calculate-0-pi-2xsinx-3-cos-2x-dx-

Question Number 57825 by maxmathsup by imad last updated on 13/Apr/19

c a l c u l a t e \int_{0}^{π} \frac{2 x s i n x}{3 + c o s (2 x)} d x .

Commented by maxmathsup by imad last updated on 13/Apr/19

l e t I = \int_{0}^{π} \frac{2 x s i n x}{3 + c o s (2 x)} d x c h a n g e m e n t x = π - t g i v e

I = \int_{0}^{π} \frac{2 (π - t) s i n t}{3 + c o s (2 t)} d t = 2 π \int_{0}^{π} \frac{s i n t}{3 + c o s (2 t)} - I \Rightarrow 2 I = 2 π \int_{0}^{π} \frac{s i n t}{3 + c o s (2 t)} d t

\Rightarrow I = π \int_{0}^{π} \frac{s i n t}{3 + c o s (2 t)} d t b u t

\int_{0}^{π} \frac{s i n t}{3 + c o s (2 t)} d t = \int_{0}^{π} \frac{s i n t}{3 + 2 {c o s}^{2} t - 1} d t = \int_{0}^{π} \frac{s i n t d t}{2 (1 + {c o s}^{2} t)}

=_{u = c o s t} - \int_{- 1}^{1} \frac{- d u}{2 (1 + u^{2})} = \frac{1}{2} \int_{- 1}^{1} \frac{d u}{1 + u^{2}} = \frac{1}{2} {[a r c t a n u]}_{- 1}^{1} = \frac{1}{2} (\frac{π}{2}) = \frac{π}{4} \Rightarrow

★ I = \frac{π^{2}}{4} ★

Answered by tanmay.chaudhury50@gmail.com last updated on 13/Apr/19

\int_{0}^{π} \frac{2 (π - x) s i n (π - x)}{3 + c o s 2 (π - x)} d x

\int_{0}^{π} \frac{2 π s i n x - 2 x s i n x}{3 + c o s 2 x} d x

2 I = \int_{0}^{π} \frac{2 π s i n x}{3 + c o s 2 x} d x

\frac{I}{π} = \int_{0}^{π} \frac{- d (c o s x)}{3 + 2 {c o s}^{2} x - 1}

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

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

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

\frac{- I}{π} = \frac{1}{2} \times - 2 \times \frac{π}{4}

I = \frac{π^{2}}{4}

Commented by maxmathsup by imad last updated on 13/Apr/19

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 13/Apr/19

m o s t w e l c o m e s i r