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Integral Calculus
6 Dec 2008 17:09:45 IST
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hari mohan
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6 Dec 2008 17:10:35 IST
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Re:find the integral of under root of sin x
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6 Dec 2008 18:58:16 IST
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(Sinx)^1/2
Let sinx=t^2
Cosx dx=2t dt
Sub
Thus eqn becomes
2t^2dt/cosx , but cos x =1-t^4
Eqn is 2t^2 dt/1-t^4
= (t^2+1/1-t^4)+ (t^2-1/1-t^4)
={t^2+1/(1-t^2)* (1+t^2)}+ {t^2-1/(1-t^2)* (1+t^2)}
= integral (1/1-t^2)- integral(1/1+t^2)
= ½ ln{1+t/1-t} - tan?1 1/t
Re substitute t as sinx
Let sinx=t^2
Cosx dx=2t dt
Sub
Thus eqn becomes
2t^2dt/cosx , but cos x =1-t^4
Eqn is 2t^2 dt/1-t^4
= (t^2+1/1-t^4)+ (t^2-1/1-t^4)
={t^2+1/(1-t^2)* (1+t^2)}+ {t^2-1/(1-t^2)* (1+t^2)}
= integral (1/1-t^2)- integral(1/1+t^2)
= ½ ln{1+t/1-t} - tan?1 1/t
Re substitute t as sinx
and not cosx.
.........dnt try
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