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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!

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

4 Answers 4

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Use fact that: $$ \cos x ~\mathrm{d}x = \mathrm{d}(\sin x )$$ Then we have: $$\begin{aligned} \int \frac{\cos x}{\sqrt{1+\sin^2 x}} \mathrm{d}x &= \int \frac{\mathrm{d}( \sin x)}{\sqrt{1+\sin^2 x}} \\ &= \int \frac{\mathrm{d}u}{\sqrt{1+u^2}} \hspace{35pt} \text{via}~u=\sin x \\ &= \int \frac{\mathrm{d}s}{\cos s}\hspace{35pt} \text{via}~ u=\tan s \\ &= \int \frac{(\sec s)(\sec s + \tan s)}{\sec s+ \tan s} \mathrm{d}s \\ &= \int \frac{\mathrm{d}r}{r} \hspace{35pt} \text{via}~ |r = \sec(s) + \tan(s)|\\ &= \ln r + C \\ &= \ln(\sec s +\tan s) + C \\ &= \ln\left(u+\sqrt{u^2+1}\right)+C \\ &= \ln\left(\sin x + \sqrt{\sin^2 x+1}\right)+C \\ &= \sinh^{-1}(\sin x) \end{aligned} $$

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

$u = \sin x$ 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

$u = \sinh t ,\mathrm{d}u = \cosh t$

$\begin{aligned} \implies \int \frac {\cosh t}{\sqrt {1+\sinh^2 t}} \mathrm{d}t &=\int 1~ \mathrm{d}t \\ &=t + C \\ &=\sinh^{-1} (\sin x) + C \end{aligned}$

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Hint: $t=\sin x$

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

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Letting $\sin x=\tan \theta$ yields $$ \begin{aligned} I & = \int \frac{\sec ^{2} \theta d \theta}{\sec \theta} \\ &=\int \sec \theta d \theta \\ &=\ln |\sec \theta+\tan \theta|+C \\ &=\ln \left|\sin x+\sqrt{1+\sin ^{2} x}\right|+C . \end{aligned} $$

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