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: Dec 28, 2023 @ 14:34