underset{3 π / 2}{\int^{2 π}} f (x) d x

\Rightarrow answer = 0

Commented by MJS last updated on 13/May/19

f (x + \frac{π}{2}) = - f (x + \frac{3 π}{2})

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