Bài 106226

a) $2^{x}.3^{2x}.5^{3x}=2^{x}.9^{x}.125^{x}=2250^{x} \Rightarrow  \int\limits 2^x.3^{2x}.5^{3x} dx=\int\limits 2250^x dx=\frac{2250^x}{\ln2250} +C$
b) $\frac{\sin 3x\sin 4x}{\tan x+\cot 2x}=\frac{\sin 3x\sin 4x}{\frac{\sin x}{\cos x}+\frac{\cos 2x}{\sin 2x}}=\frac{\sin 3x.\sin 4x}{\frac{\sin x\sin 2x+\cos x\cos 2x}{\cos x\cos 2x}}=\frac{\sin 3x\sin 4x}{\cos \left ( 2x-x \right )}\cos x\sin 2x$
                          $=\sin 4x\sin 3x\sin 2x$
                          $=\frac{1}{2}\left ( \cos x-\cos 7x \right )\sin 2x=\frac{1}{2}\left ( \sin 2x\cos x-\sin 2x\cos 7x \right )$
                          $=\frac{1}{4}\left ( \sin 3x+\sin x-\sin 9x+\sin 5x \right )$
$\Rightarrow  \int\limits \frac{\sin3x\sin4x}{\tan x+\cot x} dx=\frac{1}{4} \int\limits (\sin3x+\sin x-\sin 9x+\sin5x)dx$
$=\frac{-1}{12}\cot 3x-\frac{1}{4}\cos x+\frac{1}{36}\cos9x-\frac{1}{20} \cos 5x+C$
c) Ta có : $4\cos x.\sin4x.\cos2x$
$=2\sin4x.(\cos3x+\cos x )$
$2\sin4x\cos3x+2\sin4x\cos x$
$\sin7x+\sin x+\sin5x+\sin3x$
Vậy $\int\limits 4\cos x.\sin4x\cos2x dx=\int\limits (\sin 7x+\sin x+\sin 5x+\sin 3x)dx$
$=-\frac{1}{7} \cos 7x-\cos x-\frac{1}{5} \cos5 x-\frac{1}{3} \cos3 x +C$