How do you find the antiderivative of ${cos}^{4} (x) d x$ $cos^4 (x) dx$ ?

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How do you find the antiderivative of ${cos}^{4} (x) d x$ $cos^4 (x) dx$ ?

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Answer 1 · 2016-06-28T19:54:55

You want to split it up using trig identities to get nice, easy integrals.

Explanation:

${cos}^{4} (x) = {cos}^{2} (x) \cdot {cos}^{2} (x)$ $cos^4(x) = cos^2(x)*cos^2(x)$

We can deal with the ${cos}^{2} (x)$ $cos^2(x)$ easily enough by rearranging the double angle cosine formula.

${cos}^{4} (x) = \frac{1}{2} (1 + cos (2 x)) \cdot \frac{1}{2} (1 + cos (2 x))$ $cos^4(x) = 1/2(1 + cos(2x))*1/2(1 + cos(2x))$

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

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

${cos}^{4} (x) = \frac{3}{8} + \frac{1}{2} \cdot cos (2 x) + \frac{1}{8} \cdot cos (4 x)$ $cos^4(x) = 3/8 + 1/2*cos(2x) + 1/8*cos(4x)$

So,

$\int {cos}^{4} (x) d x = \frac{3}{8} \cdot \int d x + \frac{1}{2} \cdot \int cos (2 x) d x + \frac{1}{8} \cdot \int cos (4 x) d x$ $int cos^4(x) dx = 3/8*int dx + 1/2*int cos(2x) dx + 1/8*int cos(4x) dx$

$\int {cos}^{4} (x) d x = \frac{3}{8} x + \frac{1}{4} \cdot sin (2 x) + \frac{1}{32} \cdot sin (4 x) + C$ $int cos^4(x) dx= 3/8x + 1/4*sin(2x) + 1/32*sin(4x) + C$

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How do you find the antiderivative of ${cos}^{4} (x) d x$ $cos^4 (x) dx$ ?

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How do you find the antiderivative of ${cos}^{4} (x) d x$ $cos^4 (x) dx$ ?