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\begin{equation} \int\frac{\cos\left(\sqrt{17t}\right)}{\sqrt{17t}}\,\mathrm dt=? \end{equation}

Need the answer to this problem. Me and my friend been working on it for a while now.

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@StefanHansen: Your edit changed the question. With $\sqrt{17}\,t$ instead of $\sqrt{17t}$, there is no elementary antiderivative... –  Hans Lundmark Feb 5 '13 at 9:01
    
@HansLundmark: I'm so sorry, but thanks for noticing and editing it accordingly. –  Stefan Hansen Feb 5 '13 at 9:04

3 Answers 3

up vote 1 down vote accepted

In general: if $\,f(x)\,$ is a derivable function, then

$$\int f'(t)\cos f(t)\,dt=\sin f(t)+C\;\;(\text{differentiate using the chain rule on the RH to prove})$$

In this case,

$$f(t)=\sqrt{17 t}\,\,\,,\,\,f'(t)=\frac{\sqrt{17}}{2\sqrt t}=\frac{17}{2}\frac{1}{\sqrt{17 t}}\Longrightarrow$$

$$\int\frac{\cos\sqrt{17 t}}{\sqrt{17 t}}dt=\frac{2}{17}\int\left( \frac{17}{2}\frac{1}{\sqrt{17 t}}\,dt\right)\cos\sqrt{17 t}=\frac{2}{17}\sin\sqrt{17 t}+C$$

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Hint: Try the substitution $u = \sqrt{17t}$.

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Substituting $u = \sqrt{17t}$ will change the above integral to $\frac {2}{17}\int \cos u \,du$ which is now one line answer.

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