Tìm nguyên hàm của hàm số sau: ∫cos2x/sin3xdx

Question

\(\int \frac{\cos ^{2} x}{\sin ^{3} x} {\rm d}x\)

Answer 1

Ta có: \(I=\int \frac{\cos ^{2} x}{\sin ^{3} x} {\rm d}x =\int \frac{1-\sin ^{2} x}{\sin ^{3} x} {\rm d}x =\int \frac{1}{\sin ^{3} x} {\rm d}x -\int \frac{1}{\sin x} {\rm d}x =A+B.\)
. \(A=\int \frac{1}{\sin ^{3} x} {\rm d}x =\int \frac{\sin x}{\left(1-\cos ^{2} x\right)^{2} } {\rm d}x\)
Đặt \(t=\cos x\Rightarrow {\rm d}t=-\sin x{\rm d}x\)

 Suy ra
\(\begin{array}{l} {A=-\int \frac{1}{\left(1-t^{2} \right)^{2} } \, {\rm d}t=-\int \frac{1}{\left(t^{2} -1\right)^{2} } \, {\rm d}t=-\int \left[\frac{1}{2} \left(\frac{1}{t-1} -\frac{1}{t+1} \right)\right] ^{2} {\rm d}t} \\ {{\rm =}-\frac{1}{4} \int \left[\frac{1}{\left(t+1\right)^{2} } +\frac{1}{\left(t-1\right)^{2} } -\frac{2}{\left(t+1\right)\left(t-1\right)} \right] \, {\rm d}t} \\ {{\rm =}-\frac{1}{4} \int \left[\frac{1}{\left(t+1\right)^{2} } +\frac{1}{\left(t-1\right)^{2} } -\left(\frac{1}{t-1} -\frac{1}{t+1} \right)\right] \, {\rm d}t} \\ {=-\frac{1}{4} \left(-\frac{1}{t+1} -\frac{1}{t-1} -\ln \left|\frac{t-1}{t+1} \right|\right)+C_{1} } \\ {=-\frac{1}{4} \left(-\frac{1}{\cos x+1} -\frac{1}{\cos x-1} -\ln \left|\frac{1-\cos x}{1+\cos x} \right|\right)+C_{1} } \end{array}\)
\(B=\int \frac{1}{\sin x} {\rm d}x=\int \frac{\sin x}{\left(1-\cos ^{2} x\right)} {\rm d}x\)
Đặt \(u=\cos x\Rightarrow {\rm d}u=-\sin x{\rm d}x\)

 Suy ra

\(B=-\int \frac{{\rm d}u}{1-u^{2} } =-\frac{1}{2} \int \left(\frac{1}{1-u} +\frac{1}{1+u} \right){\rm d}u =-\frac{1}{2} \ln \left|\frac{1+u}{1-u} \right|+C_{2} =-\frac{1}{2} \ln \left|\frac{1+\cos x}{1-\cos x} \right|+C_{2}\)

 Vậy   \(I=\frac{1}{4} \left(\frac{1}{\cos x+1} +\frac{1}{\cos x-1} -\ln \left|\frac{1-\cos x}{1+\cos x} \right|\right)+C\)