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Section 6.5

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1. (a) \displaystyle{f'(x) = \frac{x}{1+x^2} + \tan^{-1}(x)}

(c) \displaystyle{g'(x) = \frac{3}{x\sqrt{1-9x^2}} - \frac{\sin^{-1}(3x)}{x^2}}

2. (a) \displaystyle{\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}}

(c) \displaystyle{\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}}

(e) \displaystyle{\sin^{-1}\left(\sin\left(\frac{3\pi}{4}\right)\right) = \frac{\pi}{4}}

(f) \displaystyle{\sin\left(\sin^{-1}\left(-\frac{1}{\sqrt{2}}\right)\right) = -\frac{1}{\sqrt{2}}}

3. (a) \displaystyle{\int \frac{1}{1 + 2x^2} \ dx = \frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}x) + c}

(c) \displaystyle{\int \frac{3}{x^2 + 4} \ dx = \frac{3}{2}\tan^{-1}\left(\frac{x}{2}\right) + c}

(e) \displaystyle{\int \frac{x}{x^2+4x+5} \ dx = \frac{1}{2}\log(x^2 + 4x + 5) - 2\tan^{-1}(x + 2) + c}

(g) \displaystyle{\int_{-1}^1 \frac{1}{1 + x^2} \ dx = \frac{\pi}{2}}

4. (a) \displaystyle{\int \frac{1}{x^3+x} \ dx = \log|x| - \frac{1}{2}\log(1 + x^2) + c}

(c) \displaystyle{\int \frac{1}{x^2(x^2+1)} \ dx = -\frac{1}{x} - \tan^{-1}(x) + c}

(e) \displaystyle{\int \sin^{-1}(x)dx = x\sin^{-1}(x) + \sqrt{1 - x^2} + c}

(g) \displaystyle{\int \frac{5}{\sqrt{1-9x^2}} \ dx = \frac{5}{3}\sin^{-1}(3x) + c}

(i) \displaystyle{\int \frac{3x}{\sqrt{1-x^2}} \ dx = -3\sqrt{1-x^2} + c}

8. \displaystyle{\int_{-\infty}^\infty \frac{1}{1 + x^2} \ dx = \pi}

9. (a) For -1 \le x \le 1 ,

\displaystyle{\tan^{-1}(x) = \sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{2n + 1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots .}

(b) \displaystyle{\pi = 4\tan^{-1}(1) \approx P_{3999}(1) = 3.1411}, rounded to 4 decimal places

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