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Question Number 113910 by bobhans last updated on 16/Sep/20

(1) \underset{0}{\int^{π}} \frac{\sin^{4} x}{{(1 + \cos x)}^{2}} d x ?

(2) \lim_{x \to \infty} \frac{\sqrt{1 - \cos (\frac{2 π}{x})}}{\frac{1}{x}} ?

Answered by bemath last updated on 16/Sep/20

(2) s e t \frac{1}{x} = z, z \to 0

\lim_{z \to 0} \frac{\sqrt{1 - \cos 2 π z}}{z} = \lim_{z \to 0} \frac{\sqrt{1 - (1 - 2 \sin^{2} π z)}}{z}

= \lim_{z \to 0} \frac{\sqrt{2} \sin π z}{z} = π \sqrt{2}

Answered by 1549442205PVT last updated on 16/Sep/20

(1) I = \underset{0}{\int^{π}} \frac{\sin^{4} x}{{(1 + \cos x)}^{2}} d x

Since 1 + cosx = {2 \cos}^{2} \frac{x}{2} and sinx = 2 sinxcosx

\frac{\sin^{4} x}{{(1 + cosx)}^{4}} = \frac{{16 \sin}^{4} \frac{x}{2} \cos^{4} \frac{x}{2}}{{4 \cos}^{4} \frac{x}{2}} = {4 \sin}^{4} \frac{x}{2}

Hence, I = 4 \int_{0}^{π} \sin^{4} \frac{x}{2} dx = 2 \int_{0}^{π} {(1 - cosx)}^{2} dx

= 2 \int_{0}^{π} (1 - 2 cosx + \cos^{2} x) dx

= 2 \int_{0}^{π} dx - 4 \int_{0}^{π} cosx + \int_{0}^{π} (1 + \cos 2 x) dx

= (3 x - 4 sinx + \frac{1}{2} \sin 2 x) ∣_{0}^{π}

= 3 π

Answered by MJS_new last updated on 16/Sep/20

\int \frac{\sin^{4} x}{{(1 + \cos x)}^{2}} d x = \int \frac{{(1 - \cos^{2} x)}^{2}}{{(1 + \cos x)}^{2}} d x =

= \int \frac{{(1 - \cos x)}^{2} {(1 + \cos x)}^{2}}{{(1 + \cos x)}^{2}} d x =

= \int (1 - 2 \cos x + \cos^{2} x) d x =

= x - 2 \sin x + \frac{x}{2} + \frac{\cos x \sin x}{2} =

= \frac{3}{2} x - \frac{1}{2} (4 - \cos x) \sin x + C

\Rightarrow \underset{0}{\int^{π}} \frac{\sin^{4} x}{{(1 + \cos x)}^{2}} d x = \frac{3 π}{2}

Answered by mathmax by abdo last updated on 17/Sep/20

1) A = \int_{0}^{π} \frac{\sin^{4} x}{{(1 + cosx)}^{2}} dx \Rightarrow A = \int_{0}^{π} \frac{{(2 \sin (\frac{x}{2}) \cos (\frac{x}{2}))}^{4}}{{({2 \cos}^{2} (\frac{x}{2}))}^{2}} dx

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

=_{\frac{x}{2} = t} 8 \int_{0}^{\frac{π}{2}} \sin^{4} (t) dt = 8 \int_{0}^{\frac{π}{2}} {(\frac{1 - \cos (2 t)}{2})}^{2} dt

= 2 \int_{0}^{\frac{π}{2}} (1 - 2 \cos (2 t) + \cos^{2} (2 t)) dt

= 2 {[t - \sin (2 t)]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} (1 + \cos (4 t)) dt

= 2 {\frac{π}{2}} + \frac{π}{2} = π + \frac{π}{2} \Rightarrow A = \frac{3 π}{2}