How do you integrate $\int \sqrt{1 - 4 x^{2}}$ $intsqrt(1-4x^2)$ by trigonometric substitution?

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Answer 1 · 2016-10-10T16:30:08

$\frac{{sin}^{2} (θ)}{2} + c$ $sin^2(theta)/2+c$

As shown in the picture below our equation matches the " $sin$ $sin$ " trigonometric substitution.
enter image source here

Therefor,

$x = \frac{sin (θ)}{2}$ $x=sin(theta)/2$

$\sqrt{1 - 4 x^{2}} = cos (θ)$ $sqrt(1-4x^2)=cos(theta)$

$d x = \frac{2 cos (θ)}{4} d θ$ $dx=(2cos(theta))/4 d theta$ = $\frac{cos (θ)}{4}$ $cos(theta)/4$

$\int cos (θ) \frac{cos (θ)}{2}$ $intcos(theta)(cos(theta))/2$ = $\frac{1}{2} \int {cos}^{2} (θ) d θ$ $1/2intcos^2(theta)d theta$

$\frac{1}{2} {sin}^{2} (θ) + c$ $1/2sin^2(theta)+c$ or $\frac{{sin}^{2} (θ)}{2} + c$ $sin^2(theta)/2+c$

Good Luck!!!

How do you integrate ∫√1−4x2intsqrt(1-4x^2) by trigonometric substitution?