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Integrate the function $\sin(ax+b)\;\cos(ax+b)$

1 Answer

  • (i)Method of substitution:
  • Given $\int f(x)dx$ can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dx}=g'(t).
  • dx=g'(t)dt.
  • Thus $ I=\int f(g(t).g'(t))dt.$
  • (ii)$\sin x\cos x=\frac{1}{2}\sin2x.$
Given $\int\sin (ax+b)\cos(ax+b)dx.$
Using the information in the tool box,let us transform the given function as,
$I=\int\frac{1}{2}\sin 2(ax+b)dx.$
$\;\;\;=\frac{1}{2}\int\sin 2(ax+b)dx.$
On integrating we get,
Hence $\int\sin(ax+b)\cos(ax+b)dx=\frac{-1}{4a}\cos(ax+b)+c.$


answered Jan 28, 2013 by sreemathi.v