1
$\begingroup$

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!

$\endgroup$
6
  • $\begingroup$ After that first substitution you did, try a new one: a hyperbolic one, putting $\;t=\sinh u\;$ and etc. $\endgroup$
    – DonAntonio
    Aug 26, 2020 at 17:56
  • $\begingroup$ I think @DanAntonio meant $u=\sinh t$ ($u=\tan t$ would work too). $\endgroup$
    – J.G.
    Aug 26, 2020 at 17:58
  • 1
    $\begingroup$ Either the answer is wrong, or you forgot to put the limits to the integral. $\endgroup$
    – Andrei
    Aug 26, 2020 at 17:58
  • 1
    $\begingroup$ $\ln(1+\sqrt{2})$ doesn't make sense for an indefinite integral. $\endgroup$
    – jjagmath
    Aug 26, 2020 at 18:00
  • $\begingroup$ Let $\sin x=t$ and proceed as followed here. $\endgroup$
    – SarGe
    Aug 26, 2020 at 18:01

3 Answers 3

3
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$u = \sin x$ 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

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

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

$\endgroup$
2
$\begingroup$

Hint: $t=\sin x$

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.