Explanation
∫f″(x)sinxdx
=sinxf′(x)−∫cosxf′(x)dx
=sinxf′(x)−[cosxf(x)+∫sinxf(x)dx]
⇒∫[f(x)+f″(x)]sinxdx=sinxf′(x)dx−cosxf(x)
⇒∫π0[f(x)+f″(x)]sinxdx=[sinxf′(x)dx−cosxf(x)]π0
⇒5=f(π)+f(0)
⇒f(0)=5−2=3
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