How do I evaluate the indefinite integral $\int {(1 + cos (x))}^{2} d x$ $int(1+cos(x))^2dx$ ?

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How do I evaluate the indefinite integral $\int {(1 + cos (x))}^{2} d x$ $int(1+cos(x))^2dx$ ?

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Answer 1 · 2014-09-08T05:02:38

$\int {(1 + cos x)}^{2} d x = \frac{3}{2} x + 2 sin x + \frac{1}{4} sin 2 x + C$ $int(1+cosx)^2dx=3/2x+2sinx+1/4sin2x+C$

Let us look at some details.
By multiplying out the integrand,
$\int {(1 + cos x)}^{2} d x = \int (1 + 2 cos x + {cos}^{2} x) d x$ $int(1+cosx)^2dx=int(1+2cosx+cos^2x) dx$
by ${cos}^{2} x = \frac{1 + cos 2 x}{2}$ $cos^2x={1+cos2x}/2$ ,
$= \int (1 + 2 cos x + \frac{1 + cos 2 x}{2}) d x$ $=int(1+2cosx+{1+cos2x}/2)dx$
by cleaning us the integrand,
$= \int (\frac{3}{2} + 2 cos x + \frac{1}{2} cos 2 x) d x$ $=int(3/2+2cosx+1/2cos2x)dx$
by finding antiderivatives,
$= \frac{3}{2} x + 2 sin x + \frac{1}{4} sin 2 x + C$ $=3/2x+2sinx+1/4sin2x+C$

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How do I evaluate the indefinite integral $\int {(1 + cos (x))}^{2} d x$ $int(1+cos(x))^2dx$ ?

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How do I evaluate the indefinite integral $\int {(1 + cos (x))}^{2} d x$ $int(1+cos(x))^2dx$ ?