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Question Number 53112 by maxmathsup by imad last updated on 21/Jan/19

c a l c u l a t e \int_{0}^{π} \frac{1 + 2 s i n x}{3 + 2 c o s x} d x

l e t A = \int_{0}^{π} \frac{1 + 2 s i n x}{3 + 2 c o s x} d x c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

A = \int_{0}^{\infty} \frac{1 + \frac{4 t}{1 + t^{2}}}{3 + 2 \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = 2 \int_{0}^{\infty} \frac{1 + t^{2} + 4 t}{{(1 + t^{2})}^{2} (\frac{3 + 3 t^{2} + 2 - 2 t^{2}}{1 + t^{2}})} d t

= 2 \int_{0}^{\infty} \frac{t^{2} + 4 t + 1}{(1 + t^{2}) (5 + t^{2})} d t l e t d e c o m p o s e F (t) = \frac{t^{2} + 4 t + 1}{(t^{2} + 1) (t^{2} + 5)}

F (t) = \frac{a t + b}{t^{2} + 1} + \frac{c t + d}{t^{2} + 5} \Rightarrow (a t + b) (t^{2} + 5) + (c t + d) (t^{2} + 1) = t^{2} + 4 t + 1 \Rightarrow

{a t}^{3} + 5 a t + {b t}^{2} + 5 b + {c t}^{3} + c t + {d t}^{2} + d = t^{2} + 4 t + 1 \Rightarrow

(a + c) t^{3} + (b + d) t^{2} + (5 a + c) t + 5 b + d = t^{2} + 4 t + 1 \Rightarrow a + c = 0 a n d b + d = 1 a n d

5 a + c = 4 a n d 5 b + d = 1 \Rightarrow c = - a \Rightarrow a = 1 \Rightarrow c = - 1

w e h a v e d = 1 - b \Rightarrow 5 b + 1 - b = 1 \Rightarrow b = 0 \Rightarrow d = 1 \Rightarrow

F (t) = \frac{t}{t^{2} + 1} + \frac{- t + 1}{t^{2} + 5} \Rightarrow A = 2 \int_{0}^{\infty} F (t) d t = \int_{0}^{\infty} \frac{2 t}{t^{2} + 1} d t + \int_{0}^{\infty} \frac{- 2 t + 2}{t^{2} + 5} d t

= {[l n (\frac{t^{2} + 1}{t^{2} + 5})]}_{0}^{+ \infty} + 2 \int_{0}^{\infty} \frac{d t}{t^{2} + 5} = l n (5) + 2 \int_{0}^{\infty} \frac{d t}{t^{2} + 5} b u t

\int_{0}^{\infty} \frac{d t}{t^{2} + 5} d t =_{t = \sqrt{5} u} \int_{0}^{\infty} \frac{\sqrt{5} d u}{5 (1 + u^{2})} = \frac{1}{\sqrt{5}} {[a r t a n u]}_{0}^{+ \infty} = \frac{π}{2 \sqrt{5}} \Rightarrow

A = l n (5) + \frac{π}{2 \sqrt{5}} .

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

\int \frac{1}{3 + 2 c o s x} d x + \int \frac{2 s i n x}{3 + 2 c o s x} d x

\int \frac{1 + {t a n}^{2} \frac{x}{2}}{3 + 3 {t a n}^{2} \frac{x}{2} + 2 - 2 {t a n}^{2} \frac{x}{2}} d x - \int \frac{d (3 + 2 c o s x)}{3 + 2 c o s x}

\int \frac{{s e c}^{2} \frac{x}{2}}{5 + {t a n}^{2} \frac{x}{2}} d x - l n (3 + 2 c o s x) + c

2 \int \frac{d (t a n \frac{x}{2})}{5 + {t a n}^{2} \frac{x}{2}} - d o

∣ 2 \times \frac{1}{\sqrt{5}} {t a n}^{- 1} (\frac{t a n \frac{x}{2}}{\sqrt{5}}) - l n (3 + 2 c o s x) ∣_{0}^{π}

[{\frac{2}{\sqrt{5}} {t a n}^{- 1} (\frac{t a n \frac{π}{2}}{\sqrt{5}}) - l n (3 + 2 c o s π)} - {\frac{2}{\sqrt{5}} {t a n}^{- 1} (\frac{t a n 0}{\sqrt{5}}) - l n (3 + 2 c o s 0)}]

= [{\frac{2}{\sqrt{5}} (\frac{π}{2}) - l n (3 - 2)} - {\frac{2}{\sqrt{5}} {t a n}^{- 1} (0) - l n 5}]

= \frac{π}{\sqrt{5}} + l n 5