# Math Help - Integration by parts and recursion formula

1. ## Integration by parts and recursion formula

I'm supposed to convert the even powers of tangent to secant using a recursion formula, repeatedly if necessary, and then evaluate.

Here are to two problems:

∫ tan^2(x)sec(x) dx

∫ tan^4(x)sec(x) dx

Here is the recursion formula:

∫ sec^n(x) dx = (tan(x)sec^(n-2)(x))/(n-1) - ((n-2)/(n-1))⋅∫ sec^(n-2)(x) dx

2. ## Re: Integration by parts and recursion formula

Use the trigonometric relation $tan^2x = sec^2x - 1 \implies tan^{2k} x = (sec^2 x - 1)^k$ now use the binomial expansion. Apply this to the above integrals.

Second integral becomes
$\int tan^4 x sec x dx = \int (sec^2 x -1)^2 sec x dx =$ $\int sec^5 x dx -2\int sec^3 x dx + \int sec x dx$ Now use the recursion relationship on each integral.

Apply the same principle to the first integral as well.

3. ## Re: Integration by parts and recursion formula

Hello, Preston019!

There is a typo in the Recursion Formua.
There should be a plus (+) between the two expressions.

I'm supposed to convert the even powers of tangent to secant using a recursion formula,
repeatedly if necessary, and then evaluate.

$\text{Recursion Formula: }\:\int\sec^n(x)\,dx \;=\;\frac{1}{n-1}\sec^{n-2}(x)\tan(x) \;{\color{red}+}\; \frac{n-2}{n-1}\int\sec^{n-2}(x)\,dx$

$\text{Problem: }\:I \;=\;\int\tan^2(x)\sec(x)\,dx$

$\text{We have: }\:I \;=\;\int\left[\sec^2(x) - 1\right]\sec(x)\,dx$
. . . . . . . . . . $=\;\int\left[\sec^3(x) - \sec(x)\right]\,dx$
. . . . . . . . . . . $=\; \underbrace{\int\sec^3(x)\,dx}_{[1]} - \underbrace{\int\sec(x)\,dx}_{[2]}$

Integral [2] has a standard formula: . $\ln|\sec(x) + \tan(x )| + C$

We will apply the Recursion Formula to integral [1].

Substitute $n = 3:$

$\int\sec^3(x)\,dx \;=\;\tfrac{1}{2}\sec(x)\tan(x) + \tfrac{1}{2}\int\sec(x)\,dx$

. . . . . . . . . $=\;\tfrac{1}{2}\sec(x)\tan(x) + \tfrac{1}{2}\ln|\sec(x) + \tan(x)|$

$\text{Then: }\:I \:=\:\tfrac{1}{2}\sec(x)\tan(x) + \tfrac{1}{2}\ln|\sec(x) + \tan(x)| - \ln|\sec(x) + \tan(x)|+C$

. . . . . . $=\;\tfrac{1}{2}\sec(x)\tan(x) - \tfrac{1}{2}\ln|\sec(x) + \tan(x)| + C$