Chủ đề này có 1 trả lời

[tritanngo99]

Tính: $I=\int_0^{\frac{\pi}{3}}tan(x)^3 dx$

[L Lawliet]

Tính: $I=\int_0^{\frac{\pi}{3}}tan(x)^3 dx$

Lời giải.

$$I=\int_{0}^{\frac{\pi}{3}}\tan ^{3}xdx=\int_{0}^{\frac{\pi}{3}}\dfrac{\sin ^{3}x}{\cos ^{3}x}dx=\int_{0}^{\frac{\pi}{3}}\dfrac{\left ( 1-\cos ^{2}x \right )\sin x}{\cos ^{3}x}dx$$

Đặt $\cos x=t\Rightarrow -\sin xdx=dt$.

$$I=\int_{\frac{1}{2}}^{1}\dfrac{1-t^{2}}{t^{3}}dt=\int_{\frac{1}{2}}^{1}\dfrac{1}{t^{3}}dt-\int_{\frac{1}{2}}^{1}\dfrac{1}{t}dt=-\dfrac{1}{2t^{2}}\bigg|_{\frac{1}{2}}^{1}-\ln \left | t \right |\bigg|_{\frac{1}{2}}^{1}$$