2-2-1-cosx-3-2sinx-dx-please-help-me-

Question Number 152572 by rexford last updated on 29/Aug/21

\int_{- \frac{Π}{2}}^{\frac{Π}{2}} \frac{1 + c o s x}{3 + 2 s i n x} d x

p l e a s e, h e l p m e

Answered by Ar Brandon last updated on 29/Aug/21

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + \cos x}{3 + 2 \sin x} d x, t = \tan \frac{x}{2} \Rightarrow 2 d t = (1 + t^{2}) d x

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

\frac{1}{(1 + t^{2}) (3 t^{2} + 4 t + 3)} = \frac{a t + b}{t^{2} + 1} + \frac{c t + d}{3 t^{2} + 4 t + 3}

= \frac{(a t + b) (3 t^{2} + 4 t + 3) + (c t + d) (t^{2} + 1)}{}

3 a + c = 0, 3 b + d = 1, 4 a + 3 b + d = 0 \Rightarrow a = - \frac{1}{4}, c = \frac{3}{4},

3 a + 4 b + c = 0 \Rightarrow b = 0, d = 1

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

= 4 {[- \frac{\ln (t^{2} + 1)}{8} + \frac{\ln (3 t^{2} + 4 t + 3)}{8}]}_{- 1}^{1} + 4 \int_{- 1}^{1} \frac{d t}{3 t^{2} + 4 t + 3}

= \ln \sqrt{5} + \frac{4}{3} \int_{- 1}^{1} \frac{d t}{t^{2} + \frac{4}{3} t + 1} = \ln \sqrt{5} + \frac{4}{3} \cdot \frac{3}{2} \cdot {[\arctan (\frac{3 t + 4}{2})]}_{- 1}^{1}

= \ln \sqrt{5} + 2 (\arctan (7) - \arctan (\frac{1}{2}))

Commented by SANOGO last updated on 30/Aug/21

Commented by puissant last updated on 30/Aug/21

{l i m}_{n \to \infty} \frac{1}{n} \underset{k = 1}{\sum^{n}} l n (1 + \frac{k^{2}}{n^{2}})

= {l i m}_{n \to \infty} \frac{b - a}{n} f (a + k \frac{b - a}{n})

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

{\begin{cases} u = l n (1 + x^{2}) \\ v^{'} = 1 \end{cases} \Rightarrow {\begin{cases} u^{'} = \frac{2 x}{1 + x^{2}} \\ v = x \end{cases}

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

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

= l n 2 - 2 {[x]}_{0}^{1} + 2 {[a r c t a n x]}_{0}^{1}

= l n 2 - 2 + \frac{π}{2}

K = l n Q \Rightarrow Q = e^{K} . .

\Rightarrow Q = e^{l n 2 - 2 + \frac{π}{2}} = 2 \frac{e^{\frac{π}{2}}}{e^{2}} . .

Commented by SANOGO last updated on 30/Aug/21

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