# Thread: reduction formula for integral

1. ## reduction formula for integral

using the identity $(sec^2 X = 1+tan^2 X)$ find a reduction formula for the integral $ln=\int tan^n X dx,$ where n is greater than or equal to 2

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

ok to express this in reduction form i made

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

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

did i do it right? and if so how would that help me to evaluate the integral.

2. Hello, sigma1!

I'll do the derivation . . .

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

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

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

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

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

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

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

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

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

3. 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.

4. Hello again, sigma1!

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

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

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

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

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

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

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

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

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

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

5. THANKS ALOT.

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