# How do you find the integral of  cos^4 (x) dx?

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#### Explanation

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#### Explanation:

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Feb 28, 2017

$\int {\cos}^{4} x \mathrm{dx} = \frac{\sin x {\cos}^{3} x}{4} + \frac{3}{8} \left(\cos x \sin x\right) + \frac{3}{8} x$

#### Explanation:

Write: ${\cos}^{4} x = {\cos}^{3} x \cdot \cos x$ and integrate by parts:

$\int {\cos}^{4} x \mathrm{dx} = \int {\cos}^{3} x \cos x \mathrm{dx} = \int {\cos}^{3} x d \left(\sin x\right)$

$\int {\cos}^{4} x \mathrm{dx} = \sin x {\cos}^{3} x - \int \sin x d \left({\cos}^{3} x\right)$

$\int {\cos}^{4} x \mathrm{dx} = \sin x {\cos}^{3} x + 3 \int {\sin}^{2} x {\cos}^{2} x \mathrm{dx}$

Now use the identity:

${\sin}^{2} x = 1 - {\cos}^{2} x$

$\int {\cos}^{4} x \mathrm{dx} = \sin x {\cos}^{3} x + 3 \int \left(1 - {\cos}^{2} x\right) {\cos}^{2} x \mathrm{dx}$

$\int {\cos}^{4} x \mathrm{dx} = \sin x {\cos}^{3} x + 3 \int {\cos}^{2} x \mathrm{dx} - 3 \int {\cos}^{4} x \mathrm{dx}$

We have now the same integral on both sides and we can solve for it:

$4 \int {\cos}^{4} x \mathrm{dx} = \sin x {\cos}^{3} x + 3 \int {\cos}^{2} x \mathrm{dx}$

$\int {\cos}^{4} x \mathrm{dx} = \frac{\sin x {\cos}^{3} x}{4} + \frac{3}{4} \int {\cos}^{2} x \mathrm{dx}$

Using the same process:

$\int {\cos}^{2} x \mathrm{dx} = \int \cos x d \left(\sin x\right) = \cos x \sin x + \int {\sin}^{2} x \mathrm{dx}$

$\int {\cos}^{2} x \mathrm{dx} = \int \cos x d \left(\sin x\right) = \cos x \sin x + \int \left(1 - {\cos}^{2} x\right) \mathrm{dx}$

$\int {\cos}^{2} x \mathrm{dx} = \int \cos x d \left(\sin x\right) = \cos x \sin x + x - \int {\cos}^{2} x \mathrm{dx}$

$\int {\cos}^{2} x \mathrm{dx} = \frac{\cos x \sin x}{2} + \frac{x}{2}$

Substituting in the expression above:

$\int {\cos}^{4} x \mathrm{dx} = \frac{\sin x {\cos}^{3} x}{4} + \frac{3}{8} \left(\cos x \sin x\right) + \frac{3}{8} x$

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