Find integral of the following?

Question

$\int {cos}^{2} x d x$ $intcos^2x dx$

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Answer 1 · 2018-04-21T17:21:24

the answer $\frac{1}{2} [x + \frac{1}{2} sin (2 x)] + c$ $1/2[x+1/2sin(2x)]+c$

$\int {cos}^{2} (x) \cdot d x$ $intcos^2(x)*dx$

$\frac{1}{2} \int (1 + cos (2 x)) \cdot d x$ $1/2int(1+cos(2x))*dx$

$\frac{1}{2} [x + \frac{1}{2} sin (2 x)] + c$ $1/2[x+1/2sin(2x)]+c$

Note That:

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

${sin}^{2} (x) = \frac{1}{2} [1 - cos (2 x)]$ $sin^2(x)=1/2[1-cos(2x)]$

Answer 2 · 2018-04-21T17:24:12

$I = \frac{1}{4} (2 x + sin 2 x) + c$ $I=1/4(2x+sin2x)+c$

Here,

$I = \int {cos}^{2} x d x$ $I=intcos^2xdx$

$= \int \frac{1 + cos 2 x}{2}$ $=int(1+cos2x)/2$

$= \frac{1}{2} [x + \frac{sin (2 x)}{2}] + c$ $=1/2[x+sin(2x)/2]+c$

$= \frac{1}{4} (2 x + sin 2 x) + c$ $=1/4(2x+sin2x)+c$