calculate-pi-4-pi-3-sinx-cosx-tanx-dx-

Question Number 42679 by maxmathsup by imad last updated on 31/Aug/18

c a l c u l a t e \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n x}{c o s x + t a n x} d x .

Commented by Meritguide1234 last updated on 03/Sep/18

Commented by maxmathsup by imad last updated on 03/Sep/18

l e t A = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n x}{c o s x + t a n x} d x \Rightarrow A = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n x c o s x}{{c o s}^{2} x + s i n x} d x

= \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{s i n x c o s x d x}{1 - {s i n}^{2} x + s i n x} =_{s i n x = t} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{t d t}{1 - t^{2} + t}

= - \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{t}{t^{2} - t - 1} d t = - \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{2 t - 1 - 1}{t^{2} - t - 1} d t

= - \frac{1}{2} {[l n ∣ t^{2} - t - 1 ∣]}_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} + \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{t^{2} - t - 1}

= - \frac{1}{2} {l n ∣ \frac{3}{4} - \frac{\sqrt{3}}{2} - 1 ∣ - l n ∣ \frac{1}{2} - \frac{1}{\sqrt{2}} - 1 ∣ + \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{t^{2} - t - 1} b u t

\int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{t^{2} - t - 1} d t = \int_{\frac{1}{\sqrt{2}}}^{\frac{\sqrt{3}}{2}} \frac{d t}{{(t - \frac{1}{2})}^{2} - \frac{3}{4}} =_{t - \frac{1}{2} = \frac{\sqrt{3}}{2} u} \int_{\frac{2}{\sqrt{3}} (\frac{1}{\sqrt{2}} - \frac{1}{2})}^{\frac{2}{\sqrt{3}} (\frac{\sqrt{3} - 1}{2})} \frac{1}{\frac{3}{4} (u^{2} - 1)} \frac{\sqrt{3}}{2} d u

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

= \frac{1}{\sqrt{3}} {[l n ∣ \frac{u - 1}{u + 1} ∣]}_{\frac{\sqrt{2} - 1}{\sqrt{3}}}^{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \frac{1}{\sqrt{3}} {l n ∣ \frac{\frac{\sqrt{3} - 1}{\sqrt{3}} - 1}{\frac{\sqrt{3} - 1}{\sqrt{3}} + 1} ∣ - l n ∣ \frac{\frac{\sqrt{2} - 1}{\sqrt{3}} - 1}{\frac{\sqrt{2} - 1}{\sqrt{3}} + 1} ∣}

= \frac{1}{\sqrt{3}} {l n ∣ \frac{1}{2 \sqrt{3} - 1} ∣ - l n ∣ \frac{\sqrt{2} - 1 - \sqrt{3}}{\sqrt{2} - 1 + \sqrt{3}} ∣ \Rightarrow

I = - \frac{1}{2} l n (\frac{1}{4} + \frac{\sqrt{3}}{2}) - l n (\frac{1}{2} + \frac{1}{\sqrt{2}}) + \frac{1}{\sqrt{3}} {l n ∣ \frac{1}{2 \sqrt{3} - 1} ∣ - l n ∣ \frac{\sqrt{2} - 1 - \sqrt{3}}{\sqrt{2} - 1 + \sqrt{3}} ∣ .