Chủ đề này có 1 trả lời

[anhtuan93]

$\int_{0}^{\dfrac{\pi }{4}}\dfrac{sin^{3}xdx}{cos^{3}x}$

[Crystal]

$$J=\int_{0}^{\dfrac{\pi }{4}}\dfrac{sin^{3}xdx}{cos^{3}x}$$

Ta có: $$J = \int\limits_0^{\dfrac{\pi }{4}} {t{g^3}xdx = \int\limits_0^{\dfrac{\pi }{4}} {tgx\left( {t{g^2}x + 1} \right)dx - } } \int\limits_0^{\dfrac{\pi }{4}} {tgxdx = {J_1} - {J_2}} $$
* Tính ${J_1} = \int\limits_0^{\dfrac{\pi }{4}} {tgx\left( {t{g^2}x + 1} \right)dx} $

Đặt $t = tgx \Rightarrow dt = \left( {t{g^2}x + 1} \right)dx$. Đổi cận: $x = 0 \Rightarrow t = 0;\,\,\,x = \dfrac{\pi }{4} \Rightarrow t = 1$

Khi đó: $$\boxed{{J_1} = \int\limits_0^1 {tdt = \left. {\dfrac{1}{2}{t^2}} \right|_0^1} = \dfrac{1}{2}}$$
* Tính $${J_2} = \int\limits_0^{\dfrac{\pi }{4}} {tgxdx = } \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\sin x}}{{\cos x}}dx = } - \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{d\left( {\cos x} \right)}}{{\cos x}} = - \left. {\ln \cos x} \right|} _0^{\dfrac{\pi }{4}} = - \ln \dfrac{{\sqrt 2 }}{2} = \boxed{\ln \sqrt 2 }$$
Vậy $$\boxed{J = \dfrac{1}{2} - \ln \sqrt 2 }$$