calculate-0-pi-x-dx-3-cosx-

Question Number 35241 by abdo mathsup 649 cc last updated on 17/May/18

c a l c u l a t e \int_{0}^{π} \frac{x d x}{3 + c o s x} .

Commented by abdo.msup.com last updated on 17/May/18

let put I=∫_0 ^π ((xdx)/(3 +cosx)) .changement tan((x/2))=t give x =2arctan(t) and I = ∫_0 ^∞ ((2arctan(t))/(3 +((1−t^2 )/(2+t^2 )))) ((2dt)/(1+t^2 )) = 4∫_0 ^∞ ((arctan(t))/(3 +3t^2 +1−t^2 ))dt =4 ∫_0 ^∞ ((arctan(t))/(4 +2t^2 ))dt=2∫_0 ^∞ ((arctan(t))/(2 +t^2 )) =_(t=x(√2)) 2∫_0 ^∞ ((arctan(x(√2)))/(2(1+x^2 )))(√2) dx =(√2) ∫_0 ^∞ ((arctan(x(√2)))/(1+x^2 ))dx let introduce the parametric function f(t) =∫_0 ^∞ ((arctan(tx))/(1+x^2 ))dx f^′ (t) = ∫_0 ^∞ (x/((1+t^2 x^2 )(1+x^2 )))dx =_(tx =u) ∫_0 ^∞ (u/(t(1+u^2 )(1+(u^2 /t^2 )))) (du/t) = ∫_0 ^∞ (u/((1+u^2 )(t^2 +u^2 )))du let decompose g(u)= (u/((u^2 +1)(u^2 +t^2 ))) g(u) = ((au +b)/(u^2 +1)) +((cu +d)/(u^2 +t^2 )) g(−u) =−g(u) ⇒b=d=0 so g(u)= ((au)/(u^2 +1)) +((cu)/(u^2 +t^2 )) lim_(u→+∞) u g(u)=0 = a+c ⇒c=−a g(u)= ((au)/(u^2 +1)) −((au)/(u^2 +t^2 )) g(1)=(1/(2(1+t^2 ))) = (a/2) −(a/(t^2 +1)) ⇒ (1/2) =(1/2)(1+t^2 )a −a ⇒ 1=(1+t^2 )a −2a =(t^2 −1)a ⇒ a= (1/(t^2 −1)) if t^2 ≠1 so g(u) = (1/(t^2 −1)) (u/(1+u^2 )) −(1/(t^2 −1)) (u/(t^2 +u^2 )) f^′ (t) =∫_0 ^∞ g(u)du =(1/(1−t^2 )){ ∫_0 ^∞ (u/(1+u^2 ))du −∫_0 ^∞ (u/(u^2 +t^2 ))du} = (1/(2(1−t^2 )))[ ln(1+u^2 )−ln(u^2 +t^2 )]_0 ^(+∞) = (1/(2(1−t^2 )))[ln(((u^2 +1)/(u^2 +t^2 )))]_0 ^(+∞) = (1/(2(1−t^2 ))) {ln(t^2 )= ((ln(t))/(1−t^2 )) f^′ (t) = ((ln(t))/(1−t^2 )) ⇒f(t)= ∫_0 ^t ((lnx)/(1−x^2 ))dx +λ λ=f(0)=0 ⇒ f(t)= ∫_0 ^t ((ln(x))/(1−x^2 ))dx I =(√2) f((√2))= (√2) ∫_0 ^(√2) ((ln(x))/(1−x^2 ))dx... be continued...

l e t p u t I = \int_{0}^{π} \frac{x d x}{3 + c o s x} . c h a n g e m e n t

t a n (\frac{x}{2}) = t g i v e x = 2 a r c t a n (t) a n d

I = \int_{0}^{\infty} \frac{2 a r c t a n (t)}{3 + \frac{1 - t^{2}}{2 + t^{2}}} \frac{2 d t}{1 + t^{2}}

= 4 \int_{0}^{\infty} \frac{a r c t a n (t)}{3 + 3 t^{2} + 1 - t^{2}} d t

= 4 \int_{0}^{\infty} \frac{a r c t a n (t)}{4 + 2 t^{2}} d t = 2 \int_{0}^{\infty} \frac{a r c t a n (t)}{2 + t^{2}}

