2006 SMT/Calculus Problems/Problem 6: Difference between revisions
Created page with "==Solution== Let <math>u=\frac{\pi}{2}-x</math> and <math>du=-dx</math>, the integral than turns into: <cmath> I = \int_{0}^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx \Rightar..." |
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==Problem 6== | |||
Evaluate | |||
<cmath> \int_{0}^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx </cmath> | |||
==Solution== | ==Solution== | ||
Latest revision as of 17:32, 14 January 2020
Problem 6
Evaluate
![\[\int_{0}^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx\]](http://latex.artofproblemsolving.com/3/8/6/386a00c7411c3466a82d7eac8b9c9bddcd2a3ba9.png)
Solution
Let 
and 
, the integral than turns into:
![\[I = \int_{0}^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx \Rightarrow -\int_{\pi/2}^{0}\frac{\sin^3(\pi/2-u)}{\sin^3(\pi/2-u)+\cos^3(\pi/2-u)}du\]](http://latex.artofproblemsolving.com/6/b/c/6bcff24f83d2a2c5fc4d36756457688dbd38d1c6.png)
Which of course equals:
![\[I =-\int_{\pi/2}^{0}\frac{\sin^3(\pi/2-u)}{\sin^3(\pi/2-u)+\cos^3(\pi/2-u)}du = \int_{0}^{\pi/2}\frac{\cos^3(u)}{\cos^3(u)+\sin^3(u)}du\]](http://latex.artofproblemsolving.com/0/8/b/08b448a8367db74f68f406995e3d56ee97b4f466.png)
We can then add the original integral, with x as the variable, to the integral with u as a variable because the variables don't matter, just the value that the integrals takes:
Therefore:
![\[I=\int_{0}^{\pi/2}\frac{\sin^3x}{\sin^3x+\cos^3x}dx = \boxed{\frac{\pi}{4}}\]](http://latex.artofproblemsolving.com/6/0/6/606a75f6b982676c1383743a0620f8d7eba9dea7.png)
