Question

integrate 3xsecxtanx

Answer 1

\[\int\limits 3x \cdot \tan(x) \sec(x) . dx\]

Integrated. :P

Answer 2

?

Answer 3

I think By parts, will work here..

Answer 4

Take algebraic as First function or \(u\) or the other as \(v\).

Answer 5

there r 3 functions !

Answer 6

Oh, thanks for telling that.. :P

Answer 7

:)

Answer 8

isn't antiderivative of secxtanx secx?

Answer 9

Let us start without knowing where actually we will reach after this..
\[\implies (3x) \int\limits(\sec(x).\tan(x)) dx - \int\limits(3) \cdot \int\limits(\sec(x)\tan(x))dx \\ \]

Answer 10

yes

Answer 11

Yes I am just applying that only here..

Answer 12

Refer to the attachment from WolframAlpha.

Answer 13

My shot at it: \[\Large \int\limits x \sec x \tan x dx = \int\limits x (\sec x)' dx \\ \Large x = u \ \ \ \ \ (\sec x)' dx = dv \\ \Large dx = du \ \ \ \ \ \sec x = v \\ \Large x \sec x - \int\limits \sec x dx \] Now how do you handle the next one? \[\Large \int\limits \sec x dx = \int\limits \sec x \left( \frac{\sec x + \tan x}{\sec x + \tan x} \right )dx \\ \Large \int\limits \frac{\sec^2x + \sec x \tan x }{\sec x + \tan x} dx \\ \Large \int\limits \frac{(\tan x)' + (\sec x) ' }{\tan x+ \sec x} dx \\ \Large \ln | \tan x + \sec x | +C\]

Answer 14

I took the wrong path in between.. :(

Answer 15

\[3x \sec(x) - 3 \int\limits (\sec(x)) \cdot dx\]