How do you find the integral of $({(sin (\frac{x}{5}))}^{2}) \cdot ({(cos (\frac{x}{5}))}^{3})$ $((sin(x/5))^2)*((cos(x/5))^3)$ ?

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How do you find the integral of $({(sin (\frac{x}{5}))}^{2}) \cdot ({(cos (\frac{x}{5}))}^{3})$ $((sin(x/5))^2)*((cos(x/5))^3)$ ?

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Answer 1 · 2016-11-27T12:14:11

$= \frac{5}{48} (6 sin (\frac{x}{5}) - 5 sin (\frac{3}{5} x) - sin x) + C$ $=5/48(6sin(x/5)-5sin(3/5x)-sinx)+C$

Explanation:

Use #sin 2A = 2 sin A cos A, sin^2 A =(1-cos 2A)/2 and cos (A + B ) +

cos ( A - B )=2 cos A cos B#.

The integrand is

${(sin (\frac{x}{5}) cos (\frac{x}{5}))}^{2} cos (\frac{x}{5})$ $(sin(x/5)cos(x/5))^2cos(x/5)$

$= \frac{1}{4} {sin}^{2} (\frac{2}{5} x) cos (\frac{x}{5})$ $=1/4sin^2(2/5x)cos(x/5)$

$= \frac{1}{8} (1 - cos (\frac{4}{5} x)) cos (\frac{x}{5})$ $=1/8(1-cos(4/5x))cos(x/5)$

$= \frac{1}{8} (cos (\frac{x}{5}) - \frac{1}{2} (cos x + cos (\frac{3}{5} x)))$ $=1/8(cos(x/5)-1/2(cosx + cos(3/5x)))$ .

So, the integral is

$\frac{1}{16} [2 \int cos (\frac{x}{5}) d x - \int cos x d x - \int cos (\frac{3}{5} x) d x]$ $1/16[2int cos(x/5) dx- int cos x dx - int cos(3/5x) dx]$

$\frac{1}{16} [10 sin (\frac{x}{5}) - sin x - \frac{5}{3} sin (\frac{3}{5} x)] + C$ $1/16[10 sin(x/5)-sin x-5/3sin(3/5x)] +C$

$= \frac{5}{48} (6 sin (\frac{x}{5}) - 5 sin (\frac{3}{5} x) - sin x) + C$ $=5/48(6sin(x/5)-5sin(3/5x)-sinx)+C$

Answer 2 · 2016-11-27T13:28:15

$= \frac{5}{3} {sin}^{3} (\frac{x}{5}) - {sin}^{5} (\frac{x}{5}) + C$ $=5/3sin^3(x/5)-sin^5(x/5)+C$

Explanation:

Let $u = sin (\frac{x}{5})$ $u=sin(x/5)$
$" "$
$d u = \frac{1}{5} cos (\frac{x}{5}) d x \Rightarrow 5 d u = cos (\frac{x}{5}) d x$ $color(blue)(du=1/5cos(x/5)dxrArr5du=cos(x/5)dx$
$" "$
Let us start computing the integral by substituting $u and d u$ $" "u" and "du$
$" "$
then , using the trigonometric identity $" "color(purple)$ " " $({cos}^{2} α = 1 - {sin}^{2} α)$ $(cos^2alpha=1-sin^2alpha)$ .
$" "$
$\int {(sin (\frac{x}{5}))}^{2} \times {(cos (\frac{x}{5}))}^{3} d x$ $int(sin(x/5))^2xx(cos(x/5))^3dx$
$" "$
$= \int {(sin (\frac{x}{5}))}^{2} \times {(cos (\frac{x}{5}))}^{2} \times cos (\frac{x}{5}) d x$ $=int(sin(x/5))^2xx(cos(x/5))^2xxcos(x/5)dx$
$" "$
$= \int {(sin (\frac{x}{5}))}^{2} \times {(cos (\frac{x}{5}))}^{2} \times 5 d u$ $=int(sin(x/5))^2xx(cos(x/5))^2xxcolor(blue)(5du)$
$" "$
$= \int u^{2} \times (1 - {sin}^{2} (\frac{x}{5})) \times 5 d u$ $=intu^2xx(1-sin^2(x/5))xx5du$
$" "$
$= \int u^{2} \times (1 - u^{2}) \times 5 d u$ $=intu^2xx(1-u^2)xx5du$
$" "$
$= 5 \int u^{2} - u^{4} d u$ $=5intu^2-u^4du$
$" "$
$= 5 (\int u^{2} d u - \int u^{4}) d u$ $=5(intu^2du-intu^4)du$
$" "$
$= 5 \int u^{2} d u - 5 \int u^{4} d u$ $=5intu^2du-5intu^4du$
$" "$
$= \frac{5}{3} u^{3} - \frac{5}{5} u^{5} + C, C$ $=5/3u^3-5/5u^5+C" ,"C " "$ is a constant.
$" "$
$= \frac{5}{3} u^{3} - u^{5} + C$ $=5/3u^3-u^5+C$
$" "$
$= \frac{5}{3} {sin}^{3} (\frac{x}{5}) - {sin}^{5} (\frac{x}{5}) + C$ $=5/3sin^3(x/5)-sin^5(x/5)+C$

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How do you find the integral of $({(sin (\frac{x}{5}))}^{2}) \cdot ({(cos (\frac{x}{5}))}^{3})$ $((sin(x/5))^2)*((cos(x/5))^3)$ ?

2 Answers

Explanation:

Explanation:

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How do you find the integral of ((sin(x/5))^2)*((cos(x/5))^3)((sin(x5))2)⋅((cos(x5))3)((sin(x/5))^2)*((cos(x/5))^3)?

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Explanation:

Explanation:

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How do you find the integral of $({(sin (\frac{x}{5}))}^{2}) \cdot ({(cos (\frac{x}{5}))}^{3})$ $((sin(x/5))^2)*((cos(x/5))^3)$ ?