bobhans-I-0-pi-2-ln-a-2-cos-2-b-2-sin-2-d-

Question Number 108401 by bobhans last updated on 16/Aug/20

\frac{b o b h a n s}{⋱ \iddots}

I = \underset{0}{\int^{π / 2}} \ln (a^{2} \cos^{2} θ + b^{2} \sin^{2} θ) d θ ?

Commented by john santu last updated on 17/Aug/20

\frac{⊸ J S ⊸}{⊛}

I = π \ln (\frac{a + b}{2})

Commented by bobhans last updated on 17/Aug/20

j o o s s . . t h a n k y o u a l l

Answered by Dwaipayan Shikari last updated on 16/Aug/20

I = \int_{0}^{\frac{π}{2}} l o g ({s i n}^{2} θ + \frac{a^{2}}{b^{2}} {c o s}^{2} θ) + 2 l o g b = I (k) + π l o g b

I (k) = \int_{0}^{\frac{π}{2}} l o g ({s i n}^{2} θ + k^{2} {c o s}^{2} θ) (k = \frac{a}{b})

I^{'} (k) = \int_{0}^{\frac{π}{2}} \frac{2 {k c o s}^{2} θ}{{s i n}^{2} θ + k^{2} {c o s}^{2} θ} d θ

I^{^{'}} (k) = \int_{0}^{\frac{π}{2}} \frac{2 k}{{t a n}^{2} θ + k^{2}} d θ

I^{^{'}} (k) = \int_{0}^{\frac{π}{2}} \frac{2 k}{({t a n}^{2} θ + k^{2}) ({t a n}^{2} θ + 1)} d t t a n θ = t {s e c}^{2} θ = \frac{d t}{d θ}

I^{^{'}} (k) = 2 k \int_{0}^{\infty} \frac{1}{t^{2} + k^{2}} - \frac{1}{t^{2} + 1} d t

I^{'} (k) = \frac{2 k}{k^{2} - 1} \int_{0}^{\infty} \frac{1}{t^{2} + 1} - \frac{1}{t^{2} + k^{2}} d t

I^{'} (k) = \frac{2 k}{k^{2} - 1} {[{t a n}^{- 1} t - \frac{1}{k} {t a n}^{- 1} \frac{t}{k}]}_{0}^{\infty}

I^{^{'}} (k) = \frac{2 k}{k^{2} - 1} . \frac{k - 1}{k} . \frac{π}{2} = \frac{π}{k + 1}

I (k) = \int \frac{π}{k + 1} = π l o g (k + 1) + C = \int_{0}^{\frac{π}{2}} l o g ({s i n}^{2} θ + k^{2} {c o s}^{2} θ)

I f k = 1, π l o g (2) + C = 0, C = - π l o g (2)

I (k) = π l o g (k + 1) - π l o g (2) = π l o g (a + b) - π l o g (2) - π l o g (b)

S o

A n s w e r i s I (k) + π l o g b

π l o g (\frac{a + b}{2})

Commented by mnjuly1970 last updated on 16/Aug/20

I = \int_{0}^{\frac{π}{2}} l o g ({s i n}^{2} θ + \frac{a^{2}}{b^{2}} {c o s}^{2} θ) + 2 l o g b = I (k) + π l o g b

I (k) = \int_{0}^{\frac{π}{2}} l o g ({s i n}^{2} θ + k^{2} {c o s}^{2} θ) (k = \frac{a}{b})

I^{'} (k) = \int_{0}^{\frac{π}{2}} \frac{2 {k c o s}^{2} θ}{{s i n}^{2} θ + k^{2} {c o s}^{2} θ} d θ

I^{^{'}} (k) = \int_{0}^{\frac{π}{2}} \frac{2 k}{{t a n}^{2} θ + k^{2}} d θ

I^{^{'}} (k) = \int_{0}^{\frac{π}{2}} \frac{2 k}{({t a n}^{2} θ + k^{2}) ({t a n}^{2} θ + 1)} d t t a n θ = t {s e c}^{2} θ = \frac{d t}{d θ}

I^{^{'}} (k) = 2 k \int_{0}^{\infty} \frac{1}{t^{2} + k^{2}} - \frac{1}{t^{2} + 1} d t

I^{'} (k) = \frac{2 k}{k^{2} - 1} \int_{0}^{\infty} \frac{1}{t^{2} + 1} - \frac{1}{t^{2} + k^{2}} d t

I^{'} (k) = \frac{2 k}{k^{2} - 1} {[{t a n}^{- 1} t - \frac{1}{k} {t a n}^{- 1} \frac{t}{k}]}_{0}^{\infty}

I^{^{'}} (k) = \frac{2 k}{k^{2} - 1} . \frac{k - 1}{k} . \frac{π}{2} = \frac{π}{k + 1}

I (k) = \int \frac{π}{k + 1} = π l o g (k + 1) + C = \int_{0}^{\frac{π}{2}} l o g ({s i n}^{2} θ + k^{2} {c o s}^{2} θ)

I f k = 1, π l o g (2) + C = 0, C = - π l o g (2)

I (k) = π l o g (k + 1) - π l o g (2) = π l o g (a + b) - π l o g (2) - π l o g (b)

S o

A n s w e r i s I (k) + π l o g b

♣ π l o g (\frac{a + b}{2}) \dots \Rightarrow \int_{0}^{\frac{π}{2}} {l o g (a^{2} {s i n}^{2} (x) + b^{2} {c o s}^{2} (x))} d x = π l o g (\frac{a + b}{2}) ♣

peace upon you . thank you

very excellent .

