How do you integrate $\sqrt{1 - x^{2}}$ $sqrt(1-x^2)$ ?

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Answer 1 · 2018-03-29T06:23:01

The answer is $= \frac{1}{2} arcsin x + \frac{1}{2} x \sqrt{1 - x^{2}} + C$ $=1/2arcsinx+1/2xsqrt(1-x^2)+C$

Let $x = sin θ$ $x=sintheta$ , $\Rightarrow$ $=>$ , $d x = cos θ d θ$ $dx=costhetad theta$

$cos θ = \sqrt{1 - x^{2}}$ $costheta=sqrt(1-x^2)$

$sin 2 θ = 2 sin θ cos θ = 2 x \sqrt{1 - x^{2}}$ $sin2theta=2sinthetacostheta=2xsqrt(1-x^2)$

Therefore, the integral is

$I = \int \sqrt{1 - x^{2}} d x = \int cos θ \cdot cos θ d θ$ $I=intsqrt(1-x^2)dx=intcostheta*costheta d theta$

$= \int {cos}^{2} θ d θ$ $=intcos^2thetad theta$

$cos 2 θ = 2 {cos}^{2} θ - 1$ $cos2theta=2cos^2theta-1$

${cos}^{2} θ = \frac{1 + cos 2 θ}{2}$ $cos^2theta=(1+cos2theta)/2$

Therefore,

$I = \frac{1}{2} \int (1 + cos 2 θ) d θ$ $I=1/2int(1+cos2theta)d theta$

$= \frac{1}{2} (θ + \frac{1}{2} sin 2 θ)$ $=1/2(theta+1/2sin2theta)$

$= \frac{1}{2} arcsin x + \frac{1}{2} x \sqrt{1 - x^{2}} + C$ $=1/2arcsinx+1/2xsqrt(1-x^2)+C$

How do you integrate √1−x2sqrt(1-x^2)?