When we try to find the integral of a certain function but fail to do so, then we use the reduction formula to integrate the function by parts. This is a method in which the exponent of the function is broken into smaller exponents and product of functions such that it becomes simpler to integrate. This is a recursive process and we keep on simplifying it until we are able to integrate it directly. 

Few of the most known reduction formula of trigonometric functions are as follows:

$\int$ $sin^n (x)\ dx$ = $\frac{-1}{n}$ $sin^{n - 1}\ (x)\ cos (x)\ +$ $\frac{n - 1}{n}$ $\int$ $sin^{n - 2}\ (x)\ dx$

$\int$ $cos^n (x)\ dx$ = $\frac{1}{n}$ $cos^{n - 1}\ (x)\ sin (x)\ +$ $\frac{n - 1}{n}$ $\int$ $cons^{n - 2}\ (x)\ dx$

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

$\int$ $cot^n (x)\ dx$ = $\frac{-1}{n - 1}$ $cot^{n - 1}\ (x)\ -$ $\int$ $cot^{n - 2}\ (x)\ dx$

$\int$ $sec^n (x)\ dx$ = $\frac{1}{n - 1}$ $sec^{n - 2}\ (x)\ tan(x)\ +$ $\frac{n - 2}{n - 1}$ $\int$ $sec^{n - 2}\ (x)\ dx$

$\int$ $csc^n (x)\ dx$ = $\frac{-1}{n - 1}$ $csc^{n - 2}\ (x)\ cot(x)\ +$ $\frac{n - 2}{n - 1}$ $\int$ $csc^{n - 2}\ (x)\ dx$

Reduction formula can be defined as a method to solve an integral problem by reducing it to a problem of easier integral problem. It is done by integration by parts and simplifying it to lower powers such that direct integration could be made. It is an iterative method until simpler integrals are carried out.
$I$ = $\int$ $sin^n x\ dx$

Breakup $sin^n x$ as a product of two functions of sine such that we can apply the integration by part method

$I$ = $\int$ $sin^{(n – 1)}\ x\ \times\ sin x\ dx$

Integrate by parts by letting $u$ = $sin^{(n – 1)}\ x$ and $dv$ = $sin\ x\ dx$

So, $du$ = $(n – 1)\ sin^{(n – 2)}\ x\ cos\ x\ dx$ and $v$ = $-\ cos x$

Integration by parts formula goes by

$u \times v\ -$ $\int$ $v\ \times\ du\ dx$

Substituting the values of $u$ and $v$, we get

$I$ = $sin^{(n – 1)}\ x \times\ -\ cos\ x\ -$ $\int$ $-\ cos\ x\ \times\ (n – 1)\ sin^{(n – 2)}\ x\ cos\ x\ dx$

On simplification

$I$ = $-\ sin^{(n – 1)}\ x\ \times\ cos x\ +\ (n – 1)$ $\int$ $sin^{(n – 2)}\ x\ cos^2\ x\ dx$

Using $sin^2\ x\ +\ cos^2\ x$ = $1$

$I$ = $-\ sin^{(n – 1)}\ x\ \times\ cos\ x\ +\ (n – 1)$ $\int$ $(1\ –\ sin^2\ x)\ sin^{(n – 2)}\ x\ dx$

Expanding

$I$ = $-\ sin^{(n – 1)}\ x\ \times\ cos\ x\ +\ (n – 1)$ $\int$ $sin^{(n – 2)}\ x\ dx\ -\ (n – 1)$ $\int$ $sin^n\ x\ dx$

We know that $I$ = $\int$ $sin^n\ x\ dx$, then

$I$ = $-\ sin^{(n – 1)}\ x\ \times\ cos\ x\ +\ (n – 1)$ $\int$ $sin^{(n – 2)}\ x\ dx\ –\ (n – 1)I$

Transposing the last integral '$I$' to the left side,

$I\ +\ (n – 1)\ I$ = $-\ sin^{(n – 1)}\ x\ \times\ cos\ x\ +\ (n – 1)$ $\int$ $sin^{(n – 2)}\ x\ dx$

$nI$ = $-\ sin^{(n – 1)}\ x\ \times\ cos\ x\ +\ (n – 1)$ $\int$ $sin^{(n – 2)}\ x\ dx$

Divide by $n$,

$\int$ $sin^n\ x\ dx$ = $-$ $\frac{sin^{(n – 1)}\ x\ \times cos\ x}{n}$ $+$ $\frac{(n – 1)}{n}$ $\int$ $sin^{(n – 2)}\ x\ dx$
$I$ = $\int$ $tan^n\ x\ dx$

Breakup $tan^n\ x$ as a product of two functions of sine such that we can apply the integration by part method

$I$ = $\int$ $tan^{(n – 2)}\ x\ \times\ tan^2\ x\ dx$

Using $sec^2\ x\ –\ tan^2\ x$ = $1$, we get

$I$ = $\int$ $tan^{(n – 2)}\ x\ \times\ (sec^2\ x – 1)\ dx$

On expanding,

$I$ = $\int$ $tan^{(n – 2)}\ x\ sec^2\ x\ dx\ -$ $\int$ $tan^{(n – 2)}\ x\ dx$ ...... Equation $1$

