find-the-value-of-0-pi-4-tan-x-dx-2-cos-x-2sin-2-x-

Question Number 40143 by maxmathsup by imad last updated on 16/Jul/18

f i n d t h e v a l u e o f \int_{0}^{\frac{π}{4}} \frac{t a n (x) d x}{\sqrt{2} c o s (x) + 2 {s i n}^{2} (x)}

Commented by math khazana by abdo last updated on 22/Jul/18

I = \int_{0}^{\frac{π}{4}} \frac{s i n x}{c o s x (\sqrt{2} c o s x + 2 (1 - {c o s}^{2} x)} d x

c h a n g e m e n t c o s x = t g i v e

I = \int_{1}^{\frac{1}{\sqrt{2}}} \frac{\sqrt{1 - t^{2}}}{t {\sqrt{2} t + 2 - 2 t^{2}}} \frac{- d t}{\sqrt{1 - t^{2}}}

= \int_{\frac{1}{\sqrt{2}}}^{1} \frac{d t}{t {- 2 t^{2} + \sqrt{2} t + 2}} l e t d e c o m p o s e

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

Δ^{} = 2 - 4 (- 2) 2 = 2 + 16 = 18 \Rightarrow

t_{1} = \frac{- \sqrt{2} + 3 \sqrt{2}}{- 2} = \frac{\sqrt{2} - 3 \sqrt{2}}{2} = - \sqrt{2}

t_{2} = \frac{- \sqrt{2} - 3 \sqrt{2}}{- 2} = \frac{\sqrt{2} + 3 \sqrt{2}}{2} = 2 \sqrt{2}

F (t) = \frac{1}{- 2 t (t - t_{1}) (t - t_{2})} = \frac{a}{t} + \frac{b}{t - t_{1}} + \frac{c}{t - t_{2}}

a = {l i m}_{t \to 0} t F (t) = \frac{1}{2}

b = {l i m}_{t \to t_{1}} (t - t_{1}) F (t) = \frac{- 1}{2 t_{1} (t_{1} - t_{2})}

= \frac{- 1}{- 2 \sqrt{2} (- 3 \sqrt{2})} = \frac{1}{12}

c = {l i m}_{t \to t_{2}} (t - t_{2}) F (t) = \frac{- 1}{2 t_{2} (t_{2} - t_{1})} = \frac{- 1}{4 \sqrt{2} (3 \sqrt{2})}

= \frac{- 1}{24} \Rightarrow

F (t) = \frac{1}{2 t} + \frac{1}{12 (t - t_{1})} - \frac{1}{24 (t - t_{2})} \Rightarrow

\int_{\frac{1}{\sqrt{2}}}^{1} F (t) d t = \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{d t}{t} + \frac{1}{12} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{d t}{t + \sqrt{2}} - \frac{1}{24} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{d t}{t - 2 \sqrt{2}}

= \frac{1}{2} {[l n ∣ t ∣]}_{\frac{1}{\sqrt{2}}}^{1} + \frac{1}{12} {[l n ∣ t + \sqrt{2} ∣]}_{\frac{1}{\sqrt{2}}}^{1} - \frac{1}{24} {[l n ∣ t - 2 \sqrt{2} ∣]}_{\frac{1}{\sqrt{2}}}^{1}

= \frac{l n (\sqrt{2)}}{2} + \frac{1}{12} {l n (1 + \sqrt{2}) - l n (\sqrt{2} + \frac{1}{\sqrt{2}})}

- \frac{1}{24} {l n (2 \sqrt{2} - 1) - l n (2 \sqrt{2} - \frac{1}{\sqrt{2}})} .