Find $\displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\sin^n(x)}{\sin^n(x)+\cos^n(x)} dx$.
I was told to switch the limits of integration then add them. How can I do that here?
Asked
Active
Viewed 78 times
5
John Ryan
- 1,269
-
1If you make the substitution $u = \frac{\pi}{2} - x$, what do you get? – Daniel Fischer Dec 18 '15 at 22:40
-
Thanks for ruining the secret Daniel! – Ron Gordon Dec 18 '15 at 22:42
-
We get $\displaystyle \int_{\frac{\pi}{2}}^{0} \dfrac{\sin^n(\frac{\pi}{2}-u)}{\sin^n((\frac{\pi}{2}-u))+\cos^n((\frac{\pi}{2}-u))}$. – John Ryan Dec 18 '15 at 22:44
-
@JohnRyan: You are missing a negative sign since $du = -dx$. After you fix that, use the identities $\sin(\tfrac{\pi}{2}-u) = \cos(u)$ and $\cos(\tfrac{\pi}{2}-u) = \sin(u)$. – JimmyK4542 Dec 18 '15 at 22:49
-
@JimmyK4542 How does that help solve the question? – John Ryan Dec 18 '15 at 22:55
-
Next, use the "then add them" hint. – grand_chat Dec 18 '15 at 22:57
-
We get $\int_0^{\pi/2}\frac{\sin^n x}{\sin^n x+\cos^n x}dx+ \int_0^{\pi/2}\frac{\cos^n u}{\cos^n u+\sin^n u}du.$ Then what? – John Ryan Dec 18 '15 at 22:59
-
Change the dummy variable of integration in the second integral to $x$. – André Nicolas Dec 18 '15 at 23:08
-
Ok, I still don't see how that helps. – John Ryan Dec 18 '15 at 23:14
-
We then have $$\int_0^{\pi/2}\frac{\sin^n x}{\sin^n x+\cos^n x}dx+ \int_0^{\pi/2}\frac{\cos^n x}{\cos^n x+\sin^n x}dx ,.$$ – John Ryan Dec 18 '15 at 23:17
-
Now add the two integrals. – Daniel Fischer Dec 18 '15 at 23:17
-
@DanielFischer Yes, then what? – John Ryan Dec 18 '15 at 23:18
-
What doe adding do? – John Ryan Dec 18 '15 at 23:30
-
@JohnRyan $I+I=\frac{\pi}{2}$. What is $I$? – Mark Viola Dec 18 '15 at 23:31
-
Why do they add to $\dfrac{\pi}{2}$? – John Ryan Dec 18 '15 at 23:32
-
What is the sum of the integrands? – Daniel Fischer Dec 18 '15 at 23:40
-
It is $\displaystyle \int_{0}^{\pi/2} 1$ – John Ryan Dec 19 '15 at 14:37