reduction formulas for integration of powers

The following reduction formulas, with integer $n$ and via integration by parts, may be used for lowing ($n>0$) or raising () the the powers:

• ${\displaystyle \int }{\sin }^{n}xdx=-{\displaystyle \frac{1}{n}}{\sin }^{n-1}x\cos x+{\displaystyle \frac{n-1}{n}}{\displaystyle \int }{\sin }^{n-2}xdx\mathit{\hspace{1em}\hspace{1em}}(n\gtrless 0)$

• ${\displaystyle \int }{\cos }^{n}xdx={\displaystyle \frac{1}{n}}{\cos }^{n-1}x\sin x+{\displaystyle \frac{n-1}{n}}{\displaystyle \int }{\cos }^{n-2}xdx\mathit{\hspace{1em}\hspace{1em}}(n\gtrless 0)$

• ${\displaystyle \int }{(\ln x)}^{n}dx=x{(\ln x)}^n-n{\displaystyle \int }{(\ln x)}^{n-1}dx\mathit{\hspace{1em}\hspace{1em}}(n\gtrless 0)$

• ${\displaystyle \int }{\displaystyle \frac{1}{{(1+x^2)}^n}}dx={\displaystyle \frac{1}{2n-2}}\cdot {\displaystyle \frac{x}{{(1+x^2)}^{n-1}}}+{\displaystyle \frac{2n-3}{2n-2}}{\displaystyle \int }{\displaystyle \frac{1}{{(1+x^2)}^{n-1}}}dx\mathit{\hspace{1em}}(n>1)$

Example.  For finding ${\displaystyle \int \frac{dx}{{\sin }^{3}x}}$, we apply the first formula with  $n:=-1$,  getting first

$$\int \frac{dx}{\sin x}=-\frac{1}{-1}\cdot \frac{\cos x}{{\sin }^{2}x}+\frac{-2}{-1}\int \frac{dx}{{\sin }^{3}x}.$$

From this we solve

$$\int \frac{dx}{{\sin }^{3}x}=-\frac{1}{2}\frac{\cos x}{{\sin }^{2}x}+\int \frac{dx}{\sin x}=-\frac{1}{2}\frac{\cos x}{{\sin }^{2}x}+\ln \left|\tan \frac{x}{2}\right|+C$$

(see integration of rational function of sine and cosine).

Note 1.  Instead of the two first formulae, it is simpler in the cases when $n$ is a positive odd or a negative even number to use the following
${\displaystyle \int {\sin }^{2m+1}xdx}={\displaystyle \int {\sin }^{2m}x\sin xdx}=-{\displaystyle \int {(1-{\cos }^{2}x)}^m(-\sin x)𝑑x}$,
${\displaystyle \int {\cos }^{2m+1}xdx}={\displaystyle \int {\cos }^{2m}x\cos xdx}={\displaystyle \int {(1-{\sin }^{2}x)}^m\cos xdx},$
${\displaystyle \int \frac{1}{{\sin }^{2m}x}𝑑x}={\displaystyle \int \frac{1}{{\sin }^{2m-2}x}\cdot \frac{1}{{\sin }^{2}x}𝑑x}=-{\displaystyle \int {(1+{\cot }^{2}x)}^{m-1}d\cot x}$,
${\displaystyle \int \frac{1}{{\cos }^{2m}x}𝑑x}={\displaystyle \int \frac{1}{{\cos }^{2m-2}x}\cdot \frac{1}{{\cos }^{2}x}𝑑x}={\displaystyle \int {(1+{\tan }^{2}x)}^{m-1}d\tan x}$,
which may be found after making the powers on the right hand sides to polynomials.

Note 2.  $\int {\tan }^{n}xdx$  ($n\in {\mathbb{Z}}_{+}$)  is obtained easily by the substitution (http://planetmath.org/IntegrationBySubstitution)  $\tan x:=t$,  $dx=\frac{dt}{t^2+1}$  and a division; e.g.

${\displaystyle \int {\tan }^{5}xdx}$$\mathrm{\hspace{0.17em}}={\displaystyle \int \frac{t^5}{t^2+1}𝑑t}={\displaystyle \int \left(t^3-t+\frac{t}{t^2+1}\right)𝑑t}$

$\mathrm{\hspace{0.17em}}={\displaystyle \frac{t^4}{4}}-{\displaystyle \frac{t^2}{2}}+{\displaystyle \frac{1}{2}}\ln (t^2+1)+C$

$\mathrm{\hspace{0.17em}}={\displaystyle \frac{{\tan }^{4}x}{4}}-{\displaystyle \frac{{\tan }^{2}x}{2}}+\ln \sqrt{{\tan }^{2}x+1}+C.$