Definite Integral evaluation and limit change of $\int_0^{\pi/2}\frac{\cos x}{\sqrt{1-\sin x}}\,dx $

Question

Evaluate the following definite integral : $$\int_0^{\pi/2}\cfrac{\cos x}{\sqrt{1-\sin x}} \qquad \qquad \qquad (1)$$

\begin{align} & = \int_0^{\pi/2}\cfrac{\cos x}{\sqrt{1-\sin x}} \ \ I \ used \ u=1-\sin x \ and \ dx= \cfrac{-du}{cosx} \\ & = -\int_1^0\cfrac{du}{\sqrt u} \\ & = \int_0^1\cfrac{du}{\sqrt u} \\ & = 2\sqrt u |_0^1 \\ & = 2-0 =2 \\ \end{align} But Symbolab says that is 0, what i have done wrong in (1) ?

Answer 1

Symbolab is wrong here, $\cos(x) \geq 0$ on the interval and $\sqrt{1-sin(x)} \geq 0$ on the interval, with regions where both functions are strictly positive.

Answer 2

I would substitute $$t=\sqrt{1-\sin(x)}$$ then we get $$dt=\frac{\cos(x)}{2\sqrt{1-\sin(x)}}dx$$

Answer 3

What you submit is correct. The error is in Symbolab; I do not know the cause. Check carefully how you enter the data in Symbolab. It could also come from a bug in the program. (?)

Answer 4

I put it into symbolab and I got the correct answer!

Answer 5

Make the substitution $w = 1 - \sin(x)$. Then you get $\mathrm{d}w = -\cos(x)\mathrm{d}x$. Thus we have \begin{align*} \int_{0}^{\pi/2}\frac{\cos(x)}{\sqrt{1-\sin(x)}}\mathrm{d}x = -\int_{1}^{0}\frac{\mathrm{d}w}{\sqrt{w}} = \int_{0}^{1}\frac{\mathrm{d}w}{\sqrt{w}} = 2\sqrt{w}\,\,\biggr|_{0}^{1} = 2 \end{align*}