Let us first carry out the first integral separately, which is $\int$ $tan^{(n – 2)}\ x\ sec^2\ x\ dx$ using $u$ substitution method

Taking up, $u$ = $tanx\ \Rightarrow\ du$ = $sec^2\ x\ dx$

$\int$ $tan^{(n – 2)}\ x\ sec^2\ x\ dx$ = $\int$ $u^{(n – 2)}\ du$ = $u^\frac{(n – 1)}{(n – 1)}$

Plugin the value of $u$ back, we get

$\int$ $tan^{(n – 2)}\ x\ sec^2\ x\ dx$ = $\frac{tan^{(n – 1) x}}{(n – 1)}$

Putting back $\int$ $tan^{(n – 2)}\ x\ sec^2\ x\ dx$ = $\frac{tan^{(n – 1) x}}{(n – 1)}$ in equation $1$ , we get the final integration result to be

$\int$ $tan^n\ x\ dx$ = $\frac{tan^{(n – 1) x}}{(n – 1)}$ $-$ $\int$ $tan^{(n – 2)}\ x\ dx$
The following are the basic reduction formula with limits of trigonometric functions:

$\rightarrow$ $\int_{\infty}^{\frac{\pi}{2}}$ $sin^2 x\ dx$ = $\int_{\infty}^{\frac{\pi}{2}}$ $cos^2 x\ dx$ = $\frac{\pi}{4}$

$\rightarrow$ $\int$ $\frac{sin (px)}{x dx}$ = $\left\{\begin{matrix} \frac{\pi}{2} & p > 0\\ 0 & p = 0\\ \frac{\pi}{2} & p < 0 \end{matrix}\right.$

$\rightarrow$ $\int$ $sin^{\frac{(px)}{x^2}}$ $dx$ = $\pi \times$ $\frac{p}{2}$

$\rightarrow$ $\int_0^1$ $-\ cos$ $⁡\frac{px}{x^2}$ $dx$ = $\pi \times$ $\frac{p}{2}$

$\rightarrow$ $\int_0^{\infty}$ $\frac{cos⁡(px)\ -\ cos⁡(qx)}{x dx}$ = $ln$ $(\frac{q}{p})$

$\rightarrow$ $\int_0^{\infty}$ $\frac{cos⁡(px)\ -\ cos⁡(qx)}{x^2\ dx}$ = $\pi$ $\frac{(q – p)}{2}$

$\rightarrow$ $\int_0^{2 \pi}$ $\frac{dx}{(a + b sin x)}$ = $\frac{2 \pi}{\sqrt{a^2\ -\ b^2}}$

$\rightarrow$ $\int_0^{2 \pi}$ $\frac{dx}{(a + b cos x)}$ = $\frac{2 \pi}{\sqrt{a^2\ -\ b^2}}$

$\rightarrow$ $\int_0^{\infty}$ $sin⁡ ax^2\ dx$  = $\int_0^{\infty}$ $cos\ ⁡ax^2\ dx$ = $\frac{1}{2}$ $\sqrt{(\frac{\pi}{2a})}$

$\rightarrow$ $\int_0^{\infty}$ $\frac{sin ⁡x}{\sqrt{x dx}}$ = $\int_0^{\infty}$ $\frac{cos ⁡x}{\sqrt{x dx}}$ = $\sqrt{(\frac{\pi}{2})}$

$\rightarrow$ $\int_0^{\infty}$ $\frac{sin^3 ⁡x}{x^3 dx}$ = $\frac{3 \pi}{8}$

$\rightarrow$ $\int_0^{\infty}$ $\frac{sin^4⁡  x}{x^4 dx}$ = $\frac{\pi}{3}$

$\rightarrow$ $\int_0^{\infty}$ $\frac{tan⁡  x}{x dx}$ = $\frac{\pi}{2}$

$\rightarrow$ $\int_0^{\frac{\pi}{2}}$ $\frac{dx}{(a + b cos x)}$ = $\frac{arcos (\frac{b}{a})}{\sqrt{(a^2\ –\ b^2)}}$
We can derive reduction formulas by applying integration by parts method. Let us derive the most common ones

$I$ = $\int$ $cos^n\ x\ dx$

Breakup $cos^n\ x$ as a product of two functions of sine such that we can apply the integration by part method

$I$ = $\int$ $cos^{(n – 1)}\ x\ \times\ cos\ x\ dx$

Integrate by parts by letting $u$ = $cos^{(n – 1)}\ x$ and $dv$ = $cos\ x\ dx$

So, $du$ = $(n – 1)\ cos^{(n – 2)}\ x\ (-sin\ x)\ dx$ and $v$ = $sin\ x$

Integration by parts formula goes by

$u \times v\ -$ $\int$ $v \times\ du\ dx$

Substituting the values of $u$ and $v$, we get

$I$ = $sin\ x\ cos^{(n – 1)}\ x\ -$ $\int$ $sin\ x\ (n – 1)\ cos^{(n – 2)}\ x\ (-sin\ x)\ dx$

