Answered: "Evaluate the integral ∫0π2sin3xsin3x+cos3xdx"

piarepm

Answered question

2021-12-30

Evaluate the integral

\int_{0}^{\frac{π}{2}} \frac{\sin^{3} x}{\sin^{3} x + \cos^{3} x} d x

Answer & Explanation

sonorous9n

Beginner2021-12-31Added 34 answers

\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x

,
If

I = \int_{0}^{\frac{π}{2}} \frac{\sin^{n} x}{\sin^{n} x + \cos^{n} x} d x

= \int_{0}^{\frac{π}{2}} \frac{\sin^{n} (\frac{π}{2} - x)}{\sin^{n} (\frac{π}{2} - x) + \cos^{n} (\frac{π}{2} - x)} d x

= \int_{0}^{\frac{π}{2}} \frac{\cos^{n} x}{\cos^{n} x + \sin^{n} x} d x

\Rightarrow I + I = \int_{0}^{\frac{π}{2}} d x

assuming

\sin^{n} x + \cos^{n} x \neq 0

which is true as

0 \leq x \leq \frac{π}{2}

Generalizition:
If

J = \int_{a}^{b} \frac{g (x)}{g (x) + g (a + b - x)} d x, J = \int_{a}^{b} \frac{g (a + b - x)}{g (x) + g (a + b - x)} d x

R i g h t \to o w J + J = \int_{a}^{b} d x

provided

g (x) + g (a + b - x) \neq 0

a = 0, b = \frac{π}{2}

and

g (x) = h (\sin x)

g (\frac{π}{2} + 0 - x) = h \sin (\frac{π}{2} + 0 - x)) = h (\cos x)

So J becomes

This is helpful 0 Flag

Corgnatiui

Beginner2022-01-01Added 35 answers

Symmetry! This is the same as the integral with

\cos^{3} x

on top.
If that is not obvious from the geometry, make the change of variable

u = \frac{π}{2} - x

Add them, you get the integral of 1.. So our integral is

\frac{π}{4}

Vasquez

Expert2022-01-09Added 669 answers

Hint: if
$I = \int_{0}^{\frac{π}{2}} \frac{\sin^{3} x}{\sin^{3} x + \cos^{3} x} d x$
and
$J = \int_{0}^{\frac{π}{2}} \frac{\cos^{3} x}{\sin^{3} x + \cos^{3} x} d x$
Then consider $I + J$ , and the effect of the substitution $y = \frac{π}{2} - x$ on the integral I.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?