calculateI-0-pi-2-ln-cosx-sinx-dx-and-J-0-pi-2-ln-cosx-sinx-dx-

Question Number 96495 by abdomathmax last updated on 01/Jun/20

calculateI = \int_{0}^{\frac{π}{2}} \ln (cosx + sinx) dx

and J = \int_{0}^{\frac{π}{2}} \ln (cosx - sinx) dx

Answered by Sourav mridha last updated on 02/Jun/20

\boldsymbol J - I = \int_{0}^{\frac{\boldsymbol π}{2}} \boldsymbol l n [\boldsymbol t a n (\frac{\boldsymbol π}{4} - \boldsymbol x)] \boldsymbol d x \dots (\boldsymbol i)

\boldsymbol a n d \boldsymbol a l s o \boldsymbol J - I = \int_{0}^{\frac{\boldsymbol π}{2}} \boldsymbol l n [\boldsymbol c o t (\frac{\boldsymbol π}{4} - \boldsymbol x)] \boldsymbol d x

\dots \dots \dots (\boldsymbol i i)

(\boldsymbol i) + (\boldsymbol i i) \boldsymbol w e \boldsymbol g e t . . \boldsymbol J - I = 0, \boldsymbol J = \boldsymbol I \dots . (\boldsymbol g)

\boldsymbol n o w J + \boldsymbol I = \int_{0}^{\frac{\boldsymbol π}{2}} \boldsymbol l n (\boldsymbol c o s 2 \boldsymbol x) \boldsymbol d x \dots (\boldsymbol v)

\boldsymbol n o w 2 \boldsymbol x = \boldsymbol t, \boldsymbol t h e n

2 \boldsymbol J = \frac{1}{2} \int_{0}^{\boldsymbol π} \boldsymbol l n (\boldsymbol c o s t) \boldsymbol d t \boldsymbol w e \boldsymbol c a n \boldsymbol a l s o

\boldsymbol w r i t e \boldsymbol t h i s \boldsymbol a s - -

2 \boldsymbol J = \int_{0}^{\frac{\boldsymbol π}{2}} \boldsymbol l n (\boldsymbol c o s t) \boldsymbol d t \dots . (\boldsymbol a)

\boldsymbol s o, 2 \boldsymbol J = \int_{0}^{\frac{\boldsymbol π}{2}} \boldsymbol l n (\boldsymbol s i n t) \boldsymbol d t \dots . (\boldsymbol b)

(\boldsymbol a) + (\boldsymbol b) 4 J = \frac{\boldsymbol π}{2} \boldsymbol l n (\frac{1}{2}) + \int_{0}^{\frac{\boldsymbol π}{2}} \boldsymbol l n (\boldsymbol s i n (2 \boldsymbol t)) \boldsymbol d t

\boldsymbol b y \boldsymbol a p p l y : the \boldsymbol s a m e \boldsymbol a p p r o c h \boldsymbol l i k e

\boldsymbol {e q}^{\boldsymbol n} (\boldsymbol v) \boldsymbol w e \boldsymbol g e t, \int_{0}^{\frac{π}{2}} \boldsymbol l n (\boldsymbol s i n 2 \boldsymbol t) \boldsymbol d t = 2 \boldsymbol J

so, 4 \boldsymbol J = \frac{\boldsymbol π}{2} \boldsymbol l n (\frac{1}{2}) + 2 \boldsymbol J

so, 2 \boldsymbol J = \frac{\boldsymbol π}{2} \ln (\frac{1}{2})

now \boldsymbol f r o m (\boldsymbol g) \boldsymbol I = \boldsymbol J = \frac{\boldsymbol π}{4} \boldsymbol l n (\frac{1}{2})

Commented by abdomathmax last updated on 02/Jun/20

thanks sir .

Commented by Sourav mridha last updated on 02/Jun/20

welcome....

Answered by mathmax by abdo last updated on 02/Jun/20

let f (a) = \int_{0}^{\frac{π}{2}} \ln (cosx + asinx) dx we have f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{sinx}{cosx + asinx} dx

= \int_{0}^{\frac{π}{2}} \frac{dx}{a + \frac{1}{tanx}} =_{tanx = u} \int_{0}^{\infty} \frac{du}{(1 + u^{2}) (a + \frac{1}{u})} = \int_{0}^{\infty} \frac{udu}{(au + 1) (u^{2} + 1)}

let decompose F (u) = \frac{u}{(au + 1) (u^{2} + 1)} \Rightarrow F (u) = \frac{α}{au + 1} + \frac{β u + λ}{u^{2} + 1}

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

\lim_{u \to + \infty} uF (u) = 0 = \frac{α}{a} + β \Rightarrow β = - \frac{α}{a} = \frac{1}{1 + a^{2}}

F (0) = 0 = α + λ \Rightarrow λ = \frac{a}{1 + a^{2}} \Rightarrow F (u) = - \frac{a}{(1 + a^{2}) (au + 1)} + \frac{1}{1 + a^{2}} \times \frac{u + a}{u^{2} + 1}

\Rightarrow \int_{0}^{\infty} F (u) du = - \frac{a}{1 + a^{2}} \int_{0}^{\infty} \frac{du}{au + 1} + \frac{1}{2 (1 + a^{2})} \int_{0}^{\infty} \frac{2 u}{u^{2} + 1} + \frac{a}{1 + a^{2}} \int_{0}^{\infty} \frac{du}{u^{2} + 1}

= \frac{1}{1 + a^{2}} {[\ln \sqrt{u^{2} + 1} - \ln (au + 1)]}_{0}^{\infty} + \frac{π a}{2 (1 + a^{2})}

= \frac{1}{1 + a^{2}} {[\ln (\frac{\sqrt{1 + u^{2}}}{au + 1})]}_{0}^{\infty} + \frac{π a}{2 (1 + a^{2})} = - \frac{\ln (a)}{1 + a^{2}} + \frac{π a}{2 (1 + a^{2})} \Rightarrow

f (a) = - \int_{0}^{a} \frac{lnt}{1 + t^{2}} dt + \frac{π}{4} \ln (1 + a^{2}) + c but

f (0) = c = - \frac{π}{2} \ln (2) \Rightarrow f (a) = \frac{π}{4} \ln (1 + a^{2}) - \frac{π}{2} \ln (2) - \int_{0}^{a} \frac{lnt}{1 + t^{2}} dt

\int_{0}^{\frac{π}{2}} \ln (cosx + sinx) dx = f (1) = \frac{π}{4} \ln (2) - \frac{π}{2} \ln (2) - \int_{0}^{1} \frac{lnt}{1 + t^{2}} dt (\to t = \tan θ)

= - \frac{π}{4} \ln (2) - \int_{0}^{\frac{π}{2}} \ln (\tan θ) d θ = - \frac{π}{4} \ln (2) - 0 (because \int_{0}^{\frac{π}{2}} \ln (\cos θ) d θ = \int_{0}^{\frac{π}{2}} \ln (\sin θ) d θ) \Rightarrow

\Rightarrow \int_{0}^{\frac{π}{2}} \ln (cosx - sinx) dx = - \frac{π}{4} \ln (2)

Commented by mathmax by abdo last updated on 02/Jun/20

error of typo \int_{0}^{\frac{π}{2}} \ln (cosx + sinx) dx = - \frac{π}{4} \ln 2

error of typo \int_{0}^{\frac{π}{2}} \ln (cosx + sinx) dx = - \frac{π}{4} \ln 2