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what is the indefinate integral of tan^7xsec^4x goodluck.

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what is the indefinate integral of tan^7xsec^4x goodluck.

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TD

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What have you tried so far?

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TD said:What have you tried so far?

i am not yet to that grade lvl yet so that is why i want some one to post this explaing how to do it and little back ground to this thx

:surprised

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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]

Somewhere...some

Last edited:

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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 thatsomewhereI made an error!....

Somewhere...somesillyerror

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TD

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No, this is correctbomba923 said:!It is very likely thatsomewhereI made an error!....

Somewhere...somesillyerror

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