Chủ đề này có 1 trả lời

[KaveZS]

Tính tích phân $\int_{0}^{\frac{\pi}{4}}\frac{tanx-2}{cosx(sinx+3cosx)}dx$

[nguyenthanhhung1985]

Ta có: \frac{\tan{x}-2}{\cos{x}(\sin{x}+3\cos{x})}=\frac{\sin{x}-2\cos{x}}{\cos^2{x}(\sin{x}+3\cos{x})}$

$=\frac{\sin{x}+3\cos{x}-5\cos{x}}{\cos^2{x}(\sin{x}+3\cos{x})}$

$=\frac{\sin{x}+3\cos{x}}{\cos^2{x}(\sin{x}+3\cos{x})}-\frac{5\cos{x}}{\cos^2{x}(\sin{x}+3\cos{x})}$

$=\frac{1}{\cos^2{x}}-\frac{5}{\cos{x}(\sin{x}+3\cos{x})}$

Khi đó: $\int_0^{\frac{\pi}{4}} \frac{\tan{x}-2}{\cos{x}(\sin{x}+3\cos{x})}dx=\int_0^{\frac{\pi}{4}} \frac{1}{\cos^2{x}}dx-5\int_0^{\frac{\pi}{4}} \frac{1}{\cos{x}(\sin{x}+3\cos{x})}$

$=\int_0^{\frac{\pi}{4}} \frac{1}{\cos^2{x}}dx-5\int_0^{\frac{\pi}{4}} \frac{\frac{1}{\cos^2{x}}}{\tan{x}+3}dx$

$=\int_0^{\frac{\pi}{4}} \frac{1}{\cos^2{x}}dx-5\int_0^{\frac{\pi}{4}} \frac{1}{\tan{x}+3}d\tan{x}$

$=\int_0^{\frac{\pi}{4}} \frac{1}{\cos^2{x}}dx-5\int_0^{\frac{\pi}{4}} \frac{1}{\tan{x}+3}d(\tan{x}+3)$

$=\int_0^{\frac{\pi}{4}} d\tan{x}-5\int_0^{\frac{\pi}{4}} d\ln|\tan{x}+3|$

$=\tan{x}|_0^{\frac{\pi}{4}}-5\ln|\tan{x}+3||_0^{\frac{\pi}{4}}$

$=1+5\ln{\frac{3}{4}}$

Bài viết đã được chỉnh sửa nội dung bởi nguyenthanhhung1985: 02-07-2017 - 12:14