Integral of (tanx)^71 * (secx)^4

**circuscircus** · October 2nd 2007, 09:19 AM

$\int tan^71x sec^4x$

$\int tan^70x sec^3x * tanx secx$

I'm stuck after this part

**TKHunny** · October 2nd 2007, 10:18 AM

I wouldn't do that.

My first impression is to take two of the secants and turn them into tangents.

After that, it is a rather obvious substitution u = tan(x) and you're done.

**Soroban** · October 2nd 2007, 03:05 PM

Hello, circuscircus!

TKHunny has the best idea . . .

$\int \left(\tan^{71}\!x\right)\left(\sec^4\!x\right)\,d x$

We have: . $\left(\tan^{71}\!x\right)\left(\sec^2\!x\right)\le ft(\sec^2\!x\right) \;=\;\left(\tan^{71}\!x\right)\left(\tan^2\!x+1\ri ght)\left(\sec^2\!x\right)$

. . Then: . $\int \left(\tan^{73}\!x + \tan^{71}\!x\right)(\sec^2\!x)\,dx$

Let: $u \:= \:\tan x\quad\Rightarrow\quad du \:=\:\sec^2\!x\,dx$

Substitute: . $\int \left(u^{73} + u^{71}\right)\,du$ . . . . Got it?

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