Trích:
| Câu 1. Tính tích phân $$I=\int_{0}^{\frac{\pi }{4}}{\frac{dx}{{{\sin }^{4}}x+{{\cos }^{4}}x}}$$ |
Giải:
$$I=\int_{0}^{\frac{\pi }{4}}{\frac{dx}{{{\sin }^{4}}x+{{\cos }^{4}}x}}=\int_{0}^{\frac{\pi}{4}}\frac{1+\tan^2x} {1+\tan^4x}\: d(\tan x)$$
$$=\int_{0}^{1}\frac{1+x^2}{1+x^4}dx=\frac{1}{2} \int_{0}^{1}\left [ \frac{1}{x^2+\sqrt{2}x+1}+\frac{1}{x^2- \sqrt{2}x+1} \right ]dx$$
$$=\frac{1}{2}\left [ \sqrt{2}\tan\left ( \sqrt{2}x+1 \right )+\sqrt{2}\tan\left ( \sqrt{2}x-1 \right ) \right ]_{0}^{1}$$
$$=\fbox{$\frac{\pi}{2\sqrt{2}}$}$$
[RIGHT][I][B]Nguồn: MathScope.ORG[/B][/I][/RIGHT]