Question

Integrate

Answer 1

\[\Huge \int\limits_{}^{}\tan^8x \sec^4x dx\]

Answer 2

Thought of letting tan^5x = t?

Answer 3

It reduces to 1/5 integration of ( tan^4x) which i know
just want to know if this method is right/or any other better method

Answer 4

\[\large \int\limits \tan^8x \sec^2x \sec^2x\;dx\]

From here we can rewrite one of our sec^2x using our `Square Identity`:\[\large \sec^2x=1+\tan^2x\]\[\large \int\limits\limits \tan^8x (1+\tan^2x)\sec^2x\;dx\]
Then we can make the substitution t=tan x.

Answer 5

my method is wrong?

Answer 6

Hmm using your method, I'm not quite sure how you replace all of the secants.

Answer 7

\[Let ~ \tan^5x=t\]
\[\frac{dt}{5\tan^4x \sec^4x}=dx\]
\[\Large \int\limits_{}^{} \tan^8x \cancel{\sec^4x} \times \frac{dt}{5\tan^4x \cancel{\sec^4x}}=\frac{1}{5}\int\limits_{}^{} \tan^4x\]
and here we can write it as (tan^2x)(sec^2x-1)
??

Answer 8

@zepdrix

Answer 9

oops a little mistake!

Answer 10

\[\tan^4x \sec^8x \]
it should be

Answer 11

So the way you did it, you end up with,\[\large \frac{1}{5}\int\limits \tan^4x\;dt\]An integral involving x but being integrated with respect to t.
Yah that doesn't quite work :(
If you want to make a substitution, you need to replace `all x's with t's`.

Answer 12

Oh that's what the original problem should be?

Answer 13

nono in my value of dx

Answer 14

okay,I got my fault :D

Answer 15

\[\LARGE \int\limits_{}^{} \tan^8x \sec^2x+\tan^8x.\tan^2x.\sec^2xdx\]
Let tanx=t
\[\LARGE \int\limits\limits_{}^{} t^8 \sec^2x+t^8.t^2.\sec^2x \times \frac{dt}{\sec^2x}\]
\[\LARGE \int\limits_{}^{} t^8+t^{10}dt\]
awesome thanks!