# 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!

• After that first substitution you did, try a new one: a hyperbolic one, putting $\;t=\sinh u\;$ and etc. Aug 26, 2020 at 17:56
• I think @DanAntonio meant $u=\sinh t$ ($u=\tan t$ would work too).
– J.G.
Aug 26, 2020 at 17:58
• Either the answer is wrong, or you forgot to put the limits to the integral. Aug 26, 2020 at 17:58
• $\ln(1+\sqrt{2})$ doesn't make sense for an indefinite integral. Aug 26, 2020 at 18:00
• Let $\sin x=t$ and proceed as followed here. Aug 26, 2020 at 18:01

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}

$$\int \frac {\cos x}{\sqrt {1+\sin^2 x}} \ \mathrm{d}x$$

$$u = \sin x$$ 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}

Hint: $$t=\sin x$$

$$(\text{arcsinh} t )’= \frac1{\sqrt{1+t^2}}$$