calculus-I-evaluate-I-0-pi-3-log-1-3-tan-x-dx-

Question Number 116284 by mnjuly1970 last updated on 02/Oct/20

\dots c a l c u l u s I \dots

e v a l u a t e :

I := \int_{0}^{\frac{π}{3}} l o g (1 + \sqrt{3} t a n (x)) d x = ? ? ?

Answered by Dwaipayan Shikari last updated on 02/Oct/20

I = \int_{0}^{\frac{π}{3}} \log (cosx + \sqrt{3} sinx) - \log (cosx) dx

= [\int_{0}^{\frac{π}{3}} \log (\sin (\frac{π}{6} + x)) - \log (cosx)] \to (I_{0}) + \int_{0}^{\frac{π}{3}} \log (2) dx

= [\int_{0}^{\frac{π}{3}} \log (sinx (\frac{π}{2} - x)) - \log (\cos (\frac{π}{3} - x))] \to (I_{0}) + \frac{π}{3} \log (2)

{2 I}_{0} = \int_{0}^{\frac{π}{3}} \log (cosx) - \log (cosx) + \frac{1}{2} cosx - \frac{1}{2} cosx + \frac{\sqrt{3}}{2} sinx - \frac{\sqrt{3}}{2} sinx

I_{0} = 0

So

I = I_{0} + \frac{π}{3} \log (2)

I = \frac{π}{3} \log (2)

Commented by mnjuly1970 last updated on 02/Oct/20

n i c e v e r y n i c e a s

a l w a y s \dots t h a n k y o u \dots

Commented by Dwaipayan Shikari last updated on 02/Oct/20

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Answered by mnjuly1970 last updated on 02/Oct/20

I \overset{\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x}{=} \int_{0}^{\frac{π}{3}} l o g (1 + \sqrt{3} (\frac{\sqrt{3} - t a n (x)}{1 + \sqrt{3} t a n (x)})) d x

= \int_{0}^{\frac{π}{3}} l o g (\frac{4}{1 + \sqrt{3} t a n (x)}) d x = \frac{π}{3} l o g (4) - I

∴ I = \frac{π}{3} l o g (2) \dots ✓

m . n . 1970

Answered by Bird last updated on 03/Oct/20

l e t g (a) = \int_{0}^{\frac{π}{3}} l n (1 + a t a n x) d x \Rightarrow

g^{^{'}} (a) = \int_{0}^{\frac{π}{3}} \frac{t a n x}{1 + a t a n x} d x

= \frac{1}{a} \int_{0}^{\frac{π}{3}} \frac{1 + a t a n x - 1}{1 + a t a n x} d x

= \frac{π}{3 a} - \frac{1}{a} \int_{0}^{\frac{π}{3}} \frac{d x}{1 + a t a n x} b u t

\int_{0}^{\frac{π}{3}} \frac{d x}{1 + a t a n x} =_{t a n x = t} \int_{0}^{\sqrt{3}} \frac{d t}{(1 + t^{2}) (1 + a t)}

l e t d e c o m p o s e F (t) = \frac{1}{(a t + 1) (t^{2} + 1)}

F (t) = \frac{α}{a t + 1} + \frac{m t + n}{t^{2} + 1}

α = \frac{1}{\frac{1}{a^{2}} + 1} = \frac{a^{2}}{1 + a^{2}}

{l i m}_{t \to + \infty} t F (t) = 0 = \frac{α}{a} + m \Rightarrow

m = - \frac{α}{a} = - \frac{a}{1 + a^{2}}

F (0) = 1 = α + n \Rightarrow n = 1 - α

= 1 - \frac{a^{2}}{1 + a^{2}} = \frac{1}{1 + a^{2}}

\Rightarrow F (t) = \frac{a^{2}}{(1 + a^{2}) (a t + 1)} + \frac{- \frac{a}{1 + a^{2}} t + \frac{1}{1 + a^{2}}}{t^{2} + 1}

\Rightarrow \int_{0}^{\frac{π}{3}} \frac{d x}{1 + a t a n x}

= \frac{a^{2}}{1 + a^{2}} \int_{0}^{\sqrt{3}} \frac{d t}{a t + 1} - \frac{a}{1 + a^{2}} \int_{0}^{\sqrt{3}} \frac{t}{t^{2} + 1} d t

+ \frac{1}{1 + a^{2}} \int_{0}^{\sqrt{3}} \frac{d t}{1 + t^{2}}

= \frac{a}{1 + a^{2}} {[l n (a t + 1)]}_{0}^{\sqrt{3}} - \frac{a}{2 (1 + a^{2})} {[l n (1 + t^{2})]}_{0}^{\sqrt{3}}

+ \frac{1}{1 + a^{2}} \times \frac{π}{3}

= \frac{a}{1 + a^{2}} l n (a \sqrt{3} + 1) - \frac{a}{(1 + a^{2})} l n 2

+ \frac{π}{3 (1 + a^{2})}

Commented by Bird last updated on 03/Oct/20

\Rightarrow g^{^{'}} (a) = \frac{π}{3 a} - \frac{1}{a} {\frac{a}{1 + a^{2}} l n (a \sqrt{3} + 1)

- \frac{a l n 2}{1 + a^{2}} + \frac{π}{3 (1 + a^{2})}}

= \frac{π}{3 a} - \frac{l n (a \sqrt{3} + 1)}{1 + a^{2}} + \frac{l n 2}{1 + a^{2}} - \frac{π}{3 a (1 + a^{2})}

\Rightarrow g (a) = \frac{π}{3} l n a - \int \frac{l n (a \sqrt{3} + 1)}{1 + a^{2}} d a

+ l n 2 a r c t a n a - \frac{π}{3} \int \frac{d a}{a (1 + a^{2})}

\dots b e c o n t i n u e d \dots

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