0-pi-2-x-sin-x-1-cos-x-dx-

Question Number 53693 by gunawan last updated on 25/Jan/19

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

Commented by maxmathsup by imad last updated on 25/Jan/19

l e t I = \int_{0}^{\frac{π}{2}} \frac{x + s i n x}{1 + c o s x} \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{x}{1 + c o s x} d x + \int_{0}^{\frac{π}{2}} \frac{s i n x}{1 + c o s x} d x b u t

\int_{0}^{\frac{π}{2}} \frac{s i n x}{1 + c o s x} d x = - \int_{0}^{\frac{π}{2}} \frac{{(c o s x)}^{,}}{1 + c o s x} d x = - {[l n ∣ 1 + c o s x ∣]}_{0}^{\frac{π}{2}} = - (- l n (2)) = l n (2)

\int_{0}^{\frac{π}{2}} \frac{x}{1 + c o s x} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} \frac{x}{{c o s}^{2} (\frac{x}{2})} d x = \frac{1}{2} \int_{0}^{\frac{π}{2}} x (1 + {t a n}^{2} (\frac{x}{2}) d x b u t

\int_{0}^{\frac{π}{2}} x (1 + {t a n}^{2} (\frac{x}{2})) d x =_{\frac{x}{2} = t} \int_{0}^{\frac{π}{4}} (2 t) (1 + {t a n}^{2} t) (2) d t

= 4 \int_{0}^{\frac{π}{4}} t (1 + {t a n}^{2} t) d t b y p a r t s u = t a n d v^{^{'}} = 1 + {t a n}^{2} t \Rightarrow

\int_{0}^{\frac{π}{4}} t (1 + {t a n}^{2} t) d t = {[t t a n t]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} t a n t d t

= \frac{π}{4} + \int_{0}^{\frac{π}{4}} \frac{- s i n t}{c o s t} d t = \frac{π}{4} + {[l n ∣ c o s t ∣]}_{0}^{\frac{π}{4}} = \frac{π}{4} + l n (\frac{1}{\sqrt{2}}) = \frac{π}{4} - \frac{1}{2} l n (2) \Rightarrow

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

I = l n (2) + \frac{1}{2} (π - 2 l n (2)) \Rightarrow I = \frac{π}{2} .

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jan/19

\int_{0}^{\frac{π}{2}} \frac{x}{1 + c o s x} d x - \int_{0}^{\frac{π}{2}} \frac{d (1 + c o s x)}{1 + c o s x}

\int_{0}^{\frac{π}{2}} x \times \frac{1}{2 {c o s}^{2} \frac{x}{2}} - \int_{0}^{\frac{π}{2}} \frac{d (1 + c o s x)}{1 + c o s x}

n o w \dots

\int {x s e c}^{2} \frac{x}{2} d x

x \times \frac{t a n \frac{x}{2}}{\frac{1}{2}} - \int [\frac{d x}{d x} \int {s e c}^{2} \frac{x}{2} d x] d x

= 2 x t a n \frac{x}{2} - \int \frac{t a n \frac{x}{2}}{\frac{1}{2}} d x

= 2 x t a n \frac{x}{2} - 2 l n s e c \frac{x}{2}

s o

\frac{1}{2} \int_{0}^{\frac{π}{2}} {x s e c}^{2} \frac{x}{2} - \int_{0}^{\frac{π}{2}} \frac{d (1 + c o s x)}{1 + c o s x}

\frac{1}{2} ∣ 2 x t a n \frac{x}{2} - 2 l n s e c \frac{x}{2} ∣_{0}^{\frac{π}{2}} - ∣ l n (1 + c o s x) ∣_{0}^{\frac{π}{2}}

= {(\frac{π}{2} t a n \frac{π}{4} - l n s e c \frac{π}{4}) - (0 \times t a n \frac{0}{2} - l n s e c 0)} - {l n (1 + 0) - l n (1 + 1)}

= {(\frac{π}{2} \times 1 - l n \sqrt{2}) - (0 - 0)} - {0 - l n 2}

= \frac{π}{2} - \frac{1}{2} l n 2 + l n 2

= \frac{π}{2} + \frac{1}{2} l n 2

p l s c h e c k s t e p s \dots