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

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

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Answer 1 · 2017-01-12T12:42:39

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

Explanation:

As the integrand function is defined for $x \in [- 1, 1]$ $x in [-1,1]$ , you can substitute:

$x = sin t$ $x= sint$ with $t \in [- \frac{π}{2}, \frac{π}{2}]$ $t in [-pi/2,pi/2]$

$d x = cos t d t$ $dx = costdt$

so the integral becomes:

$\int \sqrt{1 - x^{2}} d x = \int \sqrt{1 - {sin}^{2} t} cos t d t = \int \sqrt{{cos}^{2} t} cos t d t$ $int sqrt(1-x^2)dx = int sqrt(1-sin^2t)costdt = int sqrt(cos^2t) cost dt$

In the given interval $cos t$ $cost$ is positive, so $\sqrt{{cos}^{2} t} = cos t$ $sqrt(cos^2t) = cost$ :

$\int \sqrt{1 - x^{2}} d x = \int {cos}^{2} t d t$ $int sqrt(1-x^2)dx = int cos^2t dt$

Now we can use the identity:

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

$\int \sqrt{1 - x^{2}} d x = \int \frac{1 + cos (2 t)}{2} d t = \int \frac{d t}{2} + \frac{1}{4} \int cos (2 t) d (2 t) = \frac{1}{2} t + \frac{1}{4} sin 2 t = \frac{1}{2} (t + sin t cos t)$ $int sqrt(1-x^2)dx = int (1+cos(2t))/2 dt = int (dt)/2 + 1/4 int cos(2t)d(2t)= 1/2t+1/4sin2t = 1/2(t+sintcost)$

To substitute back $x$ $x$ we note that:

$x = sin t$ $x = sint$ for $t \in [- \frac{π}{2}, \frac{π}{2}] \Rightarrow t = arcsin x$ $t in [-pi/2,pi/2] => t = arcsinx$

$cos t = \sqrt{1 - {sin}^{2} t} = \sqrt{1 - x^{2}}$ $cost = sqrt(1-sin^2t) = sqrt(1-x^2)$

Finally:

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

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

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