Тема: интегрирование по частям

Question 1

Question 2

Решите задачу:

$9)\; \; \int\limits^{\sqrt3}_1 \frac{dx}{\sqrt{(1+x^2)^3}} =[\; x=tgt\; ,dx= \frac{dt}{cos^2t} \; ,\; t=arctgx\; ,\\\\t_1=arctg1= \frac{\pi}{4} \; ,\; t_2=arctg\sqrt3= \frac{\pi}{3} \; ]= \int\limits^{\pi /3}_{\pi /4} \frac{dt}{cos^2t\cdot \sqrt{(1+tg^2t)^3}} =\\\\=[\; 1+tg^2t= \frac{1}{cos^2t}\; ]= \int\limits^{\pi /3}_{\pi /4} \frac{dt}{cos^2t\cdot \sqrt{\frac{1}{cos^6t}}} = \int\limits^{\pi /3}_{\pi /4} \frac{cos^3t}{cos^2t} dt=$

$\int\limits^{\pi /3 }_{\pi /4}cost\, dt=sint\Big |_{\pi /4}^{\pi /3}=sin\frac{\pi}{3} -sin \frac{\pi }{4}= \frac{\sqrt3}{2} - \frac{\sqrt2}{2} =\frac{\sqrt3-\sqrt2}{2} \; ;$

$10)\; \; \int\limits^3_1 \frac{dx}{x+x^2} = \int\limits^3_1\frac{dx}{x(x+1)} = \int\limits^3_1\; \Big (\frac{1}{x}- \frac{1}{x+1} \Big )dx=\\\\=\Big (ln|x|-ln|x+1|\Big )|_1^3=(ln3-ln4)-(\underbrace {ln1}_{0}-ln2)=\\\\=ln3-2ln2+ln2=ln3-ln2=ln\frac{3}{2}\; ;$

$11)\; \; \int\limits^{\pi /2}_0sin^2x\, dx= \int\limits^{\pi /2}_0 \frac{1-cos2x}{2}dx =\frac{1}{2}\int\limits^{\pi /2}_0(1-cos2x)dx=\\\\= \frac{1}{2}\cdot (x-\frac{1}{2}sin2x)\Big |_0^{\pi /2} = \frac{1}{2}\cdot \Big ( \frac{\pi}{2} - \frac{1}{2}\underbrace {sin\pi }_{0}\Big )=\frac{1}{2} \cdot \frac{\pi }{2}=\frac{\pi }{4}\; ;$

$12)\; \; \int\limits^{\pi /2}_0\; sin^4x\; dx=\Big [\; sin^4x=(sin^2x)^2=\Big (\frac{1-cos2x}{2}\Big )^2 =\\\\=\frac{1}{4}\cdot (1-2cos2a+cos^22x)= \frac{1}{4}\cdot \Big (1-2cos2x+\frac{1+cos4x}{2}\Big )= \\\\= \frac{1}{4} \cdot \Big (\frac{3}{2}-2cos2x+\frac{1}{2} cos4x\Big )\Big ]= \frac{1}{4} \cdot \int\limits^{\pi /2}_0\Big ( \frac{3}{2} -2cos2x+ \frac{1}{2}cos4x\Big )dx=\\\\= \frac{1}{4} \cdot \Big (\frac{3}{2}x-2\cdot \frac{1}{2}sin2x+\frac{1}{2}\cdot \frac{1}{4}sin4x\Big )\Big |_0^{\pi /2}=$

$= \frac{1}{4}\cdot \Big (\frac{3}{2}\cdot \frac{\pi }{2}-\underbrace {sin\pi }_{0}+\frac{1}{2}\cdot \frac{1}{4}\cdot \underbrace {sin2\pi }_{0}\Big )= \frac{1}{4} \cdot \frac{3}{2} \cdot \frac{\pi }{2}= \frac{1\cdot 3}{2\cdot 4}\cdot \frac{\pi}{2}$