Question

Как решить интеграл x^2*dx/sqrt(2-x^2) методом замены

Answer 1

Ответ:

$\int \dfrac{x^2\, dx}{\sqrt{2-x^2}}=\Big[\ x=\sqrt2sint\ ,\ dx=\sqrt2\, cost\, dt\ \ ,\ \ sint=\dfrac{x}{\sqrt2}\ ,\\\\\\2-x^2=2(1-sin^2t)=2cos^2t\ \Big]=\int \dfrac{2\, sin^2t}{\sqrt{2\, cos^2t}}\cdot \sqrt2\, cost\, dt=\int \dfrac{2\sqrt2\, sin^2t\cdot cost}{\sqrt2\, cost}\, dt=\\\\\\=\int 2sin^2t\, dt=\int (1-cos2t)\, dt=\int \, dt-\int cos2t\, dt=t-\dfrac{1}{2}\cdot sin2t+C=\\\\\\=arcsin\dfrac{x}{\sqrt2}-\dfrac{1}{2}\cdot sin\Big(2arcsin\dfrac{x}{\sqrt2}\Big)+C=$

$=arcsin\dfrac{x}{\sqrt2}-\dfrac{1}{2}\cdot 2\, sin\Big(arcsin\dfrac{x}{\sqrt2}\Big)\cdot cos\Big(arcsin\dfrac{x}{\sqrt2}\Big)+C=\\\\\\=arcsin\dfrac{x}{\sqrt2}-\dfrac{x}{\sqrt2}\cdot \dfrac{\sqrt{2-x^2}}{\sqrt2}+C=arcsin\dfrac{x}{\sqrt2}-\dfrac{x\, \sqrt{2-x^2}}{2}+C=$