Integrate the following functions. ∫ sin (ax+b)cos (ax+b)dx.
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Solution
Let I=∫sin(ax+b)cos(ax+b)dx Let sub (ax+b)=t On differentiating w.r.t.x, we get a cos (ax+b)=dtdx ⇒dx=dtacos(ax+b)∴I=∫tcos(ax+b)dtacos(ax+b)=1a∫tdt=1a.t22+c=[sin(ax+b)]22a+C