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

= \sqrt{2} \int_{0}^{\infty} \frac{a r c t a n (x \sqrt{2})}{1 + x^{2}} d x l e t i n t r o d u c e

t h e p a r a m e t r i c f u n c t i o n

f (t) = \int_{0}^{\infty} \frac{a r c t a n (t x)}{1 + x^{2}} d x

f^{^{'}} (t) = \int_{0}^{\infty} \frac{x}{(1 + t^{2} x^{2}) (1 + x^{2})} d x

=_{t x = u} \int_{0}^{\infty} \frac{u}{t (1 + u^{2}) (1 + \frac{u^{2}}{t^{2}})} \frac{d u}{t}

= \int_{0}^{\infty} \frac{u}{(1 + u^{2}) (t^{2} + u^{2})} d u l e t

d e c o m p o s e g (u) = \frac{u}{(u^{2} + 1) (u^{2} + t^{2})}

g (u) = \frac{a u + b}{u^{2} + 1} + \frac{c u + d}{u^{2} + t^{2}}

g (- u) = - g (u) \Rightarrow b = d = 0 s o

g (u) = \frac{a u}{u^{2} + 1} + \frac{c u}{u^{2} + t^{2}}

{l i m}_{u \to + \infty} u g (u) = 0 = a + c \Rightarrow c = - a

g (u) = \frac{a u}{u^{2} + 1} - \frac{a u}{u^{2} + t^{2}}

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

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

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

a = \frac{1}{t^{2} - 1} i f t^{2} \neq 1 s o

g (u) = \frac{1}{t^{2} - 1} \frac{u}{1 + u^{2}} - \frac{1}{t^{2} - 1} \frac{u}{t^{2} + u^{2}}

f^{^{'}} (t) = \int_{0}^{\infty} g (u) d u

= \frac{1}{1 - t^{2}} {\int_{0}^{\infty} \frac{u}{1 + u^{2}} d u - \int_{0}^{\infty} \frac{u}{u^{2} + t^{2}} d u}

= \frac{1}{2 (1 - t^{2})} {[l n (1 + u^{2}) - l n (u^{2} + t^{2})]}_{0}^{+ \infty}

= \frac{1}{2 (1 - t^{2})} {[l n (\frac{u^{2} + 1}{u^{2} + t^{2}})]}_{0}^{+ \infty}

= \frac{1}{2 (1 - t^{2})} {l n (t^{2}) = \frac{l n (t)}{1 - t^{2}}

f^{^{'}} (t) = \frac{l n (t)}{1 - t^{2}} \Rightarrow f (t) = \int_{0}^{t} \frac{l n x}{1 - x^{2}} d x + λ

λ = f (0) = 0 \Rightarrow f (t) = \int_{0}^{t} \frac{l n (x)}{1 - x^{2}} d x

I = \sqrt{2} f (\sqrt{2}) = \sqrt{2} \int_{0}^{\sqrt{2}} \frac{l n (x)}{1 - x^{2}} d x \dots

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

Commented by abdo.msup.com last updated on 17/May/18

error change 1−t^2 by t^2 −1 so f(t)= ∫_0 ^t ((ln(x))/(x^2 −1))dx I =(√2) ∫_0 ^(√2) ((ln(x))/(x^2 −1))dx.

e r r o r c h a n g e 1 - t^{2} b y t^{2} - 1 s o

f (t) = \int_{0}^{t} \frac{l n (x)}{x^{2} - 1} d x

I = \sqrt{2} \int_{0}^{\sqrt{2}} \frac{l n (x)}{x^{2} - 1} d x .

