Question

What is the integral of sec (3x) tan (3x) dx?

Answer 1

`\displaystyle\int \sec 3x\tan 3x\ dx=\int(1)/(\cos 3x)\cdot(\sin 3x)/(\cos 3x)\ dx=\int(\sin 3x)/(\cos^23x)\ dx\\\\\Rightarrow \left|\begin{array}{ccc}\cos 3x=t\\-3\sin 3x\ dx=dt\\\sin 3x\ dx=-(1)/(3)\ dt\end{array}\right|\Rightarrow\int\left(-(1)/(3t^2)\right)dt=-(1)/(3)\int t^(-2)\ dt\\\\=-(1)/(3)(-t^(-1))+C=(1)/(3t)+C=(1)/(3\cos 3x)+C\\\\Answer:\boxed{\int \sec 3x\tan 3x\ dx=(1)/(3\cos 3x)+C}`

Answer 2

∫tan^3(x)*sec^3(x) dx

=∫tan^2(x)*sec^2(x)*(tanx*secx) dx

=∫(sec^2(x) - 1)*sec^2(x)*(tanx*secx) dx

=∫(sec^4(x) - sec^2(x))*(tanx*secx) dx

=∫(sec^4(x)*(tanx*secx) dx - ∫sec^2(x))*(tanx*secx) dx

=∫(sec^4(x)*d(secx) - ∫sec^2(x))d(secx)

=(1/5)sec^5(x) - (1/3)sec^3(x) + c
hope this helps