Chapter 3 - Problems Plus - Problems - Page 274: 29

$$y=\frac{x}{\sqrt{a^{2}-1}}-\frac{2}{\sqrt{a^{2}-1}} \arctan \frac{\sin x}{a+\sqrt{a^{2}-1}+\cos x}$$ Let $k=a+\sqrt{a^{2}-1}$. Then $$\begin{aligned} y^{\prime} &=\frac{1}{\sqrt{a^{2}-1}}-\frac{2}{\sqrt{a^{2}-1}} \cdot \frac{1}{1+\sin ^{2} x /(k+\cos x)^{2}} \cdot \frac{\cos x(k+\cos x)+\sin ^{2} x}{(k+\cos x)^{2}} \\ &=\frac{1}{\sqrt{a^{2}-1}}-\frac{2}{\sqrt{a^{2}-1}} \cdot \frac{k \cos x+\cos ^{2} x+\sin ^{2} x}{(k+\cos x)^{2}+\sin ^{2} x}\\ &=\frac{1}{\sqrt{a^{2}-1}}-\frac{2}{\sqrt{a^{2}-1}} \cdot \frac{k \cos x+1}{k^{2}+2 k \cos x+1} \\ &=\frac{k^{2}+2 k \cos x+1-2 k \cos x-2}{\sqrt{a^{2}-1}\left(k^{2}+2 k \cos x+1\right)}\\ &=\frac{k^{2}-1}{\sqrt{h^{2}-1}\left(k^{2}+2 k \cos x+1\right)} \end{aligned}$$ But $$k^{2}=2 a^{2}+2 a \sqrt{a^{2}-1}-1=2 a(a+\sqrt{a^{2}-1})-1=2 a k-1$$ so, $$k^{2}+1=2 a k, \text { and } k^{2}-1=2(a k-1)$$. So, $$y^{\prime}=\frac{2(a k-1)}{\sqrt{a^{2}-1}(2 a k+2 k \cos x)}=\frac{a k-1}{\sqrt{a^{2}-1} k(a+\cos x)}$$. But $$a k-1=a^{2}+a \sqrt{a^{2}-1}-1=k \sqrt{a^{2}-1}$$, so, $$y^{\prime}=1 /(a+\cos x)$$

$$y=\frac{x}{\sqrt{a^{2}-1}}-\frac{2}{\sqrt{a^{2}-1}} \arctan \frac{\sin x}{a+\sqrt{a^{2}-1}+\cos x}$$ Let $k=a+\sqrt{a^{2}-1}$. Then $$\begin{aligned} y^{\prime} &=\frac{1}{\sqrt{a^{2}-1}}-\frac{2}{\sqrt{a^{2}-1}} \cdot \frac{1}{1+\sin ^{2} x /(k+\cos x)^{2}} \cdot \frac{\cos x(k+\cos x)+\sin ^{2} x}{(k+\cos x)^{2}} \\ &=\frac{1}{\sqrt{a^{2}-1}}-\frac{2}{\sqrt{a^{2}-1}} \cdot \frac{k \cos x+\cos ^{2} x+\sin ^{2} x}{(k+\cos x)^{2}+\sin ^{2} x}\\ &=\frac{1}{\sqrt{a^{2}-1}}-\frac{2}{\sqrt{a^{2}-1}} \cdot \frac{k \cos x+1}{k^{2}+2 k \cos x+1} \\ &=\frac{k^{2}+2 k \cos x+1-2 k \cos x-2}{\sqrt{a^{2}-1}\left(k^{2}+2 k \cos x+1\right)}\\ &=\frac{k^{2}-1}{\sqrt{h^{2}-1}\left(k^{2}+2 k \cos x+1\right)} \end{aligned}$$ But $$k^{2}=2 a^{2}+2 a \sqrt{a^{2}-1}-1=2 a(a+\sqrt{a^{2}-1})-1=2 a k-1$$ so, $$k^{2}+1=2 a k, \text { and } k^{2}-1=2(a k-1)$$. So, $$y^{\prime}=\frac{2(a k-1)}{\sqrt{a^{2}-1}(2 a k+2 k \cos x)}=\frac{a k-1}{\sqrt{a^{2}-1} k(a+\cos x)}$$. But $$a k-1=a^{2}+a \sqrt{a^{2}-1}-1=k \sqrt{a^{2}-1}$$, so, $$y^{\prime}=1 /(a+\cos x)$$