`intxcos(5x)dx`
To evaluate, apply integration by parts `int udv = uv -intv du` .
So let:
`u=x ` and `dv =intcos(5x)dx`
Differentiating u and integrating dv yield:
`du=dx ` and `v=sin (5x)/5`
Plugging them to the formula, the integral becomes:
`int xcos(5x)dx`
`= x*sin(5x)/5 - int sin(5x)/5 dx`
`= (xsin(5x))/5 - 1/5int sin(5x)dx`
`= (xsin(5x))/5 - 1/5(-cos(5x)/5) + C`
`=(xsin(5x))/5 + cos(5x)/25+C`
Therefore, `intxcos(5x)dx=(xsin(5x))/5 + cos(5x)/25+C` .
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