\(\int\limits^{ }_{ }\frac{cos^2x+sin^2x}{sin^2xcos^2x}dx=\int\limits\frac{1}{sin^2x}dx+\int\limits^{ }_{ }\frac{1}{cos^2x}dx=tanx+cotgx\)
thay cân vào ta tính đc
\(\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}}\frac{Sin2x}{1-cos^4x}dx\)
\(\int\limits^{\frac{\pi}{6}}_0\frac{1}{cosx.cos\left(x+\frac{pi}{4}\right)}dx\)
\(\int\limits^{pi/2}_0\frac{sinx}{\left(sinx+\sqrt{3}cosx\right)^2}dx\)
\(\int\limits^{\frac{\pi}{4}}_0\left(tan^2x+tanx\right).e^xdx\)
\(\int\limits^{\frac{\pi}{3}}_0\frac{tanxdx}{\sqrt{1-ln^2\left(cosx\right)}}\)
\(\int\limits^{\frac{\pi}{3}}_0\frac{tanxdx}{\sqrt{1-ln^2\left(cosx\right)}}\)
Chỉ mình câu tích phân này với !!
\(\int\limits^{pi/2}_0\left(\frac{1}{cos^2\left(sinx\right)}-tan^2\left(cosx\right)\right)dx\)
\(\int\limits^{e^2}_e\left(\frac{1}{ln^2x}-\frac{1}{lnx}\right)dx\)
\(\int\limits^1_{-1}\frac{x^4}{1+2^x}dx\)