Answered by mathmax by abdo last updated on 16/Aug/20

I = \int_{0}^{\frac{π}{2}} \ln (a^{2} \frac{1 + \cos (2 θ)}{2} + b^{2} \frac{1 - \cos (2 θ)}{2}) d θ

= \int_{0}^{\frac{π}{2}} \ln (\frac{a^{2} + b^{2}}{2} + \frac{1}{2} (a^{2} - b^{2}) \cos (2 θ)) d θ

= \int_{0}^{\frac{π}{2}} \ln {(\frac{a^{2} + b^{2}}{2}) (1 + \frac{a^{2} - b^{2}}{a^{2} + b^{2}} \cos (2 θ)} d θ

= \frac{π}{2} \ln (\frac{a^{2} + b^{2}}{2}) + \int_{0}^{\frac{π}{2}} \ln {1 + \frac{a^{2} - b^{2}}{a^{2} + b^{2}} \cos (2 θ)} d θ (2 θ = x) (α = \frac{a^{2} - b^{2}}{a^{2} + b^{2}})

= \frac{π}{2} \ln (\frac{a^{2} + b^{2}}{2}) + \frac{1}{2} \int_{0}^{π} \ln (1 + α cosx) dx let explicit for o < α < 1

f (α) = \int_{0}^{π} \ln (1 + α cosx) dx \Rightarrow f^{^{'}} (α) = \int_{0}^{π} \frac{cosx}{1 + α cosx} dx

= \frac{1}{α} \int_{0}^{π} \frac{1 + α cosx - 1}{1 + α cosx} dx = \frac{π}{α} - \frac{1}{α} \int_{0}^{π} \frac{dx}{1 + α cosx}

and \int_{0}^{π} \frac{dx}{1 + α cosx} =_{\tan (\frac{x}{2}) = t} \int_{0}^{\infty} \frac{2 dt}{(1 + t^{2}) (1 + α \frac{1 - t^{2}}{1 + t^{2}})}

= \int_{0}^{\infty} \frac{2 dt}{1 + t^{2} + α - α t^{2}} = 2 \int_{0}^{\infty} \frac{dt}{(1 - α) t^{2} + 1 + α} = \frac{2}{1 - α} \int_{0}^{\infty} \frac{dt}{t^{2} + \frac{1 + α}{1 - α}}

=_{t = \sqrt{\frac{1 + α}{1 - α}} u} \frac{2}{1 - α} . \frac{1 - α}{1 + α} \int_{0}^{\infty} \frac{1}{1 + u^{2}} \times \frac{\sqrt{1 + α}}{\sqrt{1 - α}}

= \frac{2}{\sqrt{1 - α^{2}}} \times \frac{π}{2} = \frac{π}{\sqrt{1 - α^{2}}} \Rightarrow f^{^{'}} (α) = \frac{π}{α} - \frac{π}{α \sqrt{1 - α^{2}}} \Rightarrow

f (α) = π \int_{1}^{α} \frac{dt}{t} - π \int_{1}^{α} \frac{dt}{t \sqrt{1 - t^{2}}} + C = π \ln α - π \int_{1}^{α} \frac{dt}{t \sqrt{1 - t^{2}}} + C

C = f (1) = \int_{0}^{π} \ln (1 + cosx) dx we have

\int_{1}^{α} \frac{dt}{t \sqrt{1 - t^{2}}} =_{t = sinz} \int_{\frac{π}{2}}^{\arcsin α} \frac{cosz dz}{sinz cosz} = \int_{\frac{π}{2}}^{\arcsin α} \frac{dz}{sinz}

=_{\tan (\frac{z}{2}) = m} \int_{1}^{\tan (\frac{\arcsin α}{2})} \frac{2 dm}{(1 + m^{2}) \frac{2 m}{1 + m^{2}}} = \int_{1}^{\tan (\frac{\arcsin α}{2})} \frac{dm}{m}

= \ln ∣ \tan (\frac{\arcsin α}{2}) ∣ \Rightarrow f (α) = π \ln (α) - π \ln ∣ \tan (\frac{\arcsin α}{2}) ∣ \Rightarrow

I = \frac{π}{2} \ln (\frac{a^{2} + b^{2}}{2}) + \frac{1}{2} f (\frac{a^{2} - b^{2}}{a^{2} + b^{2}})

Commented by mathmax by abdo last updated on 16/Aug/20

sorry f (α) = π \ln (α) - π \ln ∣ \tan (\frac{\arcsin α}{2}) ∣ + \int_{0}^{π} \ln (1 + cosx) dx