2

The question is show that $$ \int_0^{a} f(x) dx = \int_0^{a} f(a-x) dx $$

<p>Hence or otherwise, calculate the value of the following integral </p>

<p>$$ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)} $$</p>

What I have done for the first part

$$ \int_0^{a} f(x) dx = \int_0^{a} f(a-x) dx $$

$$ \left[F(x) \right]^a_0 dx = \left[-F(a-x) \right]^a_0 $$

$$ F(a) - F(0) = \left[-F(a-a) - -F(a-0) \right] $$

$$ F(a) - F(0) =F(a) - F(0) $$

$$ LHS = RHS $$

Now I am stuck applying this to the integral. I have attempted this:

$$ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)} $$

Because $ \int_0^{a} f(x) dx = \int_0^{a} f(a-x) dx $ The integral is transformed to

$$ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(\frac{\pi}{2}-x)}{\sin^n(\frac{\pi}{2}-x) + \cos^n(\frac{\pi}{2}-x)} $$

Which then becomes

$$ \int^{\frac{\pi}{2}}_0 \frac{\cos^n(x)}{\cos^n(x) + \sin^n(x)} $$

Now I am stuck...

bigfocalchord
  • 5,597

2 Answers2

3

Let $I$ be the value of your integral.

Use the substitution $u=\pi/2-x$ to get a new integral, call it $J=I$.

Then $I+J = \int_0^{\pi/2} 1 ~ dx$ therefore $2I = \frac{\pi}{2}$.

Nicolas Bourbaki
  • 5,064
0

So you have shown that: $\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)} dx= \int^{\frac{\pi}{2}}_0 \frac{\cos^n(x)}{\cos^n(x) + \sin^n(x)}dx$

$2\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)}dx =$$\int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)}{\sin^n(x) + \cos^n(x)}dx + \int^{\frac{\pi}{2}}_0 \frac{\cos^n(x)}{\cos^n(x) + \sin^n(x)}dx\\ \int^{\frac{\pi}{2}}_0 \frac{\sin^n(x)+cos^n(x)}{\sin^n(x) + \cos^n(x)}dx\\\pi/2$

Doug M
  • 57,877