Integral [0, pi/2] [ (sin(x)) / (sin(x) + cos(x) ] dx
- Thread starter mikeyWay
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Well, since you've shown no work of your own (and you've read the Read Before Posting thread that's sticked at the top of each sub-forum), I'm forced to assume that you have no work to share with us and are stuck at the very beginning. I think a good first step would be to multiply by the conjugate of the denominator, like so:
\(\displaystyle \displaystyle \int\limits_{0}^{\frac{\pi}{2}} \left( \frac{sin(x)}{sin(x) + cos(x)} \right) dx = \int\limits_{0}^{\frac{\pi}{2}} \left( \frac{sin(x)[sin(x) - cos(x)]}{[sin(x) + cos(x)][sin(x) - cos(x)]} \right) dx\)
Where does this lead you? What do you think would be a next good step?
\(\displaystyle \displaystyle \int\limits_{0}^{\frac{\pi}{2}} \left( \frac{sin(x)}{sin(x) + cos(x)} \right) dx = \int\limits_{0}^{\frac{\pi}{2}} \left( \frac{sin(x)[sin(x) - cos(x)]}{[sin(x) + cos(x)][sin(x) - cos(x)]} \right) dx\)
Where does this lead you? What do you think would be a next good step?
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Another approach would be to write:
\(\displaystyle \displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin(x)+\cos(x)}\,dx\)
Now, if we let:
\(\displaystyle \displaystyle u=\frac{\pi}{2}-x\implies du=-dx\)
Then we may write after replacing the dummy variable u with x:
\(\displaystyle \displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos(x)}{\sin(x)+\cos(x)}\,dx\)
Add the two expressions for \(\displaystyle I\)...what do you get?
\(\displaystyle \displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\sin(x)}{\sin(x)+\cos(x)}\,dx\)
Now, if we let:
\(\displaystyle \displaystyle u=\frac{\pi}{2}-x\implies du=-dx\)
Then we may write after replacing the dummy variable u with x:
\(\displaystyle \displaystyle I=\int_{0}^{\frac{\pi}{2}}\frac{\cos(x)}{\sin(x)+\cos(x)}\,dx\)
Add the two expressions for \(\displaystyle I\)...what do you get?
Last edited:
Steven G
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Ah, you beat me to it (good job). I just learned that trick about a month ago. It just eats through these type problems.
Well, I see you've found the answer a different way (and I must admit that solution is far more elegant and nice... I only wish I'd thought of it myself
), so I'm not sure how much this is worth... but the same answer can be found via my method, just with a fair bit of messy algebra and slinging terms around. Picking up from where you left off:\(\displaystyle \displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^2(x) - \sin(x)\cos(x)}{\sin^2(x) - \cos^2(x)} \: dx\)
You've correctly identified that \(\displaystyle \sin^2(x) - \cos^2(x) = -\cos(2x)\), and we also know that \(\displaystyle \sin(2x) = 2\sin(x)\cos(x)\), hence \(\displaystyle \sin(x)\cos(x) = \displaystyle \frac{\sin(2x)}{2}\). Let's plug these in:
\(\displaystyle = \displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^2(x) - \frac{\sin(2x)}{2}}{-\cos(2x)} \: dx\)
Next, let's combine the numerator into one fraction:
\(\displaystyle = \displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{\frac{2\sin^2(x) - \sin(2x)}{2}}{-\cos(2x)} \: dx = \int\limits_{0}^{\frac{\pi}{2}} \frac{2\sin^2(x) - \sin(2x)}{-2\cos(2x)} \: dx\)
Now we can split the entire fraction into two:
\(\displaystyle = \displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{2\sin^2(x)}{-2\cos(2x)} - \frac{\sin(2x)}{-2\cos(2x)} \: dx\)
And further make use of the addition/subtraction rule for integrals to make the whole thing into two simpler integrals:
\(\displaystyle = \displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{2\sin^2(x)}{-2\cos(2x)} \: dx - \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin(2x)}{-2\cos(2x)} \: dx\)
A bit of cleanup involving pulling out some negatives and swapping signs gives:
\(\displaystyle = \displaystyle -\int\limits_{0}^{\frac{\pi}{2}} \frac{2\sin^2(x)}{2\cos(2x)} \: dx + \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin(2x)}{2\cos(2x)} \: dx\)
\(\displaystyle = \displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin(2x)}{2\cos(2x)} \: dx - \int\limits_{0}^{\frac{\pi}{2}} \frac{2\sin^2(x)}{2\cos(2x)} \: dx\)
Now let's use a few more trig identities to simplify some more:
\(\displaystyle = \displaystyle \int\limits_{0}^{\frac{\pi}{2}} \frac{\tan(2x)}{2} \: dx - \int\limits_{0}^{\frac{\pi}{2}} \frac{\sin^2(x)}{\cos(2x)} \: dx\)
\(\displaystyle = \displaystyle \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \tan(2x) \: dx - \int\limits_{0}^{\frac{\pi}{2}} \frac{\frac{1-cos(2x)}{2}}{\cos(2x)} \: dx\)
\(\displaystyle = \displaystyle \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \tan(2x) \: dx - \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \frac{1-cos(2x)}{\cos(2x)} \: dx\)
\(\displaystyle = \displaystyle \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \tan(2x) \: dx - \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \left( \frac{1}{\cos(2x)} - 1 \right) \: dx\)
\(\displaystyle = \displaystyle \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \tan(2x) \: dx - \frac{1}{2} \left( \int\limits_{0}^{\frac{\pi}{2}} \frac{1}{\cos(2x)} \: dx - \int\limits_{0}^{\frac{\pi}{2}} 1 \: dx \right)\)
\(\displaystyle = \displaystyle \frac{1}{2} \int\limits_{0}^{\frac{\pi}{2}} \tan(2x) \: dx - \frac{1}{2} \left( \int\limits_{0}^{\frac{\pi}{2}} \sec(2x) \: dx - \int\limits_{0}^{\frac{\pi}{2}} 1 \: dx \right)\)
And now we're left with only integrals we already know how to tackle:
\(\displaystyle = \displaystyle \frac{1}{2} \left[ -\frac{1}{2} \ln(|\cos(2x)|) \right]^{\frac{\pi}{2}}_0 - \frac{1}{2} \left( \left[ \frac{1}{2} \ln(|\tan(2x) + \sec(2x)|) \right]^{\frac{\pi}{2}}_0 - \frac{\pi}{2} \right)\)
By plugging in x = 0 and x = \(\displaystyle \displaystyle \frac{\pi}{2}\) respectively to each of the expressions, we discover that they all evaluate to 0, so we're left with only:
\(\displaystyle = \displaystyle -\frac{1}{2} \left( -\frac{\pi}{2} \right) = \frac{\pi}{4}\)
So, yeah, my method gets the same answer, it's just way harder. And there's no real reason to ever do it this way, as it turns out.