How do you integrate $\int {(2 + sin (\frac{x}{2}))}^{2} cos (\frac{x}{2}) d x$ $int (2+sin(x/2))^2 cos(x/2)dx$ using integration by parts?

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How do you integrate $\int {(2 + sin (\frac{x}{2}))}^{2} cos (\frac{x}{2}) d x$ $int (2+sin(x/2))^2 cos(x/2)dx$ using integration by parts?

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Answer 1 · 2016-01-09T16:36:39

$I = 10 sin (\frac{x}{2}) - 2 cos (x) - 2 sin (\frac{x}{2}) + \frac{2}{3} {sin}^{3} (\frac{x}{2}) + c$ $I = 10sin(x/2) -2cos(x) -2sin(x/2) + 2/3sin^3(x/2) + c$

Explanation:

For a non-parts approach

$I = \int {(2 + sin (\frac{x}{2}))}^{2} cos (\frac{x}{2}) d x$ $I = int(2+sin(x/2))^2cos(x/2)dx$

Expanding that

$I = \int (4 + 4 sin (\frac{x}{2}) + {sin}^{2} (\frac{x}{2})) cos (\frac{x}{2}) d x$ $I = int(4+4sin(x/2)+sin^2(x/2))cos(x/2)dx$

And that

$I = \int 4 cos (\frac{x}{2}) d x + \int 4 sin (\frac{x}{2}) cos (\frac{x}{2}) d x + \int {sin}^{2} (\frac{x}{2}) cos (\frac{x}{2}) d x$ $I = int4cos(x/2)dx + int4sin(x/2)cos(x/2)dx + intsin^2(x/2)cos(x/2)dx$

The first integral is easy

$I = 8 sin (\frac{x}{2}) + \int 4 sin (\frac{x}{2}) cos (\frac{x}{2}) d x + \int {sin}^{2} (\frac{x}{2}) cos (\frac{x}{2}) d x$ $I = 8sin(x/2) + int4sin(x/2)cos(x/2)dx + intsin^2(x/2)cos(x/2)dx$

The second is easy if you remember that

$sin (α x) cos (α x) = \frac{sin (2 α x)}{2}$ $sin(alphax)cos(alphax) = sin(2alphax)/2$ , so

$I = 8 sin (\frac{x}{2}) + \int 2 sin (x) d x + \int {sin}^{2} (\frac{x}{2}) cos (\frac{x}{2}) d x$ $I = 8sin(x/2) + int2sin(x)dx + intsin^2(x/2)cos(x/2)dx$
$I = 8 sin (\frac{x}{2}) - 2 cos (x) + \int {sin}^{2} (\frac{x}{2}) cos (\frac{x}{2}) d x$ $I = 8sin(x/2) -2cos(x) + intsin^2(x/2)cos(x/2)dx$

If you remember ${sin}^{2} (θ) = 1 - {cos}^{2} (θ)$ $sin^2(theta) = 1 -cos^2(theta)$

$I = 8 sin (\frac{x}{2}) - 2 cos (x) + \int (1 - {cos}^{2} (\frac{x}{2})) cos (\frac{x}{2}) d x$ $I = 8sin(x/2) -2cos(x) + int(1 - cos^2(x/2))cos(x/2)dx$
$I = 8 sin (\frac{x}{2}) - 2 cos (x) + \int cos (\frac{x}{2}) d x - \int {cos}^{3} (\frac{x}{2}) d x$ $I = 8sin(x/2) -2cos(x) + intcos(x/2)dx - intcos^3(x/2)dx$
$I = 8 sin (\frac{x}{2}) - 2 cos (x) + 2 sin (\frac{x}{2}) - \int {cos}^{3} (\frac{x}{2}) d x$ $I = 8sin(x/2) -2cos(x) + 2sin(x/2)- intcos^3(x/2)dx$

And now, use ${cos}^{2} (\frac{x}{2}) = 1 - {sin}^{2} (\frac{x}{2})$ $cos^2(x/2) = 1 - sin^2(x/2)$

$I = 10 sin (\frac{x}{2}) - 2 cos (x) - \int (1 - {sin}^{2} (\frac{x}{2})) cos (\frac{x}{2}) d x$ $I = 10sin(x/2) -2cos(x) - int(1-sin^2(x/2))cos(x/2)dx$

Say $u = sin (\frac{x}{2})$ $u = sin(x/2)$ so $d u = \frac{cos (\frac{x}{2})}{2}$ $du = cos(x/2)/2$

$I = 10 sin (\frac{x}{2}) - 2 cos (x) - 2 \int (1 - {sin}^{2} (\frac{x}{2})) \frac{cos (\frac{x}{2})}{2} d x$ $I = 10sin(x/2) -2cos(x) -2 int(1-sin^2(x/2))cos(x/2)/2dx$
$I = 10 sin (\frac{x}{2}) - 2 cos (x) - 2 \int (1 - u^{2}) d u$ $I = 10sin(x/2) -2cos(x) -2 int(1-u^2)du$
$I = 10 sin (\frac{x}{2}) - 2 cos (x) - 2 u + \frac{2 u^{3}}{3} + c$ $I = 10sin(x/2) -2cos(x) -2u + (2u^3)/3 + c$

Switching it back to $x$ $x$

$I = 10 sin (\frac{x}{2}) - 2 cos (x) - 2 sin (\frac{x}{2}) + \frac{2}{3} {sin}^{3} (\frac{x}{2}) + c$ $I = 10sin(x/2) -2cos(x) -2sin(x/2) + 2/3sin^3(x/2) + c$

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How do you integrate $\int {(2 + sin (\frac{x}{2}))}^{2} cos (\frac{x}{2}) d x$ $int (2+sin(x/2))^2 cos(x/2)dx$ using integration by parts?

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