Tìm nguyên hàm của hàm số sau: I=∫tan5xdx

Question

\(I=\int \tan ^{5} x{\rm d}x\)

Answer 1

Đặt \(\tan x=t\Rightarrow \frac{{\rm d}x}{\cos ^{2} x} ={\rm d}t\Rightarrow (\tan ^{2} x+1){\rm d}x={\rm d}t\Rightarrow {\rm d}x=\frac{1}{t^{2} +1} {\rm d}t.\)

 Khi đó: 
\(\begin{array}{l} {I=\int t^{5} \frac{{\rm d}t}{t^{2} +1} =\int \left(t^{3} -t+\frac{t}{t^{2} +1} \right) {\rm d}t=\int t^{3} {\rm d}t-\int t {\rm d}t+\int \frac{t}{t^{2} +1} {\rm d}t} \\ {=\frac{1}{4} t^{4} -\frac{1}{2} t^{2} +\frac{1}{2} \int \frac{{\rm d}\left(t^{2} +1\right)}{t^{2} +1} =\frac{1}{4} t^{4} -\frac{1}{2} t^{2} +\frac{1}{2} \ln \left|t^{2} +1\right|+C} \\ {=\frac{1}{4} \tan ^{4} x-\frac{1}{2} \tan ^{2} x+\frac{1}{2} \ln \left(\tan ^{2} x+1\right)+C} \\ {=\frac{1}{4} \tan ^{4} x-\frac{1}{2} \tan ^{2} x+\frac{1}{2} \ln \left(\frac{1}{\cos ^{2} x} \right)+C} \\ {=\frac{1}{4} \tan ^{4} x-\frac{1}{2} \tan ^{2} x-\ln \left|\cos x\right|+C} \end{array}\)