MyRank: Indefinite Integrals

Explanation Of Indefinite Integrals.......

A function $\phi \left( x \right)$ is called a primitive (or) an anti-derivative of a function f(x) of

$\phi '\left( x \right)=f\left( x \right)$

$\Rightarrow \,\,\,\,\,\,\int_{{}}^{{}}{f\left( x \right)}=\phi \left( x \right)+C$

ere f(x) = integrand

$\frac{d\left( \int_{{}}^{{}}{f\left( x \right)}\,dx \right)}{dx}=f\left( x \right)$

Algorithm to solve integrals of form:

$\int_{{}}^{{}}{{{\sin }^{m}}x}\,{{\cos }^{n}}x\,dx ,$

$\int_{{}}^{{}}{{{\sin }^{m}}x\,dx}$

$\int_{{}}^{{}}{{{\cos }^{n}}x}dx,$ where $m,n\in N$

Step 1: Find m,n

Step 2:

* If m is odd i.e. power or index of sin x is odd, put cos x = t and reduced the integrand in terms of it.

* If n is odd then put sin x = t and reduce the integrand.

* If both are odd you can use any of above two methods.

To Evaluate integrals of form

$\int_{{}}^{{}}{{{\sin }^{m}}x}\,{{\cos }^{n}}x\,dx$ where $m,n\in Q$ and m + n is negative even integer then put

tan x = t.

To evaluate integrals of form $\int_{{}}^{{}}{{{\sin }^{m}}x}\,{{\cos }^{n}}x\,dx$ where m,n are positive even integers then.
Let Z = cos x + i sin x

$\frac{1}{Z}=cosx-i\text{ }sinx$

$Z+\frac{1}{Z}=2\,\cos x$

$Z-\frac{1}{Z}=2i\,\operatorname{sinx}$
$cos\,nx=\frac{1}{2}\left( {{Z}^{n}}+\frac{1}{{{Z}^{n}}} \right)$

$\sin nx=\frac{1}{2i}\left( {{Z}^{n}}-\frac{1}{{{Z}^{n}}} \right)$

Some important standard integrals :

$\int_{{}}^{{}}{\frac{1}{{{x}^{2}}-{{a}^{2}}}dx=}\frac{1}{2a}\log \left| \frac{x-a}{x+a} \right|+C$

$\int_{{}}^{{}}{\frac{1}{{{a}^{2}}-{{x}^{2}}}dx=}\frac{1}{2a}\log \left| \frac{a+x}{a-x} \right|+C$

$\int_{{}}^{{}}{\frac{1}{\sqrt{{{a}^{2}}+{{x}^{2}}}}}dx=\log \left| x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right|+C$
$\int_{{}}^{{}}{\frac{1}{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx=}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C$

Integrals of form

$\int_{{}}^{{}}{\frac{dx}{{{\left( a+b\cos x \right)}^{2}}}},$ $\int_{{}}^{{}}{\frac{dx}{{{\left( a+b\sin x \right)}^{2}}}}$ to evaluate this type of integrals

Define $P=\frac{\sin x}{a+b\cos x}$ (or) $\frac{\cos x}{a+b\sin x}$

Depending on integral then find $\frac{dP}{dx}$ in terms of $\frac{1}{a+b\cos x}$ (or) $\frac{1}{a+b\sin x}$

Now integrate both sides of expression to get required integral.

Integration by parts :

$\int_{{}}^{{}}{u\nu }dx=u\left( \int_{{}}^{{}}{\nu dx} \right)-\int_{{}}^{{}}{\left\{ \frac{du}{dx}.\int_{{}}^{{}}{\nu dx} \right\}}dx$

Very IMP Note:
Choose first function as the function which come first in word “ILATE”

(1) I – Inverse trigonometric

L – logarithmic functions

A – algebraic functions

T for Trigonometric functions

E for exponential functions.

è Integrals of the form

$\int_{{}}^{{}}{{{e}^{x}}}\left( f\left( x \right)+f'\left( x \right) \right)dx={{e}^{x}}f\left( x \right)+C$
Note:
   $\int_{{}}^{{}}{{{e}^{kx}}}\left\{ k\,f\left( x \right)+f'\left( x \right) \right\}={{e}^{kx}}f\left( x \right)+C$
è    $\int_{{}}^{{}}{{{e}^{ax}}}\sin bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left( a\sin \,bx+b\cos \,bx \right)+C$
è    $\int_{{}}^{{}}{{{e}^{ax}}}\,\cos \,bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left( a\,cos\,bx+b\sin \,bx \right)+C$

Some Very Important Integrals :

      $\int_{{}}^{{}}{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{1}{2}{{a}^{2}}{{\sin }^{-1}}\left( \frac{x}{a} \right)+C$
$\int_{{}}^{{}}{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx=\frac{x}{2}\sqrt{{{a}^{2}}+{{x}^{2}}}+\frac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{a}^{2}}+{{x}^{2}}} \right|+C$
   $\int_{{}}^{{}}{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}-\frac{{{a}^{2}}}{2}\log \left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|+C$

Reduction Formulae :
$\int_{{}}^{{}}{{{\sin }^{n}}x\,dx}={{I}_{n}}$
${{I}_{n}}=\frac{-{{\sin }^{n-1}}x\,\cos \,x}{n}+\frac{n-1}{n}{{I}_{n-2}}$

$\int_{{}}^{{}}{{{\cos }^{n}}x\,dx={{I}_{n}}}$
${{I}_{n}}=\frac{{{\cos }^{n-1}}x\,\sin \,x}{n}+\frac{n-1}{n}{{I}_{n-2}}$

$\int_{{}}^{{}}{{{\tan }^{n}}x\,dx={{I}_{n}}}$
${{I}_{n}}=\frac{{{\tan }^{n-1}}x}{n-1}-{{I}_{n-2}}$

$\int_{{}}^{{}}{\cos e{{c}^{n}}x\,dx}={{I}_{n}}$
   ${{I}_{n}}=\frac{-\cos e{{c}^{n-2}}\,\cot x}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}}$

   $\int_{{}}^{{}}{{{\sec }^{n}}x\,dx}={{I}_{n}}$
${{I}_{n}}=\frac{{{\sec }^{n-2}}x\,\tan \,x}{n-1}+\frac{n-2}{n-1}{{I}_{n-2}}$

   $\int_{{}}^{{}}{{{\cot }^{n}}x\,dx}={{I}_{n}}$
   ${{I}_{n}}=\frac{-{{\cot }^{n-1}}x}{n-1}-{{I}_{n-2}}$

MyRank

Friday, April 24, 2015

Indefinite Integrals

No comments:

Post a Comment