I'm revising for an exam and came across this question:
$I = \int{\tan^2{x}\sec{x}} dx$,
and I got to a point where I just ended up with the original question:
$$I = \int{\sec^3{x}-\sec x}dx$$
$$I = \int{\sec x \sec^2x}dx – \int{\sec x}dx$$
$$I = \int{u\frac{dv}{dx}}dx – \int{\sec x}dx$$
let $u = \sec x$, $\frac{du}{dx} = \sec x\tan x$
and $\frac{dv}{dx} = \sec^2x$, $v = \tan x$
$$I = uv – \int{v\frac{du}{dx}}dx – \int{\sec x}dx$$
$$I = \sec x\tan x – \int{\tan x\sec x\tan x}dx – \ln{|\sec x + \tan x|}$$
$$I = \sec x\tan x – \int{\tan^2x\sec x}dx – \ln{|\sec x + \tan x|}$$
$$I = \sec x\tan x – I + k – \ln{|\sec x + \tan x|}$$
$$I + I = 2I = \sec x\tan x – \ln{|\sec x + \tan x|} + k$$
$$I = \frac{\sec x\tan x – \ln{|\sec x + \tan x|} + k}{2}$$
$$I = \frac{\sec x\tan x – \ln{|\sec x + \tan x|}}{2} + c$$
Is it ok to do this? I tried doing it on wolframalpha but it used somtehing called the reduction formula which I don't know,and I tried it on my calculator, which can only do definite integration, with the interval $[1,0.1]$ and I got the same answer with my result, but this doesn't proove it is correct.
(Also I didn't really know when to put in the constant ($k$)?)
Best Answer
You can always decide whether an indefinite integral is correct by differentiating the answer to see whether you get back the original function. So, differentiate your answer: do you get $\tan^2x\sec x$? If yes, then what you did was (almost certainly) OK; if not, then not.