Reduction Formulas

# Reduction Formulas

Sometimes we may be interested in deriving a reduction formula for an integral, or a general identity for a seemingly complex integral. The list below outlines the most common reduction formulas:

• Reduction Formula for Sine: $\int \sin ^n x \: dx = -\frac{1}{n} \cos x \sin ^{n-1} x + \frac{n-1}{n} \int \sin ^{n-2} x \: dx$.
• Reduction Formula for Cosine: $\int \cos ^n x \: dx = \frac{1}{n} \sin x \cos ^{n-1} x + \frac{n-1}{n} \int \cos ^{n-2} x \: dx$.
• Reduction Formula for the Natural Logarithm: $\int (\ln x)^n \: dx = x(\ln x)^n - n \int (\ln x)^{n-1} \: dx$.

For example, let's prove the reduction formula for $\int \sin ^n x \: dx$. We can accomplish this by substitution, but first let's rewrite this integral by splitting it up in the following way such that $\int \sin ^n x \: dx = \int sin^{n-1} x \sin x \: dx$. Now let's try to evaluate this integral using Indefinite Integration by Parts.

Let's first let $u = \sin^{n-1} x$ and let $dv = \sin x \: dx$. It thus follows that $du = (n-1)\sin^{n-2} x \cos x \: dx$ and $v = - \cos x$. Recall that for integration by parts, $\int u \: dv = uv - \int v \: du$. It thus follows that:

(1)
\begin{align} \quad \int \sin ^n x \: dx = - \sin^{n-1} x \cos x - \int - \cos x \cdot (n-1) \sin ^{n-2} x \cos x \: dx \\ \int \sin ^n x \: dx = - \sin^{n-1} x \cos x + \int \cos x \cdot (n-1) \sin ^{n-2} x \cos x \: dx \\ \int \sin ^n x \: dx = - \sin^{n-1} x \cos x + (n - 1)\int \cos ^2 x \sin ^{n-2} x \: dx \end{align}

Now recall that $\cos ^2 x = 1 - \sin ^2 x$. Applying this substitution we obtain:

(2)
\begin{align} \int \sin ^n x \: dx = - \sin^{n-1} x \cos x + (n - 1)\int (1 - \sin ^2 x) \sin ^{n-2} x \: dx \\ \int \sin ^n x \: dx = - \sin^{n-1} x \cos x + (n - 1)\int \sin ^{n-2} x - \sin ^n x \: dx \\ \quad \int \sin ^n x \: dx = - \sin^{n-1} x \cos x + (n - 1)[ \int \sin ^{n-2} x \: dx - \int \sin ^n x \: dx ] \\ \quad \int \sin ^n x \: dx = - \sin^{n-1} x \cos x + (n - 1)\int \sin ^{n-2} x \: dx \: - (n-1) \int \sin ^n x \: dx \\ \quad \int \sin ^n x \: dx = - \sin^{n-1} x \cos x + (n - 1)\int \sin ^{n-2} x \: dx \: - n \int \sin ^n x \: dx + n \int \sin ^n x \: dx \\ n \int \sin ^n x \: dx = - \sin^{n-1} x \cos x + (n - 1)\int \sin ^{n-2} x \: dx \\ \int \sin ^n x \: dx = \frac{1}{n} [ -\sin^{n-1} x \cos x + (n - 1)\int \sin ^{n-2} x \: dx ] \\ \end{align}

So for $n ≥ 2$, then $\int \sin ^n \: dx = \frac{1}{n} [ -\sin^{n-1} x \cos x + (n - 1)\int \sin ^{n-2} x \: dx ]$

## Example 1

Prove the following reduction formula: $\int x^n e^x \: dx = x^n e^x - n \int x^{n-1} e^x \: dx$.

Let's let $u = x^n$ and $dv = e^x \: dx$. It thus follows that $du = nx^{n-1} \: dx$ and $v = e^x$. By substituting these back into the integration by parts identity, we obtain:

(3)
\begin{align} \int x^n e^x \: dx = x^n e^x - \int e^x \cdot nx^{n-1} \: dx \\ \int x^n e^x \: dx = x^n e^x - n \int x^{n-1} e^x\: dx \\ \end{align}

And we are done, it is that simple. Reduction formulas are not always simple though as we'll see in example 2.

## Example 2

Prove the following reduction formula: $\int \sec ^n x \: dx = \frac{\tan x \sec ^{n - 2} x}{n - 1} + \frac{n-2}{n-1} \cdot \int \sec ^{n-2} x \: dx$ for n ≠ 1.

Let's first separate $\int \sec ^n x \: dx$, that is $\int \sec ^n x \: dx = \int \sec^{n-2} x \sec^2 x \: dx$.

And let's let $u = \sec ^2 x$ and $dv = \sec ^{n - 2} x \: dx$. It thus follows that $du = (n-2) \sec ^ {n - 3} x \cdot \sec x \tan x = (n-2) \sec ^{n - 2} x \tan x \: dx$, and $v = \tan x$.

Making the appropriate substitutions for integration by parts and using the fact that $\tan ^2 x = \sec ^2 x - 1$, we obtain:

(4)
\begin{align} \int \sec ^ n x \: dx = \sec ^ {n - 2} x \tan x - \int \tan x (n-2) \sec ^{n - 2} x \tan x \: dx \\ \int \sec ^ n x \: dx = \sec ^ {n - 2} x \tan x - (n - 2) \int \tan^2 x \sec ^{n - 2} x \: dx \\ \int \sec ^ n x \: dx = \sec ^ {n - 2} x \tan x - (n - 2) \int (\sec ^2 x - 1) \sec ^{n - 2} x \: dx \\ \int \sec ^ n x \: dx = \sec ^ {n - 2} x \tan x - (n - 2) \int \sec ^n x - \sec ^{n - 2} x \: dx \\ \quad \int \sec ^ n x \: dx = \sec ^ {n - 2} x \tan x - (n - 2) \int \sec ^n x \: dx + (n - 2) \int \sec ^{n - 2} x \: dx \\ \quad \int \sec ^ n x \: dx + (n - 2) \int \sec ^n x \: dx= \sec ^ {n - 2} x \tan x + (n - 2) \int \sec ^{n - 2} x \: dx \\ \quad (n - 1) \int \sec ^n x \: dx= \sec ^ {n - 2} x \tan x + (n - 2) \int \sec ^{n - 2} x \: dx \\ \quad \int \sec ^n x \: dx = \frac{1}{(n-1)} [ \sec ^ {n - 2} x \tan x + (n - 2) \int \sec ^{n - 2} x \: dx ] \\ \quad \int \sec ^n x \: dx = \frac{\sec ^ {n - 2} x \tan x}{n - 1}+ \frac{n - 2}{n-1} \int \sec ^{n - 2} x \: dx \\ \quad \int \sec ^n x \: dx = \frac{\tan x \sec ^ {n - 2} x}{n - 1}+ \frac{n - 2}{n-1} \int \sec ^{n - 2} x \: dx \\ \end{align}

So we are done. For $n ≠ 1$, $\int \sec ^n x \: dx = \frac{\tan x \sec ^ {n - 2} x}{n - 1}+ \frac{n - 2}{n-1} \int \sec ^{n - 2} x \: dx$