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Question Number 68596 by Abdo msup. last updated on 14/Sep/19

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

Commented by mathmax by abdo last updated on 15/Sep/19

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

I = \int_{\frac{π}{2}}^{\frac{π}{3}} \frac{x d x}{3 + c o s x} = \int_{1}^{\frac{1}{\sqrt{3}}} \frac{2 a r c t a n (t)}{3 + \frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}}

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

l e t f (α) = \int_{1}^{\frac{1}{\sqrt{3}}} \frac{a r c t a n (α t)}{t^{2} + 2} d t w i t h α > 0 \Rightarrow

f^{^{'}} (α) = \int_{1}^{\frac{1}{\sqrt{3}}} \frac{t}{(α^{2} t^{2} + 1) (t^{2} + 2)} d t l e t d e c o m p o s e

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

F (- t) = - F (t) \Rightarrow b = d = 0 \Rightarrow F (t) = \frac{a t}{α^{2} t^{2} + 1} + \frac{c t}{t^{2} + 2}

{l i m}_{t \to + \infty} t F (t) = 0 = \frac{a}{α^{2}} + c \Rightarrow c = - \frac{a}{α^{2}} \Rightarrow

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

F (1) = \frac{1}{3 (α^{2} + 1)} = \frac{a}{α^{2} + 1} - \frac{a}{3 α^{2}} \Rightarrow \frac{1}{3} = a - \frac{a}{3 α^{2}} (α^{2} + 1) \Rightarrow

1 = 3 a - \frac{a (α^{2} + 1)}{α^{2}} = \frac{(3 α^{2} - α^{2} - 1) a}{α^{2}} \Rightarrow (2 α^{2} - 1) a = α^{2} \Rightarrow

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

f^{^{'}} (α) = \frac{1}{2 (2 α^{2} - 1)} \int_{1}^{\frac{1}{\sqrt{3}}} \frac{2 α^{2} t}{α^{2} t^{2} + 1} d t - \frac{1}{2 (2 α^{2} - 1)} \int_{1}^{\frac{1}{\sqrt{3}}} \frac{2 t d t}{t^{2} + 2}

= \frac{1}{2 (2 α^{2} - 1)} {[l n (α^{2} t^{2} + 1)]}_{1}^{\frac{1}{\sqrt{3}}} - \frac{1}{2 (2 α^{2} - 1)} {[l n (t^{2} + 2)]}_{1}^{\frac{1}{\sqrt{3}}}

= \frac{1}{2 (2 α^{2} - 1)} {l n (\frac{α^{2}}{3} + 1) - l n (α^{2} + 1)} - \frac{1}{2 (2 α^{2} - 1)} {l n (\frac{1}{3} + 2) - l n (3)}

\Rightarrow f (α) = \int \frac{l n (α^{2} + 3) - l n (α^{2} + 1) - l n (3)}{2 (2 α^{2} - 1)} d α

- (l n (\frac{7}{3}) - l n (3)) \int \frac{d α}{2 (2 α^{2} - 1)} + c \dots . b e c o n t i n u e d \dots .