ДАЮ 50 БАЛОВ СРОЧНО !!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Решите задачу:

$1)\; \; \int \limits _2^7 \frac{4\; dx}{\sqrt{x+2}} =4\cdot \int \limits _2^7\, \frac{d(x+2)}{\sqrt{x+2}} =4\cdot 2\sqrt{x+2}\Big |_2^7=\\\\=8\cdot (\sqrt9-\sqrt4)=8\, (3-1)=8\\\\2)\; \; \int\limits^{3}_{-1}\frac{1}{\sqrt{2x+3}} \, dx = \frac{1}{2} \int\limits^3_{-1} \frac{d(2x+3)}{\sqrt{2x+3}} = \frac{1}{2} \cdot 2\sqrt{2x+3}\Big |_{-1}^3 =\\\\= \sqrt{9} - \sqrt{1} =3-1=2$

$3)\; \; \int\limits^{3\pi }_0\, \frac{dx}{cos^2\frac{x}{9}} =9\cdot \int\limits^{3\pi }_0 \frac{d(\frac{x}{9})}{cos^2\frac{x}{9}} \, dx =9\cdot tg\frac{x}{9}\Big |_0^{3\pi }=\\\\=9\cdot (tg\frac{\pi}{3}-tg\, 0)=9\cdot (\sqrt3-0)=9\sqrt3\\\\\\P.S.\; \; d(x+2)=(x+2)'dx=dx\\\\d(2x+3)=(2x+3)'dx=2\, dx\\\\d( \frac{x}{9} )=(\frac{x}{9})'dx= \frac{1}{9}\, dx$