Commented by abdo mathsup 649 cc last updated on 20/May/18

∫_0 ^(√2) ((lnx)/(x^2 −1))dx = −∫_0 ^1 ((lnx)/(1−x^2 ))dx +∫_1 ^(√2) ((lnx)/(x^2 −1))dx but ∫_0 ^1 ((ln(x))/(1−x^2 ))dx =∫_0 ^1 (Σ_(n=0) ^∞ x^(2n) )lnxdx =Σ_(n=0) ^∞ ∫_0 ^1 x^(2n) ln(x)dx by parts A_n = ∫_0 ^1 x^(2n) ln(x)dx =[(1/(2n+1))x^(2n+1) ln(x)]_0 ^1 − ∫_0 ^1 (1/(2n+1))x^(2n+1) (1/x)dx = −(1/((2n+1)^2 )) ∫_0 ^1 x^(2n) dx =−(1/((2n+1)^2 )) ⇒ ∫_0 ^1 ((ln(x))/(1−x^2 ))dx =−Σ_(n=0) ^∞ (1/((2n+1)^2 )) we have (π^2 /6) =Σ_(n=1) ^∞ (1/n^2 ) =Σ_(n=1) ^∞ (1/((2n)^2 )) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) =(1/4) Σ_(n=1) ^∞ (1/n^2 ) +Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ (1/((2n+1)^2 )) = (3/4) Σ_(n=1) ^∞ (1/n^2 ) =(3/4) (π^2 /6) = (π^2 /8)

\int_{0}^{\sqrt{2}} \frac{l n x}{x^{2} - 1} d x = - \int_{0}^{1} \frac{l n x}{1 - x^{2}} d x + \int_{1}^{\sqrt{2}} \frac{l n x}{x^{2} - 1} d x b u t

\int_{0}^{1} \frac{l n (x)}{1 - x^{2}} d x = \int_{0}^{1} (\sum_{n = 0}^{\infty} x^{2 n}) l n x d x

= \sum_{n = 0}^{\infty} \int_{0}^{1} x^{2 n} l n (x) d x b y p a r t s

A_{n} = \int_{0}^{1} x^{2 n} l n (x) d x = {[\frac{1}{2 n + 1} x^{2 n + 1} l n (x)]}_{0}^{1}

- \int_{0}^{1} \frac{1}{2 n + 1} x^{2 n + 1} \frac{1}{x} d x = - \frac{1}{{(2 n + 1)}^{2}} \int_{0}^{1} x^{2 n} d x

= - \frac{1}{{(2 n + 1)}^{2}} \Rightarrow \int_{0}^{1} \frac{l n (x)}{1 - x^{2}} d x = - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}

w e h a v e \frac{π^{2}}{6} = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \sum_{n = 1}^{\infty} \frac{1}{{(2 n)}^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}

= \frac{1}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} + \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} \Rightarrow

\sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} = \frac{3}{4} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{3}{4} \frac{π^{2}}{6} = \frac{π^{2}}{8}

Commented by abdo mathsup 649 cc last updated on 20/May/18

⇒ ∫_0 ^1 ((ln(x))/(x^2 −1))dx =−(π^2 /8) changement x=(1/t) give ∫_1 ^(√2) ((ln(x))/(x^2 −1))dx = ∫_1 ^(1/( (√2))) ((−ln(t))/(( (1/t^2 ) −1))) (−(dt/t^2 )) = −∫_(1/( (√2))) ^1 ((ln(t))/(1−t^2 ))dt =−∫_(1/( (√2))) ^1 (Σ_(n=0) ^∞ t^(2n) )ln(t)dt =−Σ_(n=0) ^∞ ∫_(1/( (√2))) ^1 t^(2n) ln(t)dt =−Σ_(n=0) ^∞ A_n by parts A_(n ) = [ (1/(2n+1))t^(2n+1) ln(t)]_(1/( (√2))) ^1 − ∫_(1/( (√2))) ^1 (1/(2n+1))t^(2n) dt =(1/(2n+1))((1/( (√2))))^(2n+1) ln((√2)) −(1/((2n+1)^2 )) ⇒ Σ_(n=0) ^∞ A_n = ln((√2))Σ_(n=0) ^∞ (1/(2n+1))((1/( (√2))))^(2n+1) −Σ_(n=0) ^∞ (1/((2n+1)^2 )) ⇒ ∫_1 ^(√2) ((ln(x))/(x^2 −1))dx = ln((√2))Σ_(n=0) ^∞ (1/((2n+1)^2 ))((1/( (√2))))^(2n+1) −(π^2 /8)

\Rightarrow \int_{0}^{1} \frac{l n (x)}{x^{2} - 1} d x = - \frac{π^{2}}{8}

c h a n g e m e n t x = \frac{1}{t} g i v e

\int_{1}^{\sqrt{2}} \frac{l n (x)}{x^{2} - 1} d x = \int_{1}^{\frac{1}{\sqrt{2}}} \frac{- l n (t)}{(\frac{1}{t^{2}} - 1)} (- \frac{d t}{t^{2}})

