Once while solving problems from Romanian Mathematical Olympiads I encountered a mathematical gem. The problem I found was asking to calcualte the following integral
\[I=\int\frac{\cos x}{a\sin x+b\cos x}\,dx.\]
What do I find especially beautiful about this problem? It rewards the desire to examine a bit more than is asked initially!
Solution. Let's try to evaluate one more integral
J=∫asinx+bcosxsinxdx.
Can you guess the next step?
Yes, we will take advantage of the fact that the sum of integrals is the integral of sum. In other words:
aJ+bI=a∫asinx+bcosxsinxdx+b∫asinx+bcosxcosxdx=∫asinx+bcosxasinx+bcosxdx=x+C.
On other hand we can easily calculate integrals of the form ∫f(x)f′(x)dx. The derivative of (asinx+bcosx) is equal to (acosx−bsinx), which can also be expressed as a linear combination of our two integrals, i.e.
aI−bJ=∫asinx+bcosxacosx−bsinxdx=ln∣asinx+bcosx∣+C.
Now we simply have to solve a system of linear equations. We obtain I=a2+b2aln∣asinx+bcosx∣+bx+C.
and
J=a2+b2ax−ln∣asinx+bcosx∣+C.
Has anyone of you seen problems with the similar ideas? How often while solving a problem you investigate possible variations of the conditions?
#Calculus
#Goldbach'sConjurersGroup
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Try solving this the same way: I=∫csinx+dcosxasinx+bcosxdx
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Invitation (click the problem) ∫04π[(1−x4)(1+x2)(1−x2)ln(1+x2)+(1+x2)−(1−x2)ln(1−x2)]xexp[x2+1x2−1]dx.
This looks very useful, thanks for sharing Nicolae. :)
Thanks for sharing !!
Invitation (click the problem) ∫04π[(1−x4)(1+x2)(1−x2)ln(1+x2)+(1+x2)−(1−x2)ln(1−x2)]xexp[x2+1x2−1]dx.
Thanks for sharing