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$\int_{- \cos x}^{\sin x} \frac{1}{\sqrt{ 1-t^{2}}} dt$ my solution is put $t=\sin {\theta}$ then $$\int_{- \cos x}^{\sin x} \frac{1}{\sqrt{ 1-t^{2}}} dt = \int_{- \cos x}^{\sin x} \frac{1}{\cos{\theta}}\cos{\theta} d\theta = \int_{- \cos x}^{\sin x} d\theta = \theta $$ since $dt=\cos {\theta} d\theta$ , evaluating from ${- \cos x}$ to ${\sin x}$ we have $$\int_{- \cos x}^{\sin x} \frac{1}{\sqrt{ 1-t^{2}}} dt = \sin(x) + \cos (x)$$

Is this correct? please help.

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  • $\begingroup$ What is "sen"? There is $\sin, \cos, \sec, \csc, \tan, \cot$... but no $\mathrm{sen}$. $\endgroup$ – AlexR Jan 2 '14 at 21:19
  • $\begingroup$ yes i edited sorry.... $\endgroup$ – Rachel Jan 2 '14 at 21:21
  • $\begingroup$ To me, there appear no changes at all. $\endgroup$ – AlexR Jan 2 '14 at 21:21
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    $\begingroup$ "sen" is the spelling in some languages. $\endgroup$ – GEdgar Jan 2 '14 at 21:25
  • $\begingroup$ Isn't it immediate? At least for the primitive.. $\endgroup$ – Matheman Jan 2 '14 at 21:30
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You're almost there. When I do a substitution, I like to record the variable of integration in the limits, so I'd write your work as $$ \int_{- \cos x}^{\sin x} \frac{1}{\sqrt{ 1-t^{2}}} dt = \int_{t = - \cos x}^{t = \sin x} \frac{1}{\cos{\theta}}\cos{\theta} d\theta = \int_{t = - \cos x}^{t = \sin x} d\theta = \left. \theta \right|_{t = - \cos x}^{t = \sin x} $$ Now in that last term, you can't replace $\theta$ by $t$. You either have to say "$\theta$ is just a name for $\sin^{-1} t$", and write $$ \left. \theta \right|_{t = - \cos x}^{t = \sin x} = \left. \sin^{-1} t \right|_{t = - \cos x}^{t = \sin x}\\ = \sin^{-1} (\sin x) - \sin^{-1} (-\cos x), $$ and then simplify, or you can say "$t$ is is a name for $\sin \theta$", and write $$ \left. \theta \right|_{t = - \cos x}^{t = \sin x} = \\ \left. \theta \right|_{\sin \theta = - \cos x}^{\sin \theta = \sin x} = \\ \left. \theta \right|_{\theta = -\sqrt{1 - x^2}}^{\theta = x} = \\ x + \sqrt{1 - x^2}, $$ where the simplification of the lower limit uses some properties of sin/arcsin, etc.

The latter approach isn't much use in this instance, but can be helpful in other substitution problems.

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  • $\begingroup$ th question that i have now is that the final answer depends of x why? $\endgroup$ – Rachel Jan 2 '14 at 21:46
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    $\begingroup$ Look at the original problem: what's the value of $x$ there? Answer: it's not specified. That means that your final answer will be different as the value of $x$ changes...so of course it has to appear in the answer. Think about $\int_1^x 2t ~dt$. The answer is $x^2 - 1$, right? And it depends on $x$. $\endgroup$ – John Hughes Jan 2 '14 at 22:20
  • $\begingroup$ Yes, now i understand really thanks. $\endgroup$ – Rachel Jan 2 '14 at 22:56
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You should evaluate the indefinite integral, then evaluate it between the bounds you are given, this will prevent you from getting confused with substituting into the bounds. The answer is $$\sqrt{1-x^2} + x,$$ but you should work it out yourself.

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Because I can't write comments/suggestions yet (not enough reputation) I have to add this as an answer.

I don't know about the results, but your substitution is dubious as you'd have to adjust your bounds. ($t = \sin(\omega) = \sin(x) \implies$ your upper bound can not be $\sin(x)$ anymore).

For example, take the integral of $x$ from $5$ to $6$. let's substitute $x$ with $t^2$ and not adjust our bounds. Then you'd get $dx=2t*dt \implies$ the integral of $2t^3$ from $5$ to $6$ and that can't obviously be the same$\dots$

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