Question

Помогите с решением интегралов

Answer 1

$1)\int (1-sin2x)^5cos2xdx=[t=1-sin2x,dt=-2cos2xdx]=\\\\=-\frac{1}{2}\int t^5dt=-\frac{1}{2}\cdot \frac{t^6}{6}+C=-\frac{1}{12}(1-sin2x)^6+C\\\\2)\; \int cos^5xdx=\int cos^4x\cdot cosxdx=\int (cos^2x)^2\cdot cosxdx=\\\\=\int (1-sin^2x)^2\cdot cosxdx=[t=sinx,dt=cosxdx]=\int (1-t^2)^2dt=\\\\=\int (1-2t^2+t^4)dt=t-2\frac{t^3}{3}+\frac{t^5}{5}+C=sinx-\frac{2}{3}sin^3x+\frac{1}{5}sin^5x+C\\\\3)\int \frac{xdx}{\sqrt{1-x^2}}=[t=1-x^2,dt=-2xdx]=-\frac{1}{2}\int \frac{dt}{\sqrt{t}}=\\\\=-\frac{1}{2}\cdot 2\sqrt{t}+C$

$=-\sqrt{1-x^2}+C$