calculus - Evaluating $int {sin theta + cos theta over sqrt{sin 2theta} }dtheta$

$$\int {\sin \theta + \cos \theta \over \sqrt{\sin 2\theta} }d\theta$$

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After simplifying I get,

$${1\over \sqrt{2}}\int {\tan \theta + 1 \over \sqrt{\tan \theta}} d\theta$$

Substituting $u = \tan \theta$

$${1\over \sqrt{2}}\int {u + 1\over \sqrt{u}(1 + u^2)} du$$

Substituting $t^2 = u$

$${\sqrt{2}}\int {t^2 + 1\over (1 + t^4)} dt = \sqrt{2} \int {1 + 1/t^2 \over(t - 1/t)^2 + 2 }dt$$

Substituting $z = t - 1/t$

$$\sqrt{2} \int {1\over z^2 + 2 }dz = \arctan\left(\tan \theta - 1\over \sqrt{2\tan \theta} \right) + C = {\arcsin(2\sqrt{\sin2\theta} (\sin\theta - \cos \theta))\over 2} + C$$

Which is far from the right answer of $\arcsin(\sin \theta - \cos \theta) + C$.

Where did I go wrong ?

Edit :

I would like to know where I am wrong rather than knowing how to solve the problem.

Answer

Your factor before $\arctan$ should be $1$ rather than $1/2$; except for that, your antiderivative is correct. The error is in the conversion to $\arcsin$. If you use $\arctan \alpha = \arcsin(\alpha/\sqrt{\alpha^2+1})$, you will get the right answer.

Update: Now that the factor 1/2 has been edited away, things are fine up to that point. And actually the last step is almost OK too! (I dismissed it a bit too quickly before.)

The question is only meaningful on an interval where $\sin 2\theta \ge 0$, say $0 \le \theta \le \pi/2$, and the answer $A=\arcsin(\sin\theta-\cos\theta)$ is correct on that whole interval, as is easily verified by differentiation (and note also that $|\sin\theta-\cos\theta| \le 1$ on that interval).

Now it seems that your expression $B$ actually agrees with the expression $A$ on the subinterval $\pi/12 \le \theta \le 5\pi/12$, whereas $B=\text{const.}-A$ on the intervals $0\le \theta \le \pi/12$ and $\le 5\pi/12 \le \theta \le \pi/2$. I haven't really bothered to analyze why this happens, I was just lazy and looked at plots of $A \pm B$. But surely (as Barry Cipra pointed out) it has something to do with the properties of the inverse trig functions; for example $\arcsin(\sin x) = \pi-x$ if $\pi/2 \le x \le \pi$.