find-0-pi-sinx-2-cos-2x-dx-

Question Number 46854 by maxmathsup by imad last updated on 01/Nov/18

f i n d = \int_{0}^{π} \frac{s i n x}{2 + c o s (2 x)} d x

Commented by maxmathsup by imad last updated on 01/Nov/18

e t A = \int_{0}^{π} \frac{s i n x}{2 + c o s (2 x)} d x \Rightarrow A = \int_{0}^{π} \frac{s i n x}{2 + 2 {c o s}^{2} x - 1} d x

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

= \frac{1}{\sqrt{2}} \int_{- \sqrt{2}}^{\sqrt{2}} \frac{d t}{1 + t^{2}} = \sqrt{2} \int_{0}^{\sqrt{2}} \frac{d t}{1 + t^{2}} = \sqrt{2} a r c t a n (\sqrt{2}) .

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Nov/18

\int_{0}^{π} \frac{s i n x}{2 + 2 {c o s}^{2} x - 1} d x

= - 1 \times \int_{0}^{π} \frac{d (c o s x)}{1 + 2 {c o s}^{2} x}

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

= \frac{- 1}{2} \times \sqrt{2} ∣ {t a n}^{- 1} (\frac{\sqrt{2} c o s x}{1}) ∣_{0}^{π}

= \frac{- 1}{\sqrt{2}} \times {{t a n}^{- 1} (- \sqrt{2} [) - {t a n}^{- 1} (\sqrt{2})}

= \frac{1}{\sqrt{2}} \times 2 {t a n}^{- 1} (\sqrt{2})

Commented by maxmathsup by imad last updated on 01/Nov/18

s i r T a n m a y y o u r a n s w e r i s c o r r e c t t h a n k s .

Commented by tanmay.chaudhury50@gmail.com last updated on 02/Nov/18

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