Тема интегрирование по частям

Question 1

Question 2

Решите задачу:

$1)\; \; \int\limits^a_0 {(x^2-ax)} \, dx =(\frac{x^3}{3}-a\cdot \frac{x^2}{2})\Big |_0^{a}=\frac{a^3}{3}-\frac{a^3}{2}=-\frac{a^3}{6}\\\\2)\; \; \int\limits^3_2 \frac{dx}{x^2} =\frac{x^{-2+1}}{-2+1}\Big |_2^3=- \frac{1}{x}\Big |_2^3= -\frac{1}{3}+\frac{1}{2}=\frac{1}{6}\\\\3)\; \; \int\limits^{\sqrt3}_0 \frac{x\, dx}{\sqrt{4-x^2}}=\Big [t=4-x^2\; ,\; dt=-2x\, dx\; ,\; t_1=4\; ,\; t_2=1\Big ]=$

$= - \frac{1}{2} \int\limits^4_1 \frac{dt}{\sqrt{t}}=- \frac{1}{2}\cdot 2\sqrt{t}\Big |_4^1= -(\sqrt1-\sqrt4)=-(1-2)=1$

$4)\; \; \int\limits^{\frac{\pi}{6}}_{\frac{\pi}{8}} {\frac{dx}{cos^22x}} =\frac{1}{2}\cdot tg2x\Big |_{\frac{\pi}{8}}^{\frac{\pi}{6}}=\frac{1}{2}\cdot (tg\frac{\pi}{3}-tg\frac{\pi}{4})=\frac{1}{2}\cdot (\sqrt3-1)= \frac{\sqrt3-1}{2}$

$5)\; \; \int \limits _0^{\frac{\pi}{2}}\, sin^6x\, dx=\int \limits _0^{\pi /2}(sin^2x)^3dx= \int\limits_0^{\pi/2}(\frac{1-cos2x}{2})^3dx=\\\\=\frac{1}{8} \int\limits^{\pi/2}_0(1-3cos2x+3cos^22x-cos^32x)dx=\\\\=\frac{1}{8} \int\limits^{\pi /2}_0(1-3cos2x+\frac{3}{2}(1+cos4x)-cos^22x\cdot cos2x)dx=$

$=\frac{1}{8}\cdot (x-\frac{3}{2}sin2x+ \frac{3}{2}x+\frac{3}{2}\cdot \frac{1}{4}sin4x)\Big |_0^{\pi /2}-\\\\-\frac{1}{8} \int\limits_0^{\pi /2}(1-sin^22x)\cdot cos2xdx=\frac{1}{8}(\frac{\pi}{2}+\frac{3}{2}\cdot \frac{\pi}{2})-\frac{1}{8} \int \limits _{0}^{\frac{\pi}{2}}cos2x\, dx+$
$+\frac{1}{8}\cdot \frac{1}{2}\int \limits _0^{\pi /2}sin^22x\cdot 2cos2x\, dx=$

$=\frac{1}{8}\cdot \frac{\pi}{2}\cdot \frac{5}{2}-\frac{1}{8}\cdot \frac{1}{2}sin2x\Big |_0^{\pi/2}+\frac{1}{16}\cdot \frac{sin^32x}{3}\Big |_0^{\pi /2}=\frac{5\pi}{32}$