Question

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

Answer 1

∫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

Answer 2

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