Question

Integrate:[ (sinx)^3]/[(sinx)^3+(cosx)^3] evaluated from 0 to Pi/2

Answer 1

My first instinct would be to try the substitution \(t=\tan\dfrac{x}{2}\), but that might be too complicated. There's probably some identity manipulation involved if there is a simpler method.

Answer 2

I tried the t=tan(x/2) but yeah, it became more complicated.

Answer 3

I wonder if you can try rewriting the expressiong using a^3+b^3=(a+b)(a^2-ab+b^2)

Answer 4

Just trying to look for way to simplify it

Answer 5

My other though is to divide both top and bottom by cos^3(x)

Answer 6

giving us something like tan^3(x)/(1+tan^3(x))
and then again trying to use the sum of cubes to do something with the bottom

Answer 7

\[\frac{ (Sinx)^{3} }{ (Sinx+Cosx)((Sinx)^{2} +SinxCosx+ (Cosx)^{2})}\]? I don't know, but I guess it is complicated to solve this way

Answer 8

I meant Sin^2x-SinxCosx lol

Answer 9

but either way it looks tragic :(


sith's way is looking better and better
but i will keep trying to think of a creative way that leads to less complication
a lot of the times you are just going to have to be creative when it comes to integrating some expressions

Answer 10

and then you could say sin^2(x)+cos^2(x)=1
or if you did the other way 1+tan^2(x)=sec^2(x)

Answer 11

the other way. 1+tan^2(x)

Answer 12

is this from a particular book and particular section of the book?

Answer 13

that could give us a hint maybe
thought it could be in the random section where you do have to think more than you want sometimes

Answer 14

It is the assignment my teacher gave me

Answer 15

ahhhhhhh I know, he said use the hint : integral of f(x)dx evaluated from 0 to Pi/2 is the same as the integral of f(x -pi/2)dx evaluated from 0 to Pi/2

Answer 16

so...we know sin(pi/2-x)=cos(x)
and cos(pi/2-x)=sin(x)

Let \[I=\int\limits_{0}^{\frac{\pi}{2} } \frac{\sin^3(x)}{\sin^3(x)+\cos^3(x)} dx\]
which also equals
\[\int\limits_{0}^{\frac{\pi}{2}}\frac{\sin^3(\frac{\pi}{2}-x)}{\sin^3(\frac{\pi}{2}-x) +\cos^3(\frac{\pi}{2}-x)} dx\]
let's use the above identity to rewrite that last integral
so we have
\[I=\int\limits_{0}^{\frac{\pi}{2}}\frac{\cos^3(x)}{\cos^3(x)+\sin^3(x)} dx\]

Answer 17

and...

Answer 18

Notice that the first integral and the last integral (third) I wrote have the same denominator

Answer 19

do you see that...

Answer 20

What would happen if you added those two equations together
I dare you to try

Answer 21

This question really is easy once you see the trick

Answer 22

Do you see where I was going with what I wrote above?

Answer 23

I used what your teacher gave that
\[\int\limits_{0}^{\frac{\pi}{2}}f(x) dx=\int\limits_{0}^{\frac{\pi}{2}}f(\frac{\pi}{2}-x) dx\]
We can prove this by doing a substitution.
Given \[\int_0^\frac{\pi}{2}f(\frac{\pi}{2}-x) dx\]

Let u=pi/2-x
du=-dx
so we have
if x=0 then u=pi/2-0=pi/2
if x=pi/2 then u=pi/2-pi/2=0
so we have that \[\int\limits_{0}^{\frac{\pi}{2}} f(\frac{\pi}{2}-x) dx=\int\limits_{\frac{\pi}{2}}^{0}-f(u) du=\int\limits_{0}^{\frac{\pi}{2}}f(u) du\]
Therefore we have
\[\int\limits_{0}^{\frac{\pi}{2}}f(x) dx=\int\limits_{0}^{\frac{\pi}{2}}f(\frac{\pi}{2}-x) dx\]

Answer 24

For another method, I was wondering if there's some foolproof way of splitting up the integrand into partial fractions. I've seen it done before, but it was on much simpler problem.