calculate-0-pi-2-x-2-sin-x-cos-x-2-dx-

Question Number 174838 by mnjuly1970 last updated on 12/Aug/22

c a l c u l a t e

Ω = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{(s i n (x) + c o s (x))}^{2}} d x = ?

Answered by princeDera last updated on 12/Aug/22

c a l c u l a t e

Ω = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{(s i n (x) + c o s (x))}^{2}} d x = ?

Ω = \int_{0}^{\frac{π}{2}} \frac{x^{2}}{1 + \sin (2 x)} d x

\sin (2 x) = \frac{1 - e^{- 4 i x}}{2 {i e}^{- 2 i x}}

Ω = 2 i \int_{0}^{\frac{π}{2}} \frac{x^{2} e^{- 2 i x}}{2 {i e}^{- 2 i x} + 1 - e^{- 4 i x}} d x

l e t z = e^{- 2 i x}

d z = - 2 {i e}^{- 2 i x} d x \Rightarrow - \frac{d z}{2 i z} = d x a n d x = - \frac{1}{2 i} \ln (z)

Ω = - \frac{1}{4 i} \int_{C} \frac{\ln^{2} (z)}{2 i z + 1 - z^{2}} d z w h e r e t h e c o n t o u r C i s i n t h e f i r s t q u a d r a n t

Ω = \frac{1}{4 i} \int_{C} \frac{\ln^{2} (z)}{{(z - i)}^{2}} d z

d o u b l e p o l e s a p p e a r a t z = i a n d t h e y a r e i n t h e c o n t o u r

Ω = \frac{1}{4 i} \times 2 π i {[\frac{d}{d z} \ln^{2} (z)]}_{z = i} = \frac{π}{2} {\frac{2 \ln (i)}{i}}

\ln (i) = i \frac{π}{2}

Ω = \frac{π}{2} \times \frac{2 i π}{2 i} = \frac{π^{2}}{2}

Ω = \frac{π^{2}}{2}

Commented by Tawa11 last updated on 12/Aug/22

Great sir

Commented by Frix last updated on 12/Aug/22

something went wrong \dots

Commented by Frix last updated on 12/Aug/22

after substituting

z = e^{- 2 i x}

I get

Ω = - \frac{1}{4} \int \frac{(z^{2} + 2 i z - 1) \ln^{2} z}{{(z^{2} + 1)}^{2}} d z

I cannot solve this but your result is

Ω \approx 4.935

and approximating I get

Ω = \underset{0}{\int^{π / 2}} \frac{x^{2}}{{(\sin x + \cos x)}^{2}} d x \approx 0.862

Commented by princeDera last updated on 12/Aug/22

h o w d i d u g e t t h a t f i r s t i n t e g r a l

Commented by Frix last updated on 13/Aug/22

\underset{0}{\int^{π / 2}} \frac{x^{2}}{1 + \sin 2 x} d x =

with \sin 2 x = \frac{e^{4 i x} - 1}{{2 ie}^{2 i x}} (= \frac{1 - e^{4 i x}}{{2 ie}^{- 2 i x}})

= 2 \underset{0}{\int^{π / 2}} \frac{x^{2} e^{2 i x} ({2 e}^{2 i x} + (e^{4 i x} - 1) i)}{{(e^{4 i x} + 1)}^{2}} d x =

with z = e^{2 i x} \Leftrightarrow x = - i \frac{\ln z}{2} \Leftrightarrow d x = - i \frac{d z}{2 z}

= - \frac{1}{4} \underset{1}{\int^{- 1}} \frac{(z^{2} - 2 i z - 1) \ln^{2} z}{{(z^{2} + 1)}^{2}} d z

Answered by mnjuly1970 last updated on 13/Aug/22

Ω = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{x^{2}}{{s i n}^{2} (x + \frac{π}{4})} d x

\overset{i . b . p}{=} \frac{1}{2} {{[- c o t (x + \frac{π}{4}) x^{2}]}_{0}^{\frac{π}{2}} + 2 \int_{0}^{\frac{π}{2}} x c o t (x + \frac{π}{4}) d x

= \frac{π^{2}}{8} + \int_{0}^{\frac{π}{2}} x . c o t (x + \frac{π}{4}) d x

\overset{i . b . p}{=} \frac{π^{2}}{8} + {{[x l n s i n (x + \frac{π}{4})]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} l n s i n (x + \frac{π}{4}) d x}

= \frac{π^{2}}{8} + \frac{π}{2} l n (\frac{\sqrt{2}}{2}) - \int_{- \frac{π}{4}}^{\frac{π}{4}} l n c o s (x) d x

= \frac{π^{2}}{8} - \frac{π}{4} l n (2) - 2 \int_{0}^{\frac{π}{4}} l n c o s (x) d x

\underset{l n {(c o s (x))}^{* *}}{\overset{f o u r i e r s e r i e s}{=}} \frac{π^{2}}{8} - \frac{π}{4} l n (2) - 2 {\frac{G}{2} - \frac{π}{4} l n (2)}

= \frac{π^{2}}{8} + \frac{π}{4} l n (2) - G

* * : l n (c o s (x)) \overset{?}{=} - l n (2) - \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} c o s (2 n x))}{n}