Question

Integrate!

Answer 1

\[\Huge \int\limits _{0}^{\pi} \sin^6x \cos^5(3x)dx\]

Answer 2

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

Answer 3

umm we don't have to actually solve it,Iguess it is based on some property,answer is 0.

Answer 4

realize its an odd function

Answer 5

@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

Answer 6

i know this property would be used but still confused

Answer 7

Where do you have confusion?

Answer 8

f(pi-x)= f(x) is not true..

Answer 9

but we want f(pi-x)=-f(x)
I'll illustrate

Answer 10

yes is that true either:?

Answer 11

Sorry I was typing and then I got "aww snap"
Just a minute, I'll retype

Answer 12

\[f(x)=\sin^6(x)\cos^5 (3x)\]
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
\[f(\pi-x)=-\sin^6(x)\cos^5(3x)=-f(x)\]

Answer 13

Do you follow this @DLS

Answer 14

Is it an even or odd function?

Answer 15

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

Answer 16

It's odd function, replace x by -x and you'll get f(x)=-f(x)
But that property is useful if we have
\[\int_{-a}^{a} f(x) dx\]

Answer 17

@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

Answer 18

\[\int\limits_{0}^{a} f(x) dx = - \int\limits_{0}^{a}f(a-x) dx\]
\[u = a-x\]
\[-\int\limits\limits_{0}^{a}f(a-x) dx\ =- \int\limits_{a}^{0} f(u) -du\]
=\[= -\int\limits_{0}^{a}f(u) du\]
\[\int\limits_{0}^{a}f(x) dx = - \int\limits_{0}^{a}f(u) du\]
0 is the only thing that equals itself

Answer 19

So the integral is 0

Answer 20

thanks!

Answer 21

You are welcome