Question #8e0d6

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Question #8e0d6

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Answer 1 · 2016-10-17T17:07:14

$= 16 (\frac{{cos}^{7} (\frac{x}{2})}{7} - \frac{{cos}^{5} (\frac{x}{2})}{5}) + C$ $=16(cos^7(x/2)/7-cos^5(x/2)/5)+C$

Explanation:

Here are the steps
$\int {sin}^{3} x cos (\frac{x}{2}) d x$ $intsin^3xcos(x/2)dx$
So we start with the substitution $u = \frac{x}{2}$ $u=x/2$ so $d u = \frac{d x}{2}$ $du=dx/2$
$\int {sin}^{3} x cos (\frac{x}{2}) d x = 2 \int {sin}^{3} (2 u) cos u d u$ $intsin^3xcos(x/2)dx=2intsin^3(2u)cosudu$
$= 2 \int {(2 sin u cos u)}^{3} cos u d u$ $=2int(2sinucosu)^3cosudu$
as $sin 2 u = 2 sin u cos u$ $sin2u=2sinucosu$
$= 16 \int {sin}^{2} u sin u {cos}^{4} u d u$ $=16intsin^2usinucos^4udu$
$= 16 \int (1 - {cos}^{2} u) {cos}^{4} u sin u d u$ $=16int(1-cos^2u)cos^4usinudu$
Now we use $v = cos u$ $v=cosu$ $d v = - sin u d u$ $dv=-sinudu$
Integral $= 16 \int (1 - v^{2}) v^{4} - d v$ $=16int(1-v^2)v^4-dv$
$= 16 \int (v^{6} - v^{4}) d v$ $=16int(v^6-v^4)dv$
$= 16 (\frac{v^{7}}{7} - \frac{v^{5}}{5})$ $=16(v^7/7-v^5/5)$
$= 16 (\frac{{cos}^{7} u}{7} - \frac{{cos}^{5} u}{5})$ $=16(cos^7u/7-cos^5u/5)$
$= 16 (\frac{{cos}^{7} (\frac{x}{2})}{7} - \frac{{cos}^{5} (\frac{x}{2})}{5}) + C$ $=16(cos^7(x/2)/7-cos^5(x/2)/5)+C$

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Question #8e0d6

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