Evaluate the integral $\int \frac{\cos(x)}{\sqrt{1+\sin^2(x)}} \, dx$ Evaluate the integral
$$\int \frac{\cos x}{\sqrt{1+\sin^2x}}  \mathrm{d}x$$
Hey guys, I've been having real problems with this one. I've tried setting $u=\sin x$ so $\mathrm{d}u=\cos x ~\mathrm{d}x$ but that didn't go anywhere... Tried subbing out $\sin^2x$ via pythagorean identity but alas, I ran into another dead end. I checked the answer and it's supposed to be $\ln(1 + \sqrt{2})$, which makes me think there is some identity involved I am not aware of... Thanks in advance!
 A: Use fact that:
$$ \cos x ~\mathrm{d}x = \mathrm{d}(\sin x )$$
Then we have:
$$\begin{aligned} \int \frac{\cos x}{\sqrt{1+\sin^2 x}} \mathrm{d}x  
&= \int \frac{\mathrm{d}( \sin x)}{\sqrt{1+\sin^2 x}} \\ 
&= \int \frac{\mathrm{d}u}{\sqrt{1+u^2}} \hspace{35pt} \text{via}~u=\sin x \\ 
&= \int \frac{\mathrm{d}s}{\cos s}\hspace{35pt} \text{via}~ u=\tan s \\ 
&= \int \frac{(\sec s)(\sec s + \tan s)}{\sec s+ \tan s} \mathrm{d}s \\
&= \int \frac{\mathrm{d}r}{r} \hspace{35pt} \text{via}~ |r = \sec(s) + \tan(s)|\\ 
&= \ln r + C \\ 
&= \ln(\sec s +\tan s) + C \\
&= \ln\left(u+\sqrt{u^2+1}\right)+C \\
&= \ln\left(\sin x + \sqrt{\sin^2 x+1}\right)+C \\
&= \sinh^{-1}(\sin x) \end{aligned} $$
A: Hint: $t=\sin x$
$$(\text{arcsinh} t )’= \frac1{\sqrt{1+t^2}}$$
A: $$\int \frac {\cos x}{\sqrt {1+\sin^2 x}} \ \mathrm{d}x$$
$u = \sin x$ seems like a reasonable place to start... but we know that that is not quite right.
$\displaystyle \int \frac {1}{\sqrt {1+u^2}} \mathrm{d}u$
When we see $1+u^2$ we should be thinking of two options.
$u = \sinh t$ or $u=\tan t.$  Ultimately, either one will work.  But, many Calc $1,2$ students never see the hyperbolics.  I will show both approaches.
$u = \tan t, \mathrm{d}u = \sec^2 t~ \mathrm{d}t$
$\begin{aligned} \displaystyle \implies \int \frac {\sec^2 t}{\sqrt {1+\tan^2 t}} \mathrm{d}t 
&=\int \frac {\sec^2 t}{|\sec t|} \mathrm{d}t \\
&=\int |\sec t| \mathrm{d}t \\
&=\ln |\sec t + \tan t| + C \end{aligned}$
And reverse the substitutions.
$\begin{aligned} \ln |\sec (\arctan u) + \tan (\arctan u)| + C 
&=\ln |\sqrt {1+u^2} + u| + C \\
&=\ln |\sqrt {1+\sin^2 x} + \sin x| + C \end{aligned}$
Or
$u = \sinh t ,\mathrm{d}u = \cosh t$
$\begin{aligned}  \implies \int \frac {\cosh t}{\sqrt {1+\sinh^2 t}} \mathrm{d}t 
&=\int 1~ \mathrm{d}t \\
&=t + C \\
&=\sinh^{-1} (\sin x) + C \end{aligned}$
A: Letting $\sin x=\tan \theta$ yields
$$
\begin{aligned}
I & = \int \frac{\sec ^{2} \theta d \theta}{\sec \theta} \\
&=\int \sec \theta d \theta \\
&=\ln |\sec \theta+\tan \theta|+C \\
&=\ln \left|\sin x+\sqrt{1+\sin ^{2} x}\right|+C .
\end{aligned}
$$
