IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ashishhawking

_{[ ][ ]} {1-sinx/2}/cosx/2(cosx/4-sinx/4)

X

sid_2222000

multiply and divide by [1+sinx/2]to get cos^2(x/2)oin trhe numerator after tht cancel the cosx/2 in the denom........then divide and multiply wid cosx/4 +sinx/4 to get cosx/2 in the denom......cosx/2 cancels out.......wat is left is (cosx/4 +sinx/4)/(1+sin x/2) whose limx-->pi is 1/root 2...rate me if u find it useful......

ruhi

hi..........lim x>pi (1-sinx/2)/cosx/2.(cosx/4 - sinx/4)
mul. num n deno. by (1+ sinx/2)(cosx/4+sinx/4).....nd on solving ull get....
lim x> pi 1/(1+sinx/2) = 1/2
hope u got it :)

iitkgp_bipin

L = _[x][pi] {1-sinx/2} / {cosx/2(cosx/4-sinx/4)}

Multiply and divide by (1 + sinx/2).(cosx/4 + sinx/4)

L = _[x][pi] (1-sin²x/2).(cosx/4+sinx/4) / {cosx/2(1+sinx/2)(cos²x/4-sin²x/4)}

L = _[x][pi] (cos²x/2).(cosx/4+sinx/4) / {cosx/2(1+sinx/2)(cosx/2)}

L = _[x][pi] (cosx/4+sinx/4) / (1+sinx/2)

L = (cos

/4+sin

/4) / (1+sin

/2)

L = 1/

2

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