Кто может помочь решить интеграл?!

Решите задачу:

$1)\; \; \int \frac{dx}{4-3cosx} =[\, t=tg\frac{x}{2},\; x=2arctgt,\; dx=\frac{2dt}{1+t^2},\; cosx=\frac{1-t^2}{1+t^2}\, ]=\\\\=\int \frac{2\, dt}{(1+t^2)(4-3\cdot \frac{1-t^2}{1+t^2})}=2\int \frac{dt}{4(1+t^2)-3(1-t^2)} =2\int \frac{dt}{7t^2+1} =\\\\=2\int \frac{dt}{(\sqrt7t)^2+1} =2\cdot \frac{1}{\sqrt7}\cdot arctg(\sqrt7t)+C=\frac{2}{\sqrt7}\cdot arctg(\sqrt7\cdot tg\frac{x}{2})+C$

$2)\; \; \int \frac{x^2\, dx}{\sqrt{49-x^2}}=[x=7sint\; ,t=arcsin\frac{x}{7},\; dx=7cost\, dt,\\\\49-x^2=49-49sin^2t=49cos^2t,\; \sqrt{49-x^2}=7cost\; ]=\\\\=\int \frac{49sin^2t\cdot 7cost\, dt}{7cost}=49\int sin^2t\, dt=49\int \frac{1-cos2t}{2} dt=\\\\=\frac{49}{2}\int (1-cos2t)dt=\frac{49}{2}(t-\frac{1}{2}sin2t)+C=\\\\=24,5\Big (arcsin\frac{x}{7} -\frac{1}{2}sin(2arcsin\frac{x}{7})\Big )+C=$

$=24,5\Big (arcsin\frac{x}{7}-\frac{x\sqrt{49-x^2}}{49 }\Big )+C=24,5arcsin\frac{x}{7} -\frac{1}{2}x\sqrt{49-x^2}+C$