3
$\begingroup$

What's the easiest way to calculate the following indefinite integral:

$$ \int \frac{\cos(x)}{\sqrt{2\sin(x)+3}} \mathrm{d}x $$

$\endgroup$
  • 2
    $\begingroup$ Notice that the numerator is (up to a constant factor) the derivative of the radicand. $\endgroup$ – Daniel Fischer Jan 4 '14 at 21:56
  • $\begingroup$ @DanielFischer : I phrased that same thought in a rather different way in my posted answer. But I wonder if it's reasonable to expect a lay reader to understand technical terms like "up to". $\endgroup$ – Michael Hardy Jan 4 '14 at 22:31
6
$\begingroup$

Hint

$$\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}+C$$

$\endgroup$
  • 1
    $\begingroup$ Thanks a lot for this interesting approach, I will accept your answer in next 5 minutes. $\endgroup$ – NullPointer Jan 4 '14 at 22:04
  • $\begingroup$ I think you're missing an integral sign there. $\endgroup$ – David Zhang Jan 5 '14 at 1:09
  • $\begingroup$ Of course, hehe. Thanks, @DavidZhang $\endgroup$ – DonAntonio Jan 5 '14 at 4:07
3
$\begingroup$

set $u=2\sin(x)+3$. then $du=2\cos(x)dx$. So it is $$ \int \frac{du}{2\sqrt{u}} $$

$\endgroup$
1
$\begingroup$

Here's a hint: $$ \int\frac{1}{\sqrt{2\sin x+3}}\Big( \cos x \, dx \Big) $$

If you don't know what that is hinting at, then that is what you need to learn about integration by substitution.

$\endgroup$
1
$\begingroup$

First, make the substitution $u = 2\sin(x) + 3$. Then, $du = 2\cos(x) \,dx$. Thus: $$\begin{align} \int \frac{\cos(x)}{\sqrt{2\sin(x) + 3}}dx &= \int \frac{du}{2\sqrt{u}}\\ &= \sqrt{u} + C\\ &= \ldots \end{align}$$

$\endgroup$

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.