Помогите с интегралом!

Решите задачу:

$1)\; \; \int \frac{dx}{2sinx+3cosx}=[\; t=tg\frac{x}{2}\; ,\; sinx=\frac{2t}{1+t^2}\; ,\; cosx=\frac{1-t^2}{1+t^2} \; ,\\\\dx=\frac{2\, dt}{1+t^2}\; ]=\int \frac{2\, dt}{(1+t^2)\cdot (2\cdot \frac{2t}{1+t^2}+3\cdot \frac{1-t^2}{1+t^2}) } =\int \frac{2\, dt}{4t+3-3t^2} =\\\\=-\frac{2}{3}\cdot \int \frac{dt}{t^2-\frac{4}{3}t-1} =- \frac{2}{3} \cdot \int \frac{dt}{(t-\frac{2}{3})^2-\frac{4}{9}-1} =- \frac{2}{3} \cdot \int \frac{dt}{(t-\frac{2}{3})^2-\frac{13}{9}} =$

$=- \frac{2}{3} \cdot \frac{1}{2\cdot \frac{\sqrt{13}}{3}} \cdot ln\Big | \frac{t-\frac{2}{3}-\frac{\sqrt{13}}{3}}{t- \frac{2}{3}+\frac{\sqrt{13}}{3} } \Big |+C=-\frac{1}{\sqrt{13}}\cdot ln\Big | \frac{3tg\frac{x}{2}-2-\sqrt{13}}{3tg\frac{x}{2}-2+\sqrt{13}} \Big |+C$

$2)\; \; \int \frac{dx}{sin^2x+5cos^2x} =\Big [\frac{:cos^2x}{:cos^2x}\Big ]=\int \frac{\frac{dx}{cos^2x}}{tg^2x+5} =\int \frac{d(tgx)}{tg^2x+5} =[t=tgx]=\\\\=\int \frac{dt}{t^2+5} =\frac{1}{\sqrt5}\cdot arctg\frac{t}{\sqrt5}+C= \frac{1}{\sqrt5}\cdot arctg\frac{tgx}{\sqrt5} +C$