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}
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1 Answers
29(\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 = ±√2/5
