prove-0-pi-2-cos-x-cos-5x-1-2sin-x-3-2-

Question Number 189325 by mnjuly1970 last updated on 14/Mar/23

p r o v e

Ω = \int_{0}^{\frac{π}{2}} \frac{c o s (x) + c o s (5 x)}{1 + 2 s i n (x)} \overset{?}{=} \frac{3}{2}

Answered by Frix last updated on 15/Mar/23

\underset{0}{\int^{\frac{π}{2}}} \frac{\cos 5 x + \cos x}{1 + 2 \sin x} d x \overset{t = \sin x}{=}

= 2 \underset{0}{\int^{1}} (4 t^{3} - 2 t^{2} - 2 t + 1) d t =

= 2 {[t^{4} - \frac{2 t^{3}}{3} - t^{2} + t]}_{0}^{1} = \frac{2}{3}

Commented by mnjuly1970 last updated on 15/Mar/23

t h a n k x a l o t s i r f r i x

Answered by cortano12 last updated on 15/Mar/23

\cos 5 x + \cos x = 2 \cos 3 x \cos 2 x

\cos 3 x = \cos 2 x \cos x - \sin 2 x \sin x

\cos 3 x = (1 - 4 \sin^{2} x) \cos x

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

= \underset{0}{\int^{π / 2}} 2 (1 - 2 \sin^{2} x) (1 - 2 \sin x) \cos x dx

= 2 \underset{0}{\int^{π / 2}} (1 - 2 \sin x - 2 \sin^{2} x + 4 \sin^{3} x) d (\sin x)

= 2 {(\sin x - \sin^{2} x - \frac{2}{3} \sin^{3} x + \sin^{4} x)}_{0}^{π / 2}

= 2 (1 - 1 - \frac{2}{3} + 1) = \frac{2}{3}

Commented by mnjuly1970 last updated on 15/Mar/23

v e r y n i c e s i r c o r t a n o . .

Commented by Spillover last updated on 15/Mar/23

n i c e s o l u t i o n