What is the indefinate integral

[sora4ever1]

i have this question can anyone help me please thx.


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

 

[TD]

What have you tried so far?

 

[sora4ever1]

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 thanks

 

[bomba923]

Basically, I believe you must simply save a factor of $$\sec ^ 2 x$$

and use $$\sec ^ 2 x = 1 + \tan ^ 2 x$$ to express the remaining factors

in terms of $\tan x$. Next, simply substitute with respect to $$\tan x$$.

(I think it's called "u"-substitution in some texts).?

As you should already know, :shy:
$$\frac{d}{dx} \tan x = \sec ^ 2 x$$

*Here's that method, put in action :
$$\int {\tan ^7 x\sec ^4 x\,dx} = \int {\tan ^7 x\left( {1 + \tan ^2 x} \right)\sec ^2 x\,dx} =$$
$$\int {\tan ^7 x\left( {1 + \tan ^2 x} \right)\,d\left( {\tan x} \right)} = \boxed{\frac{{\tan ^8 x}}{8} + \frac{{\tan ^{10} x}}{{10}} + C}$$

!It is very likely that somewhere I made an error!...
Somewhere...some silly error

 

[sora4ever1]

Basically, I believe you must simply save a factor of $$\sec ^ 2 x$$

and use $$\sec ^ 2 x = 1 + \tan ^ 2 x$$ to express the remaining factors

in terms of $\tan x$. Next, simply substitute with respect to $$\tan x$$.

(I think it's called "u"-substitution in some texts).?

As you should already know, :shy:
$$\frac{d}{dx} \tan x = \sec ^ 2 x$$

*Here's that method, put in action :
$$\int {\tan ^7 x\sec ^4 x\,dx} = \int {\tan ^7 x\left( {1 + \tan ^2 x} \right)\sec ^2 x\,dx} =$$
$$\int {\tan ^7 x\left( {1 + \tan ^2 x} \right)\,d\left( {\tan x} \right)} = \boxed{\frac{{\tan ^8 x}}{8} + \frac{{\tan ^{10} x}}{{10}} + C}$$

!It is very likely that somewhere I made an error!...
Somewhere...some silly error

thx a lot.

 

[TD]

!It is very likely that somewhere I made an error!...
Somewhere...some silly error

No, this is correct