Answer
$\frac{9\pi}{20}$
Work Step by Step
$\int_{0}^{\frac{3}{5}}\sqrt (9-25x^{2})dx$
$let$ $x=\frac{3}{5}sinθ$
$dx=\frac{3}{5}cosθdθ$
$\frac{3}{5}=\frac{3}{5}sinθ$
$1=sinθ$
$θ=\frac{\pi}{2}$
$0=\frac{3}{5}sinθ$
$0=sinθ$
$θ=0$
$\int_{0}^{\frac{\pi}{2}}\sqrt (9-25(\frac{9}{25}sin^{2}θ))(\frac{3}{5}cosθ)dθ$
$\int_{0}^{\frac{\pi}{2}}\frac{3}{5}cosθ\sqrt (9-9sin^{2}θ)dθ$
$\int_{0}^{\frac{\pi}{2}}3(\frac{3}{5}cosθ)\sqrt (1-sin^{2}θ)dθ$
Through the Pythagorean identity $sin^{2}θ+cos^{2}θ=1$, $1-sin^{2}θ=cos^{2}θ$:
$\int_{0}^{\frac{\pi}{2}}\frac{9}{5}cosθ\sqrt (cos^{2}θ)dθ$
$\int_{0}^{\frac{\pi}{2}}\frac{9}{5}cos^{2}θdθ$
Through the power reducing formula $cos^{2}θ=\frac{1}{2}+\frac{1}{2}cos2θ$:
$\frac{9}{5}\int_{0}^{\frac{\pi}{2}}(\frac{1}{2}+\frac{1}{2}cos2θ)dθ$
$let$ $u=2θ$
$du=2dθ$
$\frac{1}{2}du=2dθ$
$u(0)=0$
$u(\frac{\pi}{2})=\pi$
$(\frac{9}{5})(\frac{1}{2})\int_{0}^{\pi}(\frac{1}{2}+\frac{1}{2}cosu)du$
$\frac{9}{10}[\frac{1}{2}u+\frac{1}{2}sinu]_{0}^{\pi}$
$\frac{9}{10}[\frac{\pi}{2}+\frac{1}{2}sin\pi-(\frac{1}{2}(0)-\frac{1}{2}sin0)]$
$\frac{9}{10}[\frac{\pi}{2}]$
$\frac{9\pi}{20}$
