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I have got 2 questions I can't seem to get the same answer to as the book.

31) $\displaystyle \int \sin(\sqrt x)\,\mathrm{d}x = -2\sqrt x \cos(\sqrt x)+2\sin(\sqrt x )+c$.

I just get the integral to be $\frac{1}{2}\cos(\frac{1}{x})+c$

and

36) $\displaystyle\int \frac{\mathrm{d}x}{x^2(4+x^2)} = -\frac{1}{4}x +\frac{1}{8} \tan^{-1}\left(\frac{2}{x}\right)+c$

This one has me all over the place. The best I can come to is setting $x = 2\tan x$ but thats about it.

doraemonpaul
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1ftw1
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  • You can check whether $\frac{1}{2}\cos(\frac{1}{x})+c$ is right: differentiate it and see if youset $\sin\sqrt{x}$. You don't: there is no way you would get a square-root when you do that differentiation. – GEdgar May 26 '13 at 13:13

1 Answers1

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Hints:

For the first one, take: $ t = \sqrt{x} $ and you get $$ 2\int{t \sin(t) dx} $$

For the second:

$$ \begin{align} \displaystyle\int \frac{\mathrm{d}x}{x^2(4+x^2)} &= \dfrac{1}{4} \displaystyle\int \frac{(x^2 + 4 - x^2) \ \mathrm{d}x}{x^2(4+x^2)} \\ &= \frac{1}{4} \left[ \displaystyle\int \frac{\mathrm{d}x}{x^2} - \displaystyle\int \frac{\mathrm{d}x}{x^2 + 4} \right] \end{align} $$

hjpotter92
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