Reduction Formula

Reduction formula is regarded as a method of integration. Helps us solve the powers of elementary functions, polynomials of arbitrary degree, products of transcendental functions and the functions that cannot be integrated easily, thus, easing the process of integral problem.

Below given are some of the reduction formulas:

\[\large \int Sin^{n}(x)dx=\frac{-Sin^{n-1}(x)Cos(x)}{n}+\frac{n-1}{n}Sin^{n-2}(x)dx\]

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

\[\large \int sin^{n}(x)\: cos^{m}(x)dx=\frac{sin^{n+1}(x)cos^{m-1}(x)}{n+m}+\frac{m-1}{n+m}\: \int sin^{n}(x)\: cos^{m-2}(x)dx\]

\[\large \int x^{n}cos(x)dx=x^{n}sin(x)-n\int x^{n-1}sin(x)dx\]

\[\large \int x^{n}sin(x)dx=-x^{n}cos(x)+n\int x^{n-1}cos(x)dx\]

Solved example

Question: Evaluate the integral: $\int tan^{5}(2x)dx$

Solution:

Use:

$\int tan^{n}(u)du=\frac{1}{n-1}tan^{n-1}(u)-\int tan^{n-2}(u)du$

Substitution:

$a=2x$
$\frac{1}{2}\:da=dx$

Hence,

$\int tan^{5}(2x)dx=\frac{1}{2}\left[\int tan^{5}(a)da\right]$

$=\frac{1}{2}\left[\frac{1}{4}\:tan^{4}(a)-\int tan^{3}(a)da\right]$

$=\frac{1}{2} \left [ \frac{1}{4} \: tan^{4}(a) – \left [ \frac{1}{2} \: tan^{2}(a) – \int tan(a)da \right] \right]$

$\frac{1}{8}\:tan^{4}(a)-\frac{1}{4}\:tan^{2}(a)+\frac{1}{2}\:ln\:sec(a)+C$

$\frac{1}{8}\:tan^{4}(2x)-\frac{1}{4}\:tan^{2}(2x)+\frac{1}{2}\:ln\:sec(2x)+C$


Practise This Question

Given the area of rectangle is A=25a235a+12. The length is given as (5a3). Therefore, the width is: