$\displaystyle \int {{{\sin }^{ - 1}}(x)dx} $ I use u and v' started using u=sin^-1 (x) but I don't know what u' is v=x, v'=1
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Originally Posted by genlovesmusic09 $\displaystyle \int {{{\sin }^{ - 1}}(x)dx} $ I use u and v' started using u=sin^-1 (x) but I don't know what u' is v=x, v'=1 look it up ... it's in your text.
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