calculate-0-pi-sin-2x-2-cosx-dx-

Question Number 36439 by prof Abdo imad last updated on 02/Jun/18

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

Commented by abdo.msup.com last updated on 03/Jun/18

I = \int_{0}^{π} \frac{2 s i n x c o s x}{2 + c o s x} d x c h a n g e m e n t

c o s x = t g i v e - s i n x d x = d t

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

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

= 4 - 4 {[l n ∣ t + 2 ∣]}_{- 1}^{1}

= 4 - 4 (l n (3))

I = 4 - 4 l n (3) .

Answered by ajfour last updated on 02/Jun/18

l e t \cos x = t

I = \int_{1}^{- 1} \frac{- 2 t d t}{2 + t} = - 2 \int_{1}^{- 1} d t + 4 \int_{1}^{- 1} \frac{d t}{t + 2}

I = 4 - \ln 3 .

Commented by MJS last updated on 02/Jun/18

minor mistake at the end : I = 4 - 4 \ln 3

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