How do you find the integral of $\int [{cos}^{3} (2 x)] d x$ $int [cos^3 (2x)] dx$ ?

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How do you find the integral of $\int [{cos}^{3} (2 x)] d x$ $int [cos^3 (2x)] dx$ ?

1 Answer

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Answer 1 · 2015-09-21T13:14:48

$\frac{1}{2} sin 2 x - \frac{1}{6} {sin}^{3} 2 x + C$ $1/2sin 2x-1/6sin^3 2x+C$

Explanation:

$I = \int {cos}^{3} 2 x d x = \int cos 2 x {cos}^{2} 2 x d x$ $I=int cos^3 2xdx = int cos 2x cos^2 2x dx$
$I = \int cos 2 x (1 - {sin}^{2} 2 x) d x$ $I=int cos 2x (1-sin^2 2x)dx$

$sin 2 x = t \Rightarrow 2 cos 2 x d x = d t \Rightarrow cos 2 x d x = \frac{d t}{2}$ $sin 2x=t => 2cos 2xdx = dt => cos 2xdx = dt/2$

$I = \int (1 - t^{2}) \frac{d t}{2} = \frac{1}{2} (t - \frac{t^{3}}{3}) + C$ $I=int(1-t^2)dt/2=1/2(t-t^3/3)+C$

$I = \frac{1}{2} sin 2 x - \frac{1}{6} {sin}^{3} 2 x + C$ $I=1/2sin 2x-1/6sin^3 2x+C$

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How do you find the integral of $\int [{cos}^{3} (2 x)] d x$ $int [cos^3 (2x)] dx$ ?

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Explanation:

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How do you find the integral of int [cos^3 (2x)] dx ∫[cos3(2x)]dxint [cos^3 (2x)] dx ?

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How do you find the integral of $\int [{cos}^{3} (2 x)] d x$ $int [cos^3 (2x)] dx$ ?