How do you find the antiderivative of ${cos}^{2} (x)$ $cos^2 (x)$ ?

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How do you find the antiderivative of ${cos}^{2} (x)$ $cos^2 (x)$ ?

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Answer 1 · 2016-07-30T09:44:34

$\int {cos}^{2} (x) d x = \frac{x}{2} + \frac{cos (x) sin (x)}{2} + C$ $int cos^2(x) d x=x/2+(cos(x) sin(x))/2+C$

Explanation:

$\int {cos}^{2} (x) d x = ?$ $int cos^2(x) d x=?$

$let us use the reduction formula :$ $"let us use the reduction formula :"$

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

$Apply n=2$ $"Apply n=2"$

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

$\int {cos}^{2} (x) d x = \frac{1}{2} \int {cos}^{0} (x) d x + \frac{cos (x) sin (x)}{2}$ $int cos^2(x) d x=1/2 int cos^0 (x) d x +(cos (x) sin (x))/2$

$\int {cos}^{2} (x) d x = \frac{1}{2} \int d x + \frac{cos (x) sin (x)}{2}$ $int cos^2(x) d x=1/2 int d x+(cos (x) sin(x))/2$

$\int {cos}^{2} (x) d x = \frac{1}{2} x + \frac{cos (x) sin (x)}{2}$ $int cos^2(x) d x=1/2 x +(cos(x) sin(x))/2$

$\int {cos}^{2} (x) d x = \frac{x}{2} + \frac{cos (x) sin (x)}{2} + C$ $int cos^2(x) d x=x/2+(cos(x) sin(x))/2+C$

Answer 2 · 2016-07-30T10:37:37

Remember that $cos 2 x = {cos}^{2} x - {sin}^{2} x \Rightarrow cos 2 x = 2 {cos}^{2} x - 1 \Rightarrow {cos}^{2} x = \frac{1}{2} (1 + cos 2 x)$ $cos2x=cos^2x-sin^2x=>cos2x=2cos^2x-1=> cos^2x=1/2(1+cos2x)$

Hence

$\int {cos}^{2} x d x = \int \frac{1}{2} (1 + cos 2 x) d x = \frac{1}{2} \int 1 d x + \frac{1}{2} \int cos 2 x d x = \frac{1}{2} x + \frac{1}{4} sin 2 x + c$ $int cos^2xdx=int 1/2 (1+cos2x)dx=1/2int 1dx+1/2intcos2xdx= 1/2x+1/4sin2x+c$

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