Question

tính nguyên hàm của 1/((sinx)^2-4(cos)^2)

Answer 1

\(\int\frac{1}{sin^2-4cos^2x}dx=\int\frac{\frac{dx}{cos^2x}}{tan^2x-4}\)

\(=\int\frac{1}{tan^2x-4}d\left(tanx\right)=\int\frac{d\left(tanx\right)}{\left(tanx-2\right)\left(tanx+2\right)}\\ =\frac{1}{4}\int\left(\frac{1}{tanx-2}-\frac{1}{tanx+2}\right)d\left(tanx\right)\\ =\frac{1}{4}\left(ln\left|tanx-2\right|-ln\left|tanx+2\right|\right)+C\\ =\frac{1}{4}ln\left|\frac{tanx-2}{tanx+2}\right|+C\)