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Решите задачу:

$1)\; \int_1^2\frac{cos^7x}{sin^2x}dx=\int _1^2\frac{(cos^2x)^3\cdot cosx\cdot dx}{sin^2x}=\int _1^2\frac{(1-sin^2x)^3\cdot d(sinx)}{sin^2x}=\\\\=\int _1^2\frac{1-3sin^2x+3sin^4x-sin^6x}{sin^2x}\cdot d(sinx)=\\\\=\int_1^2(\frac{1}{sin^2x}-3+3sin^2x-sin^4x)\cdot d(sinx)=\\\\=(-\frac{1}{sinx}-3sinx+3\frac{sin^3x}{3}-\frac{sin^5x}{5})_1^2=\\\\=-\frac{1}{sin2}-3sin2+sin^32-\frac{sin^52}{5}+\frac{1}{sin1}+3sin1-sin^31+\frac{sin^51}{5}$

$2)\; \int _0^1arctg(2\sqrt{x})dx=\\\\=[\, u=arctg(2\sqrt{x}),du=\frac{2\cdot \frac{1}{2\sqrt{x}}\cdot dx}{1+4x}=\frac{dx}{\sqrt{x}(1+4x)},\; dv=dx,\; v=x\, ]=\\\\=x\cdot arctg(2\sqrt{x})|_0^1-\int _0^1\frac{\sqrt{x}dx}{1+4x}=\\\\=[\, \sqrt{x}=t,x=t^2,dx=2tdt,t_1=\sqrt0=0,t_2=\sqrt1=1\, ]=\\\\=arctg2-\int _0^1\frac{2t^2dt}{1+4t^2}=arctg2-\int _0^1(\frac{1}{2}-\frac{1}{2(4t^2+1)})dt=\\\\=arctg2-(\frac{1}{2}t-\frac{1}{2}\cdot \frac{1}{2}\cdot arctg(2t))|_0^1=\\\\=arctg2-\frac{1}{2}+\frac{1}{4}arctg2=$

$=\frac{5}{4}arctg2-\frac{1}{2}$