Answer: (a) \(\displaystyle \int \frac{\theta-1}{\sqrt{1-(\theta-1)^2}}\,d\theta=-\sqrt{1-(\theta-1)^2}+C\).
(b) \(\displaystyle y=\theta(\theta-1)\sin^{-1}(\theta-1)+\theta\sqrt{1-(\theta-1)^2}\).
(a) Let \(u=1-(\theta-1)^2\).
Then \(\displaystyle \frac{du}{d\theta}=-2(\theta-1)\), so \((\theta-1)\,d\theta=-\tfrac12\,du\).
Hence
\(\displaystyle \int \frac{\theta-1}{\sqrt{1-(\theta-1)^2}}\,d\theta=-\frac12\int u^{-1/2}\,du=-\sqrt{u}+C\).
Substituting back,
\(\displaystyle \int \frac{\theta-1}{\sqrt{1-(\theta-1)^2}}\,d\theta=-\sqrt{1-(\theta-1)^2}+C\).
(b) Start with
\(\displaystyle \theta\frac{dy}{d\theta}-y=\theta^2\sin^{-1}(\theta-1)\).
Since \(0\lt \theta\lt 2\), divide by \(\theta\):
\(\displaystyle \frac{dy}{d\theta}-\frac{y}{\theta}=\theta\sin^{-1}(\theta-1)\).
This is linear, with integrating factor
\(\displaystyle \mu(\theta)=e^{\int -1/\theta\,d\theta}=e^{-\ln \theta}=\theta^{-1}\).
Multiplying through by \(\theta^{-1}\),
\(\displaystyle \frac{d}{d\theta}(\theta^{-1}y)=\sin^{-1}(\theta-1)\).
Integrate:
\(\displaystyle \theta^{-1}y=\int \sin^{-1}(\theta-1)\,d\theta + C\).
Use integration by parts with \(u=\sin^{-1}(\theta-1)\) and \(dv=d\theta\). Then \(v=\theta\) and \(\displaystyle du=\frac{1}{\sqrt{1-(\theta-1)^2}}\,d\theta\).
So
\(\displaystyle \int \sin^{-1}(\theta-1)\,d\theta=\theta\sin^{-1}(\theta-1)-\int \frac{\theta}{\sqrt{1-(\theta-1)^2}}\,d\theta\).
Now write
\(\displaystyle \int \frac{\theta}{\sqrt{1-(\theta-1)^2}}\,d\theta=\int \frac{\theta-1}{\sqrt{1-(\theta-1)^2}}\,d\theta+\int \frac{1}{\sqrt{1-(\theta-1)^2}}\,d\theta\).
From part (a),
\(\displaystyle \int \frac{\theta-1}{\sqrt{1-(\theta-1)^2}}\,d\theta=-\sqrt{1-(\theta-1)^2}\).
Also,
\(\displaystyle \int \frac{1}{\sqrt{1-(\theta-1)^2}}\,d\theta=\sin^{-1}(\theta-1)\).
Therefore
\(\displaystyle \int \frac{\theta}{\sqrt{1-(\theta-1)^2}}\,d\theta=-\sqrt{1-(\theta-1)^2}+\sin^{-1}(\theta-1)\).
Hence
\(\displaystyle \theta^{-1}y=\theta\sin^{-1}(\theta-1)+\sqrt{1-(\theta-1)^2}-\sin^{-1}(\theta-1)+C\).
Use \(y=1\) when \(\theta=1\):
\(\displaystyle 1=1+C\), so \(C=0\).
Thus
\(\displaystyle \frac{y}{\theta}=\theta\sin^{-1}(\theta-1)-\sin^{-1}(\theta-1)+\sqrt{1-(\theta-1)^2}\).
Multiplying by \(\theta\),
\(\displaystyle y=\theta(\theta-1)\sin^{-1}(\theta-1)+\theta\sqrt{1-(\theta-1)^2}\).
