How do you Integrate ?

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How do you Integrate ?

$\int_{0}^{2 x} \sqrt{1 + sin (\frac{x}{2})} d x$ $int_0^(2x)sqrt(1+sin(x/2))dx$

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Answer 1 · 2018-05-04T19:04:59

$sin (\frac{x}{2}) + cos (\frac{x}{2}) - 1$ $sin(x/2)+cos(x/2)-1$

Explanation:

$\int_{0}^{2 x} \sqrt{1 + sin (\frac{x}{2})} \cdot d x$ $int_0^(2x)sqrt(1+sin(x/2))*dx$

$sin (\frac{x}{2}) = 2 sin (\frac{x}{4}) cos (\frac{x}{4})$ $color(green)(sin(x/2)=2sin(x/4)cos(x/4)$

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

Substitute

$\int_{0}^{2 x} \sqrt{1 + sin (\frac{x}{2})} \cdot d x$ $int_0^(2x)sqrt(1+sin(x/2))*dx$

$= \int_{0}^{2 x} \sqrt{{sin}^{2} (\frac{x}{4}) + 2 sin (\frac{x}{4}) cos (\frac{x}{4}) + {cos}^{2} (\frac{x}{4})} \cdot d x$ $=int_0^(2x)sqrt(sin^2(x/4)+2sin(x/4)cos(x/4)+cos^2(x/4))*dx$

By Completing The Square

$= \int_{0}^{2 x} \sqrt{{(sin (\frac{x}{4}) + cos (\frac{x}{4}))}^{2}} \cdot d x$ $=int_0^(2x)sqrt((sin(x/4)+cos(x/4))^2)*dx$

$= \int_{0}^{2 x} (sin (\frac{x}{4}) + cos (\frac{x}{4})) \cdot d x$ $=int_0^(2x)(sin(x/4)+cos(x/4))*dx$

$= {[sin (\frac{x}{4}) + cos (\frac{x}{4})]}_{0}^{2 x}$ $=[sin(x/4)+cos(x/4)]_0^(2x)$

$= sin (\frac{x}{2}) + cos (\frac{x}{2}) - 1$ $=sin(x/2)+cos(x/2)-1$

$if this question was like this \sqrt{1 + cos (\frac{x}{2})} it can be simplified to look like this$ $color(blue)("if this question was like this " sqrt(1+cos(x/2)) " it can be simplified to look like this"$

$\sqrt{2} \cdot cos (\frac{x}{4}) using the Double angles Formulae$ $color(blue)(sqrt2*cos(x/4) "using the Double angles Formulae"$

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How do you Integrate ?

$\int_{0}^{2 x} \sqrt{1 + sin (\frac{x}{2})} d x$ $int_0^(2x)sqrt(1+sin(x/2))dx$

1 Answer

Explanation:

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$\int_{0}^{2 x} \sqrt{1 + sin (\frac{x}{2})} d x$ $int_0^(2x)sqrt(1+sin(x/2))dx$