= - \int_{\frac{1}{\sqrt{2}}}^{1} \frac{l n (t)}{1 - t^{2}} d t = - \int_{\frac{1}{\sqrt{2}}}^{1} (\sum_{n = 0}^{\infty} t^{2 n}) l n (t) d t

= - \sum_{n = 0}^{\infty} \int_{\frac{1}{\sqrt{2}}}^{1} t^{2 n} l n (t) d t = - \sum_{n = 0}^{\infty} A_{n} b y p a r t s

A_{n} = {[\frac{1}{2 n + 1} t^{2 n + 1} l n (t)]}_{\frac{1}{\sqrt{2}}}^{1} - \int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{2 n + 1} t^{2 n} d t

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

\sum_{n = 0}^{\infty} A_{n} = l n (\sqrt{2}) \sum_{n = 0}^{\infty} \frac{1}{2 n + 1} {(\frac{1}{\sqrt{2}})}^{2 n + 1} - \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}}

\Rightarrow \int_{1}^{\sqrt{2}} \frac{l n (x)}{x^{2} - 1} d x = l n (\sqrt{2}) \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{2}} {(\frac{1}{\sqrt{2}})}^{2 n + 1}

- \frac{π^{2}}{8}

Commented by prof Abdo imad last updated on 20/May/18

∫_1 ^(√2) ((ln(x))/(x^2 −1))dx = ln((√2))Σ_(n=0) ^∞ (1/(2n+1))((1/( (√2))))^(2n+1) −(π^2 /8) let calculate the value of Σ_(n=0) ^∞ (1/(2n+1))((1/( (√2))))^(2n+1) let put S(x)= Σ_(n=0) ^∞ (1/(2n+1))x^(2n+1) with ∣x∣<1 S^′ (x) = Σ_(n=0) ^∞ x^(2n) = (1/(1−x^2 )) ⇒ S(x)= ∫ (dx/(1−x^2 )) +c S(x)= (1/2)∫ { (1/(1−x)) +(1/(1+x))}dx +c =(1/2)ln∣((1+x)/(1−x))∣ +c we have S(o)=0 ⇒c=0 ⇒ S(x) = (1/2)ln∣((1+x)/(1−x))∣ and Σ_(n=0) ^∞ (1/(2n+1))((1/( (√2))))^(2n+1) = S((1/( (√2)))) =(1/2)ln∣ ((1 +(1/( (√2))))/(1−(1/( (√2)))))∣ = (1/2)ln((((√2) +1)/( (√2) −1))) ⇒ ∫_1 ^(√2) ((ln(x))/(x^2 −1))dx =((ln((√2)))/2)ln((((√2) +1)/( (√2) −1))) −(π^2 /8)

\int_{1}^{\sqrt{2}} \frac{l n (x)}{x^{2} - 1} d x = l n (\sqrt{2}) \sum_{n = 0}^{\infty} \frac{1}{2 n + 1} {(\frac{1}{\sqrt{2}})}^{2 n + 1} - \frac{π^{2}}{8}

l e t c a l c u l a t e t h e v a l u e o f \sum_{n = 0}^{\infty} \frac{1}{2 n + 1} {(\frac{1}{\sqrt{2}})}^{2 n + 1}

l e t p u t S (x) = \sum_{n = 0}^{\infty} \frac{1}{2 n + 1} x^{2 n + 1} w i t h ∣ x ∣< 1

S^{^{'}} (x) = \sum_{n = 0}^{\infty} x^{2 n} = \frac{1}{1 - x^{2}} \Rightarrow S (x) = \int \frac{d x}{1 - x^{2}} + c

S (x) = \frac{1}{2} \int {\frac{1}{1 - x} + \frac{1}{1 + x}} d x + c

= \frac{1}{2} l n ∣ \frac{1 + x}{1 - x} ∣ + c w e h a v e S (o) = 0 \Rightarrow c = 0 \Rightarrow

S (x) = \frac{1}{2} l n ∣ \frac{1 + x}{1 - x} ∣ a n d

\sum_{n = 0}^{\infty} \frac{1}{2 n + 1} {(\frac{1}{\sqrt{2}})}^{2 n + 1} = S (\frac{1}{\sqrt{2}})

