Evaluate the integral \int_0^{\frac{\pi}{2}}\frac{\sin^3 x}{\sin^3 x+\cos^3 x}dx

piarepm

piarepm

Answered question

2021-12-30

Evaluate the integral 0π2sin3xsin3x+cos3xdx

Answer & Explanation

sonorous9n

sonorous9n

Beginner2021-12-31Added 34 answers

As
abf(x)dx=abf(a+bx)dx,
If
I=0π2sinnxsinnx+cosnxdx
=0π2sinn(π2x)sinn(π2x)+cosn(π2x)dx
=0π2cosnxcosnx+sinnxdx
I+I=0π2dx
assuming sinnx+cosnx0 which is true as 0xπ2
Generalizition:
If J=abg(x)g(x)+g(a+bx)dx, J=abg(a+bx)g(x)+g(a+bx)dx
RightowJ+J=abdx
provided g(x)+g(a+bx)0
If a=0,b=π2 and g(x)=h(sinx),
g(π2+0x)=hsin(π2+0x))=h(cosx)
So J becomes
Corgnatiui

Corgnatiui

Beginner2022-01-01Added 35 answers

Symmetry! This is the same as the integral with cos3x on top.
If that is not obvious from the geometry, make the change of variable u=π2x
Add them, you get the integral of 1.. So our integral is π4.
Vasquez

Vasquez

Expert2022-01-09Added 669 answers

Hint: if
I=0π2sin3xsin3x+cos3xdx
and
J=0π2cos3xsin3x+cos3xdx
Then consider I+J, and the effect of the substitution y=π2x on the integral I.

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