Question

∫√1-4x². dx. Mais uma vez, como calcular essa integral por substitiuição trigométrica, por gentileza.

Answer 1

Explicação:

$\int \sqrt{1-4x^2}dx \\\\\\\textsf{Seja: }x = \frac{sen\theta}{2} , \theta \in [-\pi/2,\pi/2]\\\\\\\int \sqrt{1-4.(\frac{sen\theta}{2})^2}dx = \\\\\int \sqrt{1-sen^2\theta}dx=~~~[aplique: sen^2\theta + cos^2\theta = 1]\\\\\ \int \sqrt{cos^2\theta}dx =~~~~~~~~~[cos\theta~positivo~no~intervalo~[-\pi/2,\pi/2]]\\\\\int cos\theta dx =~~~~~~~~~~~~~[dx = \frac{1}{2}.cos\theta}.d\theta]\\\\\int cos\theta.\frac{1}{2}.cos\theta.d\theta =$

$\frac{1}{2}.\int cos^2\theta.d\theta =~~~~~~~~[cos^2\theta =\frac{1+cos(2\theta)}{2}]\\\\\frac{1}{2}.\int (\frac{1+cos(2\theta)}{2})d\theta =\\\\\ \frac{1}{4}.\int 1 + cos(2\theta}).d\theta = \\\\ \frac{1}{4}.(\theta + \frac{sen(2\theta)}{2}) + C = \\\\\frac{\theta}{4} + \frac{sen(2\theta)}{8}} + C$

Sabemos que:

$x = \frac{sen\theta}{2}\\\\\boxed{sen\theta = 2x}\\\\\boxed{\theta = arcsen{(2x)}}$

Logo:

$\frac{arcsen(2x)}{4} + \frac{2.sen(\theta).cos(\theta)}{8}} + C = \\\\\frac{arcsen(2x)}{4} + \frac{2.2x.cos(\theta)}{8}} + C =~~~~[cos\theta = \sqrt{1-4x^2}] \\\\\boxed{\int \sqrt{1-4x^2}dx = \frac{arcsen(2x)}{4} + \frac{x.\sqrt{1-4x^2}}{2} + C}$