= \frac{1}{2} l n ∣ \frac{1 + \frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}} ∣ = \frac{1}{2} l n (\frac{\sqrt{2} + 1}{\sqrt{2} - 1}) \Rightarrow

\int_{1}^{\sqrt{2}} \frac{l n (x)}{x^{2} - 1} d x = \frac{l n (\sqrt{2})}{2} l n (\frac{\sqrt{2} + 1}{\sqrt{2} - 1}) - \frac{π^{2}}{8}

Commented by prof Abdo imad last updated on 20/May/18

so I =(√2) ∫_0 ^(√2) ((ln(x))/(x^2 −1))dx =(√2){ ∫_0 ^1 ((ln(x))/(x^2 −1))dx + ∫_1 ^(√2) ((ln(x))/(x^2 −1))dx} =(√2_ ) { −(π^2 /8) +((ln((√2)))/2)ln((((√2) +1)/( (√2) −1))) −(π^2 /8)} I =(√2){ ((ln((√2)))/2)ln((((√2) +1)/( (√2) −1))) −(π^2 /4)}

s o I = \sqrt{2} \int_{0}^{\sqrt{2}} \frac{l n (x)}{x^{2} - 1} d x

= \sqrt{2} {\int_{0}^{1} \frac{l n (x)}{x^{2} - 1} d x + \int_{1}^{\sqrt{2}} \frac{l n (x)}{x^{2} - 1} d x}

= \sqrt{2_{}} {- \frac{π^{2}}{8} + \frac{l n (\sqrt{2})}{2} l n (\frac{\sqrt{2} + 1}{\sqrt{2} - 1}) - \frac{π^{2}}{8}}

I = \sqrt{2} {\frac{l n (\sqrt{2})}{2} l n (\frac{\sqrt{2} + 1}{\sqrt{2} - 1}) - \frac{π^{2}}{4}}

Answered by math1967 last updated on 17/May/18

∫_(0 ) ^π ((xdx)/(3+2cos^2 (x/2)−1))=∫_(0 ) ^π ((xdx)/(2(1+cos^2 (x/2)))) now 2I=π∫_0 ^π (dx/(2(1+cos^2 (x/2)))) =(π/2)∫_0 ^π ((sec^2 (x/2)dx)/(sec^2 (x/2)+1)) =π∫_0 ^π (((1/2)sec^2 (x/2)dx)/(2+tan^2 (x/2))) =π∫_0 ^π ((d(tan(x/2)))/(((√2))^2 +tan^2 (x/2)))=π[(1/( (√2)))tan^(−1) tan (x/(2(√2)))]_0 ^π =π×[(1/( (√2)))×(π/(2(√2)))]=(π^2 /4)∴I=(π^2 /8) plz. check

\underset{0}{\int^{π}} \frac{x d x}{3 + 2 \cos^{2} \frac{x}{2} - 1} = \underset{0}{\int^{π}} \frac{x d x}{2 (1 + \cos^{2} \frac{x}{2})}

n o w 2 I = π \underset{0}{\int^{π}} \frac{d x}{2 (1 + {c o s}^{2} \frac{x}{2})}

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

= π \underset{0}{\int^{π}} \frac{\frac{1}{2} \sec^{2} \frac{x}{2} d x}{2 + {t a n}^{2} \frac{x}{2}}

= π \underset{0}{\int^{π}} \frac{d (t a n \frac{x}{2})}{{(\sqrt{2})}^{2} + {t a n}^{2} \frac{x}{2}} = π {[\frac{1}{\sqrt{2}} \tan^{- 1} \tan \frac{x}{2 \sqrt{2}}]}_{0}^{π}

= π \times [\frac{1}{\sqrt{2}} \times \frac{π}{2 \sqrt{2}}] = \frac{π^{2}}{4} ∴ I = \frac{π^{2}}{8}

p l z . c h e c k

Commented by MJS last updated on 17/May/18

in this case if f(x)=((x/2)/(1+cos^2 (x/2))) ⇒ ⇒ f(π−x)=(((π−x)/2)/(1+sin^2 (x/2))) and we get f(x)+f(π−x)= =((3π)/(8+sin^2 x))+((πcos x)/(8+sin^2 x))−((2xcos x)/(8+sin^2 x)) we can solve the 1^(st) and 2^(nd) but the 3^(rd) looks nasty again...

