Posted by Soroban on August 31, 2002 at 01:26:37:
In Reply to: Re: Need help desperately - seems like a simple integral? posted by Ryan on August 29, 2002 at 14:50:22:
: ill give your prob. a try--> I have been teaching myself calculus so I am not sure if my answer is right....use your own judgement!
: integral ( tan(x^1/2))
: u= x^1/2
: du = (1/2)x^(-1/2) dx
: 2du = x^-1/2 dx
: 2u^2 du = dx
Not quite true, Ryan.
2 du = x^(-1/2)dx
Multiply by x^(1/2):
2 x^(1/2) du = dx
Since x^(1/2) = u,
2u du = dx
: 2* integral [( u^2)*(tan u ) du ]
: 2 * (1/3)u^3 tan u + u^2 ln |sec u | + C
Did you just invent a "Product Formula" for integrals?
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