How find this integral: $\int_{- \cos x}^{\sin x} \frac{1}{\sqrt{ 1-t^{2}}} dt$ $\int_{- \cos x}^{\sin x} \frac{1}{\sqrt{ 1-t^{2}}} dt$ 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.
 A: 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. 
A: 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$\dots$
A: 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.
