calculate-0-pi-sin-2x-3-cos-4x-dx-

Question Number 128956 by mathmax by abdo last updated on 11/Jan/21

calculate \int_{0}^{π} \frac{\sin (2 x)}{3 + \cos (4 x)} dx

Answered by chengulapetrom last updated on 11/Jan/21

c o s 4 x = 2 {c o s}^{2} 2 x - 1

\int_{0}^{π} \frac{s i n (2 x)}{3 + c o s (4 x)} d x = \int_{0}^{π} \frac{s i n (2 x)}{2 + 2 {c o s}^{2} 2 x} d x

= - \frac{1}{4} \int_{0}^{π} \frac{- 2 s i n (2 x)}{1 + {c o s}^{2} 2 x} d x

l e t t = c o s 2 x \Rightarrow d t = - 2 s i n 2 x d x

x = 0 \Rightarrow t = 1 a n d x = π \Rightarrow t = 1

\int_{1}^{1} \frac{d t}{1 + t^{2}} = 0

Commented by chengulapetrom last updated on 11/Jan/21

a m I c o r r e c t ?

Answered by Olaf last updated on 11/Jan/21

Ω = \int_{0}^{π} \frac{\sin (2 x)}{3 + \cos (4 x)} d x

Ω = \int_{0}^{π} \frac{\sin (2 x)}{3 + {2 \cos}^{2} (2 x) - 1} d x

Ω = - \frac{1}{4} \int_{0}^{π} \frac{- 2 \sin (2 x)}{1 + \cos^{2} (2 x)} d x

Ω = - \frac{1}{4} {[\arctan (\cos (2 x))]}_{0}^{π} = 0

Answered by mathmax by abdo last updated on 11/Jan/21

let I = \int_{0}^{π} \frac{\sin (2 x)}{3 + \cos (4 x)} dx \Rightarrow I =_{2 x = t} \int_{0}^{2 π} \frac{sint}{3 + \cos (2 t)} \frac{dt}{2}

= \frac{1}{2} \int_{0}^{2 π} \frac{sint}{3 + 1 - {2 \sin}^{2} t} dt = \frac{1}{2} \int_{0}^{2 π} \frac{sint}{4 - {2 \sin}^{2} t} dt = \frac{1}{4} \int_{0}^{2 π} \frac{sint}{2 - \sin^{2} t} dt

\Rightarrow 4 I = \int_{0}^{π} \frac{sint}{2 - \sin^{2} t} dt + \int_{π}^{2 π} \frac{sint}{2 - \sin^{2} t} dt (\to t = x + π

= \int_{0}^{π} \frac{sint}{2 - \sin^{2} t} - \int_{0}^{π} \frac{sinx}{2 - \sin^{2} x} dx = 0 \Rightarrow 4 I = 0 \Rightarrow I = 0