IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

zehra

^{integretion of cube of cosecx}

titun

It's easy.

Take cosecx as the first function & (cosecx)^2 as the second function. Now, integrate by parts. For the second integration of cosecx(cotx)^2 write
cosecx (cotx)^2 = cosecx [(cosecx)^2 - 1]

The ans. will be,

[ -cosecx cotx + ln { cosecx - cotx } ] /2

{} represents the modulus sign.

If you still face any difficulty, I'll write the whole solution. !!

Keep Rockin !!

magiclko

let cosec x = t

then -cosec x cot x dx = dt

=> cosec x dx = -dt/ [sqrrt(t² -1)]

thrfore

_{[ ]}

^{[ ]} cosec ³ x = _{[ ]}

^{[ ]} - t²/ [sqrrt(t² -1)] dt

subtracting and adding 1 in numerator we have,

RHS =

_{[ ]}

^{[ ]} -[sqrrt(t² -1)] dt - _{[ ]}

^{[ ]} [1/{sqrrt(t² -1)}] dt

and for this we hav standard formulae.....

if any problem, do ask again

puneet

hiiii

the answer is already there .. isn;t it

cheers

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