Question

8 задание А вариант пожалуйста помогите

Answer 1

$\int\limits^{\frac{\sqrt2}{2}}_0\, \frac{x^4dx}{\sqrt{(1-x^2)^3}} \, dx=\Big [\; x=sint\; ,\; dx=cost\, dt,\; t_1=arcsin0=0,\\\\ t_2=arcsin\frac{\sqrt2}{2}=\frac{\pi}{4},\; 1-x^2=1-sin^2t=cos^2t\; \Big ]=\\\\=\int\limits^{\frac{\pi}{4}}_0\; \frac{sin^4t\cdot cost\, dt}{\sqrt{cos^6t}}= \int\limits^{\frac{\pi}{4}}_0\; \frac{sin^4t\cdot cost\, dt}{cos^3t}= \int\limits^{\frac{\pi}{4}}_0\; \frac{sin^4t\cdot dt}{cos^2t} =\\\\= \int\limits^{\frac{\pi}{4}}_0\; \frac{(1-cos^2t)^2\cdot dt}{cos^2t}=\int\limits^{\frac{\pi}{4}}_0\; \frac{1-2cos^2t+cos^4t}{cos^2t}\, dt= \int\limits^{\frac{\pi}{4}}_0(\frac{1}{cos^2t}-2+cos^2t)dt=$

$=(tgt-2t)\Big |_0^{\frac{\pi}{4}}+\int\limits^{\frac{\pi}{4}}_0 \frac{1+cos2t}{2}dt=tg\frac{\pi}{4}-2\cdot \frac{\pi }{4}+\frac{1}{2}\cdot (t+\frac{1}{2}sin2t)\Big |_0^{\frac{\pi}{4}}=\\\\=1-\frac{\pi}{2}+\frac{1}{2}\cdot (\frac{\pi}{4}+\frac{1}{2}\cdot sin\frac{\pi}{2})= 1-\frac{\pi}{2}+\frac{\pi}{8}+\frac{1}{4}=\frac{10-3\pi }{8}$