IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ramkumar_november

given integral is ........

I = $\int\frac{cos8x-cos7x}{1+2cos5x}\;dx$

now consider the expression .......

$\frac{cos8x-cos7x}{1+2cos5x}\;$

in this expression multiply and divide by 2sin5x

$\frac{cos8x-cos7x}{1+2cos5x}\;\dot\;\frac{2sin5x}{2sin5x}$

= $\frac{sin13x-sin3x-sin12x+sin2x}{2(sin5x+sin10x)}$

= $\frac{2sin(\frac{15x}{2})cos(\frac{11x}{2})-2sin(\frac{15x}{2})cos(\frac{9x}{2})}{4sin(\frac{15x}{2})cos(\frac{5x}{2})}$

= $\frac{cos(\frac{11x}{2})-cos(\frac{9x}{2})}{2cos(\frac{5x}{2})}$

= $\frac{-2sin5xsin(x/2)}{2cos(\frac{5x}{2})}$

= $-2sin(\frac{5x}{2})sin(\frac{x}{2})$

= $cos3x\;-\;cos2x$

so the integral becomes .................

I = $\int(cos3x\;-\;cos2x)\;dx$

I = $\frac{sin3x}{3}\;-\;\frac{sin2x}{2}\;+\;C$

i hope you are clear with my simplification......................