0-pi-2-log-4-3-sin-x-4-3-cos-x-dx-

Question Number 91493 by Zainal Arifin last updated on 01/May/20

\underset{0}{\int^{π / 2}} \log (\frac{4 + 3 \sin x}{4 + 3 \cos x}) d x =

Commented by mathmax by abdo last updated on 01/May/20

I = \int_{0}^{\frac{π}{2}} l n (\frac{1 + \frac{3}{4} s i n x}{1 + \frac{3}{4} c o s x}) = \int_{0}^{\frac{π}{2}} l n (1 + \frac{3}{4} s i n x) - \int_{0}^{\frac{π}{2}} l n (1 + \frac{3}{4} c o s x)

l e t f (a) = \int_{0}^{\frac{π}{2}} l n (1 + a s i n x) w i t h 0 < a < 1

f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{s i n x}{1 + a s i n x} d x = \frac{1}{a} \int_{0}^{\frac{π}{2}} \frac{1 + a s i n x - 1}{1 + a s i n x} d x

= \frac{π}{2 a} - \frac{1}{a} \int_{0}^{\frac{π}{2}} \frac{d x}{1 + a s i n x} b u t \int_{0}^{\frac{π}{2}} \frac{d x}{1 + a s i n x} =_{t a n (\frac{x}{2}) = t} \int_{0}^{1} \frac{2 d t}{(1 + t^{2}) (1 + a \frac{2 t}{1 + t^{2}})}

= 2 \int_{0}^{1} \frac{d t}{1 + t^{2} + 2 a t} = 2 \int_{0}^{1} \frac{d t}{t^{2} + 2 a t + 1} = 2 \int_{0}^{1} \frac{d t}{{(t + a)}^{2} + 1 - a^{2}}

=_{t + a = \sqrt{1 - a^{2}} u} 2 \int_{\frac{a}{\sqrt{1 - a^{2}}}}^{\frac{1 + a}{\sqrt{1 - a^{2}}}} \frac{\sqrt{1 - a^{2}} d u}{(1 - a^{2}) (1 + u^{2})}

= \frac{2}{\sqrt{1 - a^{2}}} (a r c t a n (\frac{1 + a}{\sqrt{1 - a^{2}}}) - a r c t a n (\frac{a}{\sqrt{1 - a^{2}}})) \Rightarrow

f (a) = \frac{π}{2} l n (a) - 2 \int \frac{1}{a \sqrt{1 - a^{2}}} a r c t a n (\frac{1 + a}{\sqrt{1 - a^{2}}}) d a

+ 2 \int \frac{1}{a \sqrt{1 - a^{2}}} a r c t a n (\frac{a}{\sqrt{1 - a^{2}}}) + c w e h a v e

\int \frac{1}{a \sqrt{1 - a^{2}}} a r c t a n (\frac{a}{\sqrt{1 - a^{2}}}) =_{a = s i n x} \int \frac{1}{s i n x c o s x} a r c t a n (\frac{s i n x}{c o s x}) c o s x d x

= \int \frac{x}{s i n x} d x

\int \frac{1}{a \sqrt{1 - a^{2}}} a r c t a n (\frac{1 + a}{\sqrt{1 - a^{2}}}) d a =_{a = c o s x} - \int \frac{1}{c o s x s i n x} a r c t a n (\frac{1 + c o s x}{s i n x}) s i n x d x

= - \int \frac{1}{c o s x} a r c t a n (\frac{2 {c o s}^{2} (\frac{x}{2})}{2 s i n (\frac{x}{2}) c o s (\frac{x}{2})}) d x

= - \int \frac{1}{c o s x} a r c t a n (\frac{1}{t a n x}) d x = - \int \frac{1}{c o s x} (\frac{π}{2} - x)

= - \frac{π}{2} \int \frac{d x}{c o s x} + \int \frac{x}{c o s x} d x \dots . b e c o n t i n u e d \dots

Answered by MJS last updated on 01/May/20

\ln \frac{4 + 3 \sin x}{4 + 3 \cos x} = - \ln \frac{4 + 3 \sin (\frac{π}{2} - x)}{4 + 3 \cos (\frac{π}{2} - x)} \Rightarrow

\Rightarrow answer is 0

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