# reduction formula for integral

• Apr 8th 2010, 12:18 PM
sigma1
reduction formula for integral
using the identity $\displaystyle (sec^2 X = 1+tan^2 X)$ find a reduction formula for the integral $\displaystyle ln=\int tan^n X dx,$ where n is greater than or equal to 2

hence evaluate $\displaystyle \int tan^4 X dx$ from 0 to pie/4 ( cant really type this in the mathematical terms.

ok to express this in reduction form i made

$\displaystyle tan^2 X = sec^2X-1$ (for the identity)

then substituted it in the expression $\displaystyle ln=\int sec^2X -1$$\displaystyle dx$

did i do it right? and if so how would that help me to evaluate the integral.
• Apr 8th 2010, 12:42 PM
Soroban
Hello, sigma1!

I'll do the derivation . . .

Quote:

Using the identity $\displaystyle \sec^2\!x \:= \:1+\tan^2\!x$, find a reduction formula for the integral:

. . $\displaystyle \int \tan^n\!x\,dx \;\;\text{ where }n \geq 2$

We have: .$\displaystyle \int(\tan x)^n\,dx$

. . . . . $\displaystyle =\;\int \tan^2\!x\,(\tan x)^{n-2}\,dx$

. . . . . $\displaystyle =\;\int\left(\sec^2\!x-1\right)(\tan x)^{n-2}\,dx$

. . . . . $\displaystyle =\;\int \bigg[\sec^2\!x(\tan x)^{n-2} - (\tan x)^{n-2}\bigg]\,dx$

. . . . . $\displaystyle =\;\int(\tan x)^{n-2}\sec^2\!x\,dx \;- \int(\tan x)^{n-2}\,dx$

. . . . . $\displaystyle =\;\frac{1}{n-1}(\tan x)^{n-1} - \int(\tan x)^{n-2}\,dx$

Therefore: .$\displaystyle \int(\tan x)^n\,dx \;=\;\frac{1}{n-1}(\tan x)^{n-1} - \int(\tan x)^{n-2}\,dx$

• Apr 8th 2010, 01:04 PM
sigma1
thanks for the reply , but i have been trying to follow your steps and i have to admit i got bit confused after the first step. do you mind explaining to me the steps.
• Apr 8th 2010, 01:35 PM
Soroban
Hello again, sigma1!

We have: .$\displaystyle \int(\tan x)^n\,dx$

Factor: . $\displaystyle \int \tan^2\!x\,(\tan x)^{n-2}\,dx$

. . . . $\displaystyle \int\overbrace{\left(\sec^2\!x-1\right)}(\tan x)^{n-2}\,dx$ . .
Trig identity: $\displaystyle \tan^2\!x \:=\:\sec^2\!x-1$

Multiply: .$\displaystyle \int \bigg[\sec^2\!x(\tan x)^{n-2} - (\tan x)^{n-2}\bigg]\,dx$

Make two integrals: .$\displaystyle \int(\tan x)^{n-2}\sec^2\!x\,dx \;- \int(\tan x)^{n-2}\,dx$

. . For the first integral, let: .$\displaystyle u\,=\,\tan x \quad\Rightarrow\quad du \,=\,\sec^2\!x\,dx$

. . Substitute: .$\displaystyle \int u^{n-2}\,du \;=\;\frac{1}{n-1}u^{n-1}$

. . Back-substitute: .$\displaystyle \frac{1}{n-1}(\tan x)^{n-1}$

Hence, we have: . $\displaystyle =\;\frac{1}{n-1}(\tan x)^{n-1} - \int(\tan x)^{n-2}\,dx$

Therefore: .$\displaystyle \int(\tan x)^n\,dx \;=\;\frac{1}{n-1}(\tan x)^{n-1} - \int(\tan x)^{n-2}\,dx$

• Apr 8th 2010, 01:47 PM
sigma1
THANKS ALOT.

now that its reduced how would i use that to evaluate $\displaystyle \int tan^4x dx$ using pie/4 as the upper limit and 0 as the lower limit.