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Integral Calculus
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In the first 1 u need to brk up the cubic term and integrate.......The second 1 is also damn easy............
Write cotx=cosx/sinx.....
So it becums-sinx/[sinx+cosx]...
in numerator write-[1/2][sinx+cosx-(cosx-sinx)]....
divide the two into two parts and u will get
1/2[sinx+cosx-ln|sinx+cos|.....Answer......
RATE IF USEFUL.....
In the first 1 u need to brk up the cubic term and integrate.......The second 1 is also damn easy............
Write cotx=cosx/sinx.....
So it becums-sinx/[sinx+cosx]...
in numerator write-[1/2][sinx+cosx-(cosx-sinx)]....
divide the two into two parts and u will get
1/2[sinx+cosx-ln|sinx+cos|.....Answer......
RATE IF USEFUL.....
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2 . 1 / (1+cotx)
= sin(x) / ( cosx +sin(x))
= 1/ 2 ( 2 sin(x) / (sin(x) +cos(x))
= 1/2 ( sin(x)+cos(x) +sin(x) - cos(x)) / ( sinx +cosx)
=1/2 * ( 1 + (sinx-cosx)/(sinx+cosx))
integration = x/2 - log (sin(x) +cos(x)) /2
1 . Just resolve it .