# We have only one function, but one we can differentiate

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One trick that might work is to let the other function be the unity function 1.  This is best illustrated by an example.

$I = \int {{{\sin }^{ - 1}}x\;dx} = \int {1 \cdot {{\sin }^{ - 1}}x\;dx} =$

$\left| {\begin{array}{*{20}{c}} {u = {{\sin }^{ - 1}}x} \\ {\frac{{dv}}{{dx}} = 1} \\ {\frac{{du}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }}} \\ {v = x} \end{array}} \right.$

$= x{\sin ^{ - 1}}x - \int {\frac{x}{{\sqrt {1 - {x^2}} }}dx}$

Now we have an integral that we can solve by substitution

$\left| {\begin{array}{*{20}{c}} {w = 1 - {x^2}} \\ {dw = - 2dx} \\ { - \frac{{dw}}{2} = dx} \end{array}} \right.$

$\begin{gathered} \int {\frac{x}{{\sqrt {1 - {x^2}} }}dx} = - \frac{1}{2}\int {\frac{{dw}}{{\sqrt w }}} = - \frac{1}{2}\int {{w^{ - 1/2}}dw} \hfill \\ \quad = - \frac{1}{2}\frac{{{w^{1/2}}}}{{(1/2)}} + C = - {w^{1/2}} + C = - \sqrt {1 - {x^2}} + C \hfill \\ \end{gathered}$

Back to our original integral we get

$\begin{gathered} I = x{\sin ^{ - 1}}x - \int {\frac{x}{{\sqrt {1 - {x^2}} }}dx} \hfill \\ \quad = x{\sin ^{ - 1}}x + \sqrt {1 - {x^2}} + C \hfill \\ \end{gathered}$

Up a level : Integrals
Previous page : One of the functions is sin(bx) or cos(bx) or a linear combination of them, the other is an exponential function
Next page : One is a logarithmLast modified: May 20, 2018 @ 14:54