- 4
- 0
what is the indefinate integral of tan^7xsec^4x goodluck.
TD said:What have you tried so far?
thx a lot.bomba923 said:Basically, I believe you must simply save a factor of [tex] \sec ^ 2 x[/tex]
and use [tex]\sec ^ 2 x = 1 + \tan ^ 2 x [/tex] to express the remaining factors
in terms of [itex] \tan x [/itex]. Next, simply substitute with respect to [tex] \tan x [/tex].
(I think it's called "u"-substitution in some texts).?
As you should already know, :shy:
[tex] \frac{d}{dx} \tan x = \sec ^ 2 x [/tex]
*Here's that method, put in action:
[tex] \int {\tan ^7 x\sec ^4 x\,dx} = \int {\tan ^7 x\left( {1 + \tan ^2 x} \right)\sec ^2 x\,dx} = [/tex]
[tex] \int {\tan ^7 x\left( {1 + \tan ^2 x} \right)\,d\left( {\tan x} \right)} = \boxed{\frac{{\tan ^8 x}}{8} + \frac{{\tan ^{10} x}}{{10}} + C} [/tex]
!It is very likely that somewhere I made an error!....![]()
![]()
Somewhere...some silly error![]()
No, this is correctbomba923 said:!It is very likely that somewhere I made an error!....![]()
![]()
Somewhere...some silly error![]()