On simplification,

$I$ = $sin\ x\ cos^{(n – 1)}\ x\ +\ (n – 1)$ $\int$ $sin^2\ x\ cos^{(n – 2)}\ x\ dx$

Using $sin^2\ x\ +\ cos^2\ x$ = $1$,

$I$ = $sin\ x\ cos^{(n – 1)}\ x\ +\ (n – 1)$ $\int$ $(1\ –\ cos^2\ x)\ cos^{(n – 2)}\ x\ dx$

Expanding,

$I$ = $sin\ x\ cos^{(n – 1)}\ x\ +\ (n – 1)$ $\int$ $cos^{(n – 2)}\ x\ dx\ –\ (n – 1)$ $\int$ $cos^n\ x\ dx$

We know that $I$ = $\int$ $cos^n\ x\ dx$, then

$I$ = $sin\ x\ cos^{(n – 1)}\ x\ +\ (n – 1)$ $\int$ $cos^{(n – 2)}\ x\ dx\ –\ (n – 1)I$

Transposing the last integral '$I$' to the left side,

$I\ +\ (n – 1)\ I$ = $sin\ x\ cos^{(n – 1)}\ x\ +\ (n – 1)$ $\int$ $cos^{(n – 2)}\ x\ dx$

$nI$ = $sin\ x\ cos^{(n – 1)}\ x\ +\ (n – 1)$ $\int$ $cos^{(n – 2)}\ x\ dx$

Divide by $n$,

$\int$ $cos^n\ x\ dx$ = $\frac{sin\ x\ cos^{(n – 1)}\ x}{n}$ $+$ $\frac{(n – 1)}{n}$ $\int$ $cos^{(n – 2)}\ x\ dx$
Example 1: 

Find out $\int$ sin^3 x dx

Solution: 

It is known that,

$\int$ $sin^n\ x\ dx$ = $-$ $\frac{sin^{(n – 1)}\ x\ \times\ cos\ x}{n}$ $+$ $\frac{(n – 1)}{n}$ $\int$ $sin^{(n – 2)}\ x\ dx$

Here $n$ = $3$, so

$\int$ $sin^3\ x\ dx$ = $-$ $\frac{sin^{(3 – 1)}\ x\ \times\ cos\ x}{3}$ $+$ $\frac{(3 – 1)}{3}$ $\int$ $sin^{(3 – 2)}\ x\ dx$

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

$\int$ $sin^3\ x\ dx$ = $-$ $\frac{sin^2\ x\ \times\ cos\ x}{3}$ $+$ $\frac{2}{3}$ $(-cos\ x)$

$\int$ $sin^3\ x\ dx$ = $-$ $\frac{sin^2\ x\ \times\ cos\ x}{3}$ $–$ $\frac{2cos\ x}{3}$
Example 2: 

Find out $\int$ $cos^3\ x\ dx$

Solution: 

It is known that 

$\int$ $cos^n\ x\ dx$ = $\frac{sin\ x\ cos^{(n – 1)} x}{n}$ $+$ $\frac{(n – 1)}{n}$ $\int$ $cos^{(n – 2)}\ x\ dx$

Here, $n$ = $3$. So,

$\int$ $cos^3\ x\ dx$ = $\frac{sin\ x\ cos^{(3 – 1)}\ x}{3}$ $+$ $\frac{(3 – 1)}{3}$ $\int$ $cos^{(3 – 2)}\ x\ dx$

$\int$ $cos^3\ x\ dx$ = $\frac{sin\ x\ cos^2\ x}{3}$ $+$ $\frac{2}{3}$ $\int$ $cos\ x\ dx$

$\int$ $cos^3\ x\ dx$ = $\frac{sin\ x\ cos^2\ x}{3}$ $+$ $\frac{2}{3}$ $sin\ x$
Example 3: 

Find out $\int$ $tan^3\ x\ dx$

Solution: 

It is known that,

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

Here $n$ = $3$, so

$\int$ $tan^3\ x\ dx$ = $\frac{tan^{(3 – 1)}\ x}{(3 – 1)}$ $-$ $\int$ $tan^{(3 – 2)}\ x\ dx$

$\int$ $tan^3\ x\ dx$ = $\frac{tan^2\ x}{2}$ $-$ $\int$ $tan\ x\ dx$

$\int$ $tan^3\ x\ dx$ = $\frac{tan^2\ x}{2}$ $+$ $ln|cos x|$ 
Example 4: 

Find out $\int$ $sec^4\ x\ dx$

Solution: 

It is known that
 
Here, $n$ = $4$. So,

$\int$ $sec^4\ x\ dx$ = $\frac{1}{(4 – 1)}$ $sec^{(4 – 2)}\ x\ tan\ x\ +$ $\frac{(4 – 2)}{(4 – 1)}$ $\int$ $sec^{(4 – 2)}\ dx$

$\int$ $sec^4\ x\ dx$ = $\frac{1}{3}$ $sec^2\ x\ tan\ x\ +$ $\frac{2}{3}$ $\int$ $sec^2\ x\ dx$

$\int$ $sec^4\ x\ dx$ = $\frac{1}{3}$ $sec^2\ x\ tan\ x\ +$ $\frac{2}{3}$ $tan\ x$