how do I integrate sqrt(1+4tan^(2)(theta)) sec^(2)(theta) d(theta)

Question

anonymous · Answer

Make a substitution to remove the secant term: $u=\tan \theta$, then $du=\sec^2\theta\,d\theta$.
$$\int\sqrt{1+4\tan^2\theta}\sec^2\theta\,d\theta=\int\sqrt{1+4u^2}\,du$$
Make another substitution, introducing a trigonometric form: $u=\dfrac{1}{2}\tan t$, then $du=\dfrac{1}{2}\sec^2t\,dt$.
$$\int\sqrt{1+4u^2}\,du=\frac{1}{2}\int\sqrt{1+4\left(\frac{1}{2}\tan t\right)^2}\sec^2t\,dt=\frac{1}{2}\int\sqrt{1+\tan^2t}\sec^2t\,dt$$
Recall the identity, $1+\tan^2t=\sec^2t$. Then.
$$\frac{1}{2}\int\sqrt{1+\tan^2t}\sec^2t\,dt=\frac{1}{2}\int\sec^3t\,dt$$
You can deal with this using integration by parts, or refer to the power reduction formula for secant.

anonymous · Answer

Of course, you could have made the trig sub right away, setting $\dfrac{1}{2}\tan t=\tan\theta$, but sometimes subbing one step at a time makes the next step more apparent.