Tích phân \(\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\cos2x}\text{d}x\) bằng
\(-\frac{5\sqrt{3}}{9}+\frac{1}{2}\ln\left(2+\sqrt{3}\right)\). \(-\frac{10\sqrt{3}}{27}+\frac{1}{2}\ln\left(2+\sqrt{3}\right)\). \(\frac{10\sqrt{3}}{9}+\frac{1}{2}\ln\left(2+\sqrt{3}\right)\). \(-\frac{10\sqrt{3}}{9}+\frac{1}{2}\ln\left(2+\sqrt{3}\right)\). Hướng dẫn giải:Ta có:
\(\cos2x=\cos^2x-\sin^2x=\cos^2x\left(1-\tan^2x\right)\)
\(I=\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\cos2x}\text{d}x=\int\limits^{\frac{\pi}{6}}_0\frac{\tan^4x}{\left(1-\tan^2x\right)}.\frac{1}{\cos^2x}\text{d}x\)
Đặt \(t=\tan x\Rightarrow\text{d}t=\frac{1}{\cos^2x}\text{d}x\)
\(I=\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{t^4}{1-t^2}\text{d}t\)
\(=\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{t^4-1+1}{1-t^2}\text{d}t\)
\(=\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{\left(t^2-1\right)\left(t^2+1\right)+1}{1-t^2}\text{d}t\)
\(=-\int\limits^{\frac{1}{\sqrt{3}}}_0\left(t^2+1\right)\text{d}t+\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{1}{\left(1-t\right)\left(1+t\right)}\text{d}t\)
\(=-\left(\frac{t^3}{3}+t\right)|^{\frac{1}{\sqrt{3}}}_0+\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{1}{2}\left[\frac{1}{1-t}+\frac{1}{1+t}\right]\text{d}t\)
\(=-\left(\frac{t^3}{3}+t\right)|^{\frac{1}{\sqrt{3}}}_0+\frac{1}{2}\left[\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{-\text{d}\left(1-t\right)}{1-t}+\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{\text{d}\left(1+t\right)}{1+t}\right]\)
\(=-\left(\frac{t^3}{3}+t\right)|^{\frac{1}{\sqrt{3}}}_0+\frac{1}{2}\left[-\\ln\left(1-t\right)+\\ln\left(1+t\right)\right]|^{\frac{1}{\sqrt{3}}}_0\)
\(=\left[-\left(\frac{t^3}{3}+t\right)+\frac{1}{2}\\ln\frac{1+t}{1-t}\right]|^{\frac{1}{\sqrt{3}}}_0\)
\(=-\frac{10\sqrt{3}}{27}+\frac{1}{2}\ln\left(2+\sqrt{3}\right)\)