Using Integration By Parts for $\int\sec^2(x)\tan(x)\,\mathrm dx$

Question

Evaluate $\int\sec^2(x)\tan(x)\,\mathrm dx$

I know we can use $\text{substitution}$, but I instinctively used Integration By Parts and am not getting the right answer. What am I missing?

Answer 1

$\frac {\tan^2 \theta} 2 + C$ is not incorrect! Try differentiating it again, you'll get the same thing.

We have $1 + \tan^2 \theta = \sec^2 \theta$ (divide the identity $\sin^2 \theta + \cos^2 \theta = 1$ by $\cos^2 \theta$ if you haven't seen this before) so we could write:

$$\frac {\tan^2 \theta} 2 + C = \frac {\sec^2 \theta} 2 + C - \frac 1 2$$

and writing a new arbitrary constant $C' = C - \frac 1 2$ gives you the cleaner:

$$\int \sec^2 \theta \tan \theta d\theta =\frac {\tan^2 \theta} 2 + C = \frac {\sec^2 \theta} 2 + C'$$

This is a quirk of "indefinite" integration, you are not just looking at one function, but a family of functions that all differ by a constant. You are free to shift this constant without really changing anything, because the constant will always just differentiate away to zero. Formally speaking we are saying that the set of functions $\{f_C(\theta) =\sec^2 \theta + C : C \in \mathbb R\}$ is equal to the set of functions $\{g_C(\theta) = \tan^2 \theta + C : C \in \mathbb R\}$.

Answer 2

HINT

$$\text{Since } \int u\; du = u^2/2 +c_1, \text{we write }\int \sec x \;d( \sec x) = \frac{\sec^2 x }{2}+ c_1 $$

Note that

$$ \frac{\tan^2 x }{2}+ c_2 \;\text{is also correct due to trig identity} $$

$ \sec^2x -\tan^2 x=1. $

Answer 3

Hint: U-Substitution is the most efficient way to approach this integral.

Another hint: What is the derivative of $tan(x)$?