in this case if f (x) = \frac{\frac{x}{2}}{1 + \cos^{2} \frac{x}{2}} \Rightarrow

\Rightarrow f (π - x) = \frac{\frac{π - x}{2}}{1 + \sin^{2} \frac{x}{2}}

and we get f (x) + f (π - x) =

= \frac{3 π}{8 + \sin^{2} x} + \frac{π \cos x}{8 + \sin^{2} x} - \frac{2 x \cos x}{8 + \sin^{2} x}

we can solve the 1^{st} and 2^{nd} but the 3^{rd} looks

nasty again \dots

Commented by math1967 last updated on 17/May/18

O K t h a n k s

Answered by math1967 last updated on 17/May/18

2I=∫_0 ^π ((xdx)/(3+cosx))+∫_0 ^π (((π−x)dx)/(3+cos(π−x))) =π∫_0 ^π ((6dx)/((3+cosx)(3−cosx))) =6π∫_0 ^π (dx/(9−cos^2 x))=6π∫_0 ^π ((sec^2 xdx)/(9sec^2 x−1)) =6π∫_0 ^π ((sec^2 xdx)/(8+9tan^2 x))=2π∫_0 ^π ((3sec^2 xdx)/((2(√(2 )))^2 +(3tan x)^2 )) =2π∫_0 ^π ((d(3tanx))/((2(√2))^2 +(3tan x)^2 )) =2π[(1/(2(√2)))tan^(−1) ((3tan x)/(2(√2)))]_0 ^π =2π[(1/(2(√2)))×(π/2) −0] =(π^2 /(2(√2))) ∴ I=(π^2 /(4(√2)))

2 I = \underset{0}{\int^{π}} \frac{x d x}{3 + c o s x} + \underset{0}{\int^{π}} \frac{(π - x) d x}{3 + c o s (π - x)}

= π \underset{0}{\int^{π}} \frac{6 d x}{(3 + c o s x) (3 - c o s x)}

= 6 π \underset{0}{\int^{π}} \frac{d x}{9 - {c o s}^{2} x} = 6 π \underset{0}{\int^{π}} \frac{\sec^{2} x d x}{9 {s e c}^{2} x - 1}

= 6 π \underset{0}{\int^{π}} \frac{{s e c}^{2} x d x}{8 + 9 \tan^{2} x} = 2 π \underset{0}{\int^{π}} \frac{3 \sec^{2} x d x}{{(2 \sqrt{2})}^{2} + {(3 \tan x)}^{2}}

= 2 π \underset{0}{\int^{π}} \frac{d (3 t a n x)}{{(2 \sqrt{2})}^{2} + {(3 \tan x)}^{2}}

= 2 π {[\frac{1}{2 \sqrt{2}} \tan^{- 1} \frac{3 \tan x}{2 \sqrt{2}}]}_{0}^{π}

= 2 π [\frac{1}{2 \sqrt{2}} \times \frac{π}{2} - 0]

= \frac{π^{2}}{2 \sqrt{2}} ∴ I = \frac{π^{2}}{4 \sqrt{2}}

Commented by MJS last updated on 17/May/18

this seems wrong: (x/(3+cos x))+((π−x)/(3+cos (π−x)))=(x/(3+cos x))+((π−x)/(3−cos x))= =((3π+(π−2x)cos x)/(9−cos^2 x))≠((6π)/(9−cos^2 x)) please explain (π^2 /(4(√2)))≈1.744716 but I get ∫_0 ^π (x/(3+cos x))dx≈1.988159

this seems wrong :

\frac{x}{3 + \cos x} + \frac{π - x}{3 + \cos (π - x)} = \frac{x}{3 + \cos x} + \frac{π - x}{3 - \cos x} =

= \frac{3 π + (π - 2 x) \cos x}{9 - \cos^{2} x} \neq \frac{6 π}{9 - \cos^{2} x}

please explain

\frac{π^{2}}{4 \sqrt{2}} \approx 1.744716

but I get \underset{0}{\int^{π}} \frac{x}{3 + \cos x} d x \approx 1.988159

Commented by math1967 last updated on 17/May/18

Y e s i t i s m y m i s t a k e

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