Chủ đề này có 1 trả lời

[hihi2zz]

$\int_{0}^{\frac{\pi}{4}}\frac{sinxdx}{5sinx.cos^{2}x+2cosx}$

[chinhanh9]

$\int_{0}^{\frac{\pi}{4}}\frac{sinxdx}{5sinx.cos^{2}x+2cosx}$

 

$I=\int_{0}^{\frac{\pi}{4}}\frac{\tan xdx}{5\sin x\cos x+2}= \int_{0}^{\frac{\pi}{4}}\frac{\tan xdx}{\frac{5}{2}\sin 2x+2}$

Đặt $t=\tan x$ thì $dt=\frac{dx}{\cos ^{2}x}\Rightarrow dx=\frac{dt}{t^{2}+1}$

Khi $x=0$ thì $t=0$, khi $x=\frac{\pi}{4}$ thì $t=1$

Mặt khác: $\sin 2x=\frac{2t}{t^{2}+1}$

Do đó:

$I=\int_{0}^{1}\frac{tdt}{(t^{2}+1)\left ( \frac{5t}{t^{2}+1} +2\right )}= \int_{0}^{1}\frac{tdt}{2t^{2}+5t+2}$

$=\frac{1}{4}\int_{0}^{1}\frac{4t+5-5}{2t^{2}+5t+2}dt=\frac{1}{4}\int_{0}^{1}\frac{d\left ( 2t^{2} +5t+2\right )}{2t^{2}+5t+2}-\frac{5}{4}\int_{0}^{1}\frac{dt}{2t^{2}+5t+2}=...=\frac{1}{4}\ln 9-\frac{1}{4}\ln 2+\frac{5}{12}\ln \frac{1}{2}$