Question

7) Calcule a integral

$\int\ {} \frac{ \sqrt{16-t^2} }{t^2} \, dt$

Answer 1

$I=\displaystyle\int\!\frac{\sqrt{16-t^2}}{t^2}\,dt~~~~~~\mathbf{(i)}$


Faça a seguinte substituição (trigonométrica):

$t=4\,\mathrm{sen\,}u~~~(-\pi/2<u<\pi/2)~~\Rightarrow~~\left\{ \begin{array}{l} dt=4\cos u\,du\\\\ u=\mathrm{arcsen}\left(\dfrac{t}{4} \right ) \end{array} \right.\\\\\\ 16-t^2=16-(4\,\mathrm{sen\,}u)^2\\\\ 16-t^2=16-16\,\mathrm{sen^2\,}u\\\\ 16-t^2=16\cdot (1-\mathrm{sen^2\,}u)\\\\ 16-t^2=16\cos^2 u\\\\ \sqrt{16-t^2}=\sqrt{16\cos^2 u}\\\\ \sqrt{16-t^2}=4\cos u~~~~(\text{pois }-\pi/2<u<\pi/2)$


Substituindo, a integral fica

$=\displaystyle\int\!\frac{4\cos u}{(4\,\mathrm{sen\,}u)^2}\cdot 4\cos u\,du\\\\\\ =\int\!\frac{16\cos^2 u}{16\,\mathrm{sen^2\,}u}\,du\\\\\\ =\int\!\left(\frac{\cos u}{\mathrm{sen\,}u} \right )^{\!\!2}du\\\\\\ =\int\!\mathrm{cotg^2\,}u\,du\\\\\\ =\int\!(\mathrm{cossec^2\,}u-1)\,du\\\\\\ =-\,\mathrm{cotg\,}u-u+C\\\\ =-\,\frac{\cos u}{\mathrm{sen\,}u}-u+C~~~~~~\mathbf{(ii)}$


Para voltar para a variável original $t,$ basta observarmos as equações que usamos na substituição.

(Caso queira, pode observar o triângulo retângulo em anexo)

$=-\,\dfrac{\left(\frac{\sqrt{16-t^2}}{4} \right )}{\frac{t}{4}}-\mathrm{arcsen}\left(\dfrac{t}{4} \right )+C\\\\\\ =-\,\dfrac{\sqrt{16-t^2}}{\diagup\!\!\!\! 4}\cdot \dfrac{\diagup\!\!\!\! 4}{t}-\mathrm{arcsen}\left(\dfrac{t}{4} \right )+C\\\\\\ =-\,\dfrac{\sqrt{16-t^2}}{t}-\mathrm{arcsen}\left(\dfrac{t}{4} \right )+C\\\\\\\\ \therefore~~\boxed{\begin{array}{c} \displaystyle\int\!\frac{\sqrt{16-t^2}}{t^2}\,dt=-\,\dfrac{\sqrt{16-t^2}}{t}-\mathrm{arcsen}\left(\dfrac{t}{4} \right )+C \end{array}}$


Bons estudos! :-)

Answer 2

Oi Lucas

∫ √(16 - t²)/t² dt 

t = 4sen(u) e dt = 4cos(u) 

√(16 - 14sen²(u)) = 4cos(u) e u = sen^(-1)(t/4) 

∫ √(16 - t²)/t² dt  = 4 ∫ cotg²(u)/4 du 

∫ cotg²(u) du = ∫ csc²(u) du - ∫ 1 du = -cotg(u) - u 

então 

∫ √(16 - t²)/t² dt = - (√(16 - t²)  + t*arcsen(t/4))/t + C