Evaluate the integral of (√siny) cos y dy from 3 to π/2
Solution.
Make a replacement:
t=siny,t≥0,t=\sqrt{\sin{y}}, t\geq0,t=siny,t≥0, from here siny=t2.\sin{y}=t^2.siny=t2.
Then cosydy=2tdt.\cos{y}dy=2tdt.cosydy=2tdt.
We will have
∫3π2t⋅2tdt=∫3π22t2dt=2t33∣3π2==2sin32y3∣3π2=2sin32π23−2sin3233==23−2sin3233≈0.63.\int\limits_3^{\frac{\pi}{2}}t\cdot 2tdt=\int\limits_3^{\frac{\pi}{2}}2t^2dt=\frac{2t^3}{3}|_3^{\frac{\pi}{2}}=\newline =\frac{2\sin^{\frac{3}{2}}y}{3}|_3^{\frac{\pi}{2}}= \frac{2\sin^{\frac{3}{2}}\frac{\pi}{2}}{3}-\frac{2\sin^{\frac{3}{2}}3}{3}=\newline =\frac{2}{3}-\frac{2\sin^{\frac{3}{2}}3}{3} \approx 0.63.3∫2πt⋅2tdt=3∫2π2t2dt=32t3∣32π==32sin23y∣32π=32sin232π−32sin233==32−32sin233≈0.63.
Answer.
∫3π2sinycosydy≈0.63.\int\limits_3^{\frac{\pi} {2}}\sqrt{sin{y}}\cos{y}dy\approx 0.63.3∫2πsinycosydy≈0.63.
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