area of the curve f is :
∫0π312(f(θ))2dθ\intop_0^{\frac{\pi}{3}} \frac{1}{2}{(f(\theta))}^2d\theta∫03π21(f(θ))2dθ
here, f=1+2sinθf=1+2sin\thetaf=1+2sinθ
=∫0π312(1+2sinθ)2dθ=\intop_0^{\frac{\pi}{3}} \frac{1}{2}{(1+2sin\theta)}^2d\theta=∫03π21(1+2sinθ)2dθ
=∫0π312(1+4sin2θ+4sinθ)dθ=\intop_0^{\frac{\pi}{3}} \frac{1}{2}{(1+4sin^2\theta+4sin\theta)}d\theta=∫03π21(1+4sin2θ+4sinθ)dθ
=12[π3−4cosθ∣0π3+4∣θ2−sin2θ4∣0π3]=\frac{1}{2} [\frac{\pi}{3}-4cos\theta|_0^{\frac{\pi}{3}}+4 |\frac{\theta}{2}-\frac{sin2\theta}{4}|_0^{\frac{\pi}{3}}]=21[3π−4cosθ∣03π+4∣2θ−4sin2θ∣03π]
because ∫sin2xdx=x2−sin2x4\int sin^2 x dx=\frac{x}{2}-\frac{sin2x}{4}∫sin2xdx=2x−4sin2x
=12[π3+4.12+4(π6−32)]= \frac{1}{2}[\frac{\pi}{3}+4.\frac{1}{2}+4(\frac{\pi}{6}-\frac{\sqrt3}{2})]=21[3π+4.21+4(6π−23)]
=π2+1−34=\frac{\pi}{2}+1-\frac{\sqrt3}{4}=2π+1−43
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