1÷D2+a2× sinax is equal to
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Step-by-step explanation:
My approach:
PI=1D2+a2(sinax).
Knowing that 1D(ex)=xex and sinax=eiax−e−iax2i. I tried solving this.
Here is where I face difficulty,
12i(D2+a2)(eiax)=eiax12i((D+ia)2+a2)(1)=eiax12iD(D+2ai)(1)=eiax1−4a(1+D2ai)−1(x)=eiax1−4a(x−12ai)
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