calculate-pi-6-pi-6-1-tan-x-1-sin-2x-dx-

Question Number 40142 by maxmathsup by imad last updated on 16/Jul/18

c a l c u l a t e \int_{- \frac{π}{6}}^{\frac{π}{6}} \frac{1 + t a n (x)}{1 + s i n (2 x)} d x

Commented by maxmathsup by imad last updated on 19/Jul/18

l e t I = \int_{- \frac{π}{6}}^{\frac{π}{6}} \frac{1 + t a n (x)}{1 + s i n (2 x)} d x c h a n g e m e n t t a n x = t g i v e

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

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

= l n (1 + \frac{1}{\sqrt{3}}) - l n (1 - \frac{1}{\sqrt{3}}) = l n (\sqrt{3} + 1) - l n (\sqrt{3}) - l n (\sqrt{3} - 1) + l n (\sqrt{3})

I = l n (\sqrt{3} + 1) - l n (\sqrt{3} - 1)

Answered by tanmay.chaudhury50@gmail.com last updated on 17/Jul/18

t = t a n x d t = {s e c}^{2} x d x

\int_{- \frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{(1 + t) (1 + t^{2}) (1 + t^{2}) d t}{1 + t^{2} + 2 t}

\int_{- \frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{t^{4} + 2 t^{2} + 1}{t + 1} d t

\int_{\frac{- 1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} \frac{t^{4} + t^{3} - t^{3} - t^{2} + 3 t^{2} + 3 t - 3 t - 3 + 4}{t + 1}

\int_{\frac{- 1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} t^{3} - t^{2} + 3 t - 3 + \frac{4}{t + 1} d t

=∣ \frac{t^{4}}{4} - \frac{t^{3}}{3} + 3 \frac{t^{2}}{2} - 3 t + 4 l n (t + 1) ∣_{\frac{- 1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}}

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

= \frac{- 2}{9 \sqrt{3}} - \frac{6}{\sqrt{3}} + 4 l n (\frac{\sqrt{3} + 1}{\sqrt{3} - 1})

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