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Question Number 10241 by FilupSmith last updated on 31/Jan/17
∫aa+δsin(x)cos(x+δ)dx=???
Commented by prakash jain last updated on 31/Jan/17
sin(x)cos(x+δ)=sin(x+δ−δ)cos(x+δ)=sin(x+δ)cosδ−cos(x+δ)sinδcos(x+δ)=tan(x+δ)cosδ−sinδ∫aa+δ(tan(x+δ)cosδ−sinδ)dx=[−cosδlncos(x+δ)−xsinδ]aa+δ=[−cosδlncos(a+δ)cosa−δsinδ]
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