integration freaked me .doubt

x^{2}\left(\int_{0}^{\pi /2}{(2sint+3cost)dt} \right)-x\left(\int_{-3}^{3}{\frac{t^{2}sin2t}{t^{2}+1}}dt \right)-2=0, then x= ......

ans----->\pm \sqrt{2/5}

1 Answers

(\int_{0}^{\pi /2}{(2sint+3cost)dt} \right) = 5\int_{0}^{\pi /2}{costdt} ..

by using the property of definite integrals

(\int_{0}^{\pi /2}{(2sint+3cost)dt} \right) = 5\int_{0}^{\pi /2}{costdt}..

\left\int_{-3}^{3}{\frac{t^{2}sin2t}{t^{2}+1}}dt \right=0

coz...sin2t is an odd function...

so the equation reduces to 5x² = 2...

hence x = Â±√2/5

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