IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

rtiit

Evaluate

rtiit

ne1 plz

ziauddin75

PUT X=ATAN@

THEN FURTHER SOLVING YOU WILL GET (COSX)^12 NOW YOU CAN SOLVE USING TRIGO

ramkumar_november

this reduction formula will be useful for you.......

when $I_n=\int\frac{dx}{(x^2+a^2)^n}$

$I_n\;\;=$ $\frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}$ + $\frac{(2n-3)I_{n-1}}{2a^2(n-1)}$

put n=7 .....

use this reduction formula again and again and then

at last you will get I₁ = dx/x²+a² = 1/a tan^-1(x/a)

clear..???

rtiit

thnx

but wont it be too lengthy ?????

karthik2789

ans:231 *pi /2048 *a^13

bladeX

substitute X= A TAN t
THEN YOU GET A^-14 AND COS^14
THEN EXPAND IT BY USING TRIGONOMETCIAL SERIES

rtiit

no substituting tan is not working.......

@karthick
can u plz give explaination??

mathwiz

how can u evaluate integral cos^14x dx???

ramkumar_november

I = $\int\limits_0^{\infty}\frac{dx}{(a^2+x^2)^7}$

$now\;\;put\;\;x=atan\theta$

$dx=asec^2\theta.d\theta$

I = $\int\limits_0^{\frac{\pi}{2}}\frac{cos^{12}\theta.d\theta}{a^{13}}$

now use walli's formula.....

I = $\frac{1}{a^{13}}.\frac{11*9*7*5*3*1}{12*10*8*6*4*2}.\frac{\pi}{2}$

I = $\frac{231.\pi}{a^{13}.2048}$