\(\int \tan 3 x \,d x=\int \tan x \cdot \tan ^2 \cdot d x=\int \tan x .\left[\sec ^{2} x-1\right] d x \)
\( I=\int \tan \sec ^{2} x d x-\int \tan x\, d x=I_{1}-\log \sec x\)
\(I_{1}=\int \tan x \sec ^{2} x \,d x\)
\(\text {Put } \tan x=1\) तो \(\frac{d \tan x}{d x}=\frac{d t}{d x}\)
\(\sec ^{2} x \,d x=d t\)
\(I_{1}=\int t \,d t=\frac{t^{2}}{2}+c=\frac{\tan ^{2} x}{2}+c\)
अत: \(\mathrm{I}=\frac{\tan ^{2} x}{2}-\log \sec x+c\)
