1) $\int_{0}^{\pi/3} \frac{cosx}{cosx.\sqrt{3+sin^2x}}dx$
2)$\int \frac{sin^6x+ cos^6 x}{sin^4x+cos^4x-1/2}dx$
ai giải hộ e với T.T
Cách khác:
ta có $cos^6x+sin^6x=\frac{5}{8}+\frac{3}{8}cos4x=\frac{5}{8}+\frac{3}{8}.\frac{1-tan^22x}{1+tan^22x}$
$$ sin^4x+cos^4x-1/2=cos^2 2x$$
Nên: $\int \frac{sin^6x+ cos^6 x}{sin^4x+cos^4x-1/2}dx=\int \frac{\frac{5}{8}+\frac{3}{8}.\frac{1-tan^22x}{1+tan^22x}}{cos^22x}dx=\int (\frac{5}{8}+\frac{3}{8}.\frac{1-tan^22x}{1+tan^22x})d(tan2x)$
Xong....