DLS
  • DLS
Integrate!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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DLS
  • DLS
\[\Huge \int\limits _{0}^{\pi} \sin^6x \cos^5(3x)dx\]
anonymous
  • anonymous
see cos3x = 4cos^3x - 3cosx ( plz check formula for cos3x before) now cos3x = 4cos^3x - 3cosx=(4cos^2x-3)cosx=(4-4sin^2 x -3)cosx=(1-4sin^2 x)cosx now take sinx =t and hence cosxdx =dt and so on
DLS
  • DLS
umm we don't have to actually solve it,Iguess it is based on some property,answer is 0.

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DLS
  • DLS
realize its an odd function
ash2326
  • ash2326
@DLS We could use this property \[\int_{0} ^{a} f(x) dx\] if f(a-x)=f(x) \[2\int_0^{\frac{a}{2}} f(x) dx\] if f(a-x)=-f(x) then I=0
DLS
  • DLS
i know this property would be used but still confused
ash2326
  • ash2326
Where do you have confusion?
DLS
  • DLS
f(pi-x)= f(x) is not true..
ash2326
  • ash2326
but we want f(pi-x)=-f(x) I'll illustrate
DLS
  • DLS
yes is that true either:?
ash2326
  • ash2326
Sorry I was typing and then I got "aww snap" Just a minute, I'll retype
ash2326
  • ash2326
\[f(x)=\sin^6(x)\cos^5 (3x)\] let's find f(pi-x) \[f(\pi-x)=\sin^6(\pi-x)\cos^5(3\pi-3x)\] We know \[\sin(\pi-x)=\sin x\] and \[\cos (\pi-x)=-\cos x\] therfore \[\cos(2\pi+\pi-x)=-\cos x\] \[\cos (3\pi-x)=-\cos x\] \[f(\pi-x)=\sin^6(x)(-\cos 3x)^5\] so \[f(\pi-x)=-\sin^6(x)\cos^5(3x)=-f(x)\]
ash2326
  • ash2326
Do you follow this @DLS
anonymous
  • anonymous
Is it an even or odd function?
DLS
  • DLS
okay..i got it :D thanks! in our class we were told something like at some midpoint in the interval like pi/2 or smt it becomes 0,maybe odd function like if we integrate the same function from -pi/2 to pi/2 dunno smt like this
ash2326
  • ash2326
It's odd function, replace x by -x and you'll get f(x)=-f(x) But that property is useful if we have \[\int_{-a}^{a} f(x) dx\]
ash2326
  • ash2326
@DLS Yes we could do that, we can change the limits \[u=x-\frac{\pi}{2}\] then lower limit would be -pi/2 upper limit would be pi/2 dx=-du there would be one more minus sign so the integral will become \[\large \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\sin^6u\ \cos^5 3u\ du \] using odd function property we can prove this 0
anonymous
  • anonymous
\[\int\limits_{0}^{a} f(x) dx = - \int\limits_{0}^{a}f(a-x) dx\] \[u = a-x\] \[-\int\limits\limits_{0}^{a}f(a-x) dx\ =- \int\limits_{a}^{0} f(u) -du\] =\[= -\int\limits_{0}^{a}f(u) du\] \[\int\limits_{0}^{a}f(x) dx = - \int\limits_{0}^{a}f(u) du\] 0 is the only thing that equals itself
anonymous
  • anonymous
So the integral is 0
DLS
  • DLS
thanks!
anonymous
  • anonymous
You are welcome

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