So I have this problem $${\int^{\pi/2}_0} {{\cos\theta \sin\theta}\over \sqrt{\cos^{2}\theta +8}}d\theta $$

and I'm not sure if this is the right direction to begin. If I have $u = \cos\theta$ then, $du = -\sin\theta$ can I then do this:

$$-du=\sin\theta$$ $$-\cos\theta du = \cos\theta \sin\theta dx$$ and then: $$-\cos\theta {{\int^{0}_1}} {1\over \sqrt{u}}du$$ $$=\cos\theta {{\int^{1}_0}} {1\over \sqrt{u}}du$$

And then do the rest from there? Also, what other way could I do this?

  • 1
    $\begingroup$ You have a $\sin 2\theta$ in the numerator and a $\cos 2\theta = 2\cos^2 \theta - 1$ expression in the denominator .. (btw is it a typo or did you just pull out a $\cos \theta$ from inside the integral ? ! :O ) $\endgroup$
    – r9m
    Commented Dec 18, 2014 at 20:05
  • 2
    $\begingroup$ Let $u=\cos^2\theta+8$. $\endgroup$ Commented Dec 18, 2014 at 20:07
  • $\begingroup$ @r9m yea the $cos\theta$ that was pulled out would later be multiplied back to all the terms inside. $\endgroup$
    – Nolohice
    Commented Dec 18, 2014 at 20:09
  • $\begingroup$ @stuntstool ah ! fine then :) $\endgroup$
    – r9m
    Commented Dec 18, 2014 at 20:19
  • $\begingroup$ final answer i have after doing what @AndréNicolas suggested is $9^{1/2} - 8^{1/2}$ $\endgroup$
    – Nolohice
    Commented Dec 18, 2014 at 20:45

1 Answer 1


Another way is to note that the numerator is a constant times the derivative of expression within the square root in the denominator, i.e. $$\frac{d}{d\theta}(\cos^2\theta+8)=-2\cos\theta\sin\theta$$ so that the indefinite integral is simply $$-(\cos^2\theta+8)^{\frac{1}{2}}$$ You can verify this by differentiating the above, using the chain rule.

  • $\begingroup$ Good answer. +1 $\endgroup$
    – Timbuc
    Commented Dec 18, 2014 at 20:18

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