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Calculating with Mathematica, one can have $$\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}\,\mathrm dt=\frac{\pi}{4}.$$

  • How can I get this formula by hand? Is there any simpler idea than using $u = \sin t$?
  • Is there a simple way to calculate $$ \int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}\,\mathrm dt $$ for $n>3$?
  • Could anyone come up with a reference for this exercise?
Jam
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  • 1
    If I had to solve the indefinite case, I'd probably divide top and bottom by $\cos^3t$, make the substitution $t=\tan^{-1}x$, then evaluate with partial fractions. – Mike Mar 15 '12 at 14:05

4 Answers4

26

The substitution $y=\frac{\pi}{2}-t$ solves it... If you do this substitution, you get:

$$\int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}dt= \int_0^{\pi/2}\frac{\cos^n y}{\cos^n y+\sin^n y}dy \,.$$

N. S.
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20

Use the Calculus identity that $$f(x)=f(a-x),$$ and let $$I=\int_0^\frac{\pi}{2} \frac{\sin^3t}{\sin^3t+\cos^3t}dt.$$ Then, $$f(t)=f(\frac{\pi}{2}-t)=\frac{\sin^3(\frac{\pi}{2}-t)}{\sin^3(\frac{\pi}{2}-t)+\cos^3(\frac{\pi}{2}-t)}=\frac{\cos^3t}{\cos^3t+\sin^3t}$$Thus, $$I=\int_0^\frac{\pi}{2} \frac{\cos^3t}{\cos^3t+\sin^3t}dt.$$ So we have $$2I=\int_0^\frac{\pi}{2} \frac{\sin^3t}{\sin^3t+\cos^3t}dt+\int_0^\frac{\pi}{2} \frac{\cos^3t}{\cos^3t+\sin^3t}dt=\int_0^\frac{\pi}{2}dt=\frac{\pi}{2}.$$ So $$I=\frac{\pi}{4}.$$ Note that this is true for any natural number $n$.

smanoos
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2

In general

Dividing both the numerator and denominator by $\cos ^n x$ converts $$ \begin{aligned} I_{n} =&\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan ^{n} t} d t \\ \begin{aligned} \\ \end{aligned} \\\stackrel{t\mapsto\frac{\pi}{2}-t}{=} &\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot ^{n} t} d t \\ =&\int_{0}^{\frac{\pi}{2}} \frac{\tan ^{n} t}{\tan ^{n} t+1} d t \\ 2 I_{n} =&\int_{0}^{\frac{\pi}{2}}\left(\frac{1}{1+\tan ^{n} t}+\frac{\tan ^{n} t}{\tan ^{n} t+1}\right) d t = \int_{0}^{\frac{\pi}{2}} 1 d t \\ I_{n} =&\frac{\pi}{4} \end{aligned} $$

Lai
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1

Substitute

$$u = \sec t + \tan t \implies t = 2 \arctan \frac{u-1}{u+1} \implies dt = \frac{2\,du}{1+u^2} \\ \implies \sin t = \frac{u^2-1}{u^2+1} \text{ and } \cos t = \frac{2u}{u^2+1}\quad.$$

Then for $0 \le \theta < \dfrac\pi2$, we have

$$\begin{align*} I (\theta) &= \int_0^\theta \frac{\sin^3t}{\sin^3t+\cos^3t} \, dt \\ &= \int_1^{\phi:=\sec\theta+\tan\theta} \frac{\frac{\left(u^2-1\right)^3}{\left(u^2+1\right)^3}}{\frac{\left(u^2-1\right)^3}{\left(u^2+1\right)^3} + \frac{8u^3}{\left(u^2+1\right)^3}} \, \frac{2\,du}{1+u^2} \\ &= \int_1^\phi \frac{2\left(u^2-1\right)^3}{u^8-2u^6+8u^5+8u^3+2u^2-1} \, du \\ &= \int_1^\phi \left(\frac23 \frac{2u^3-3u^2+2u+1}{u^4-2u^3+2u^2+2u+1} - \frac13 \frac{u+1}{u^2+2u-1} - \frac{u-1}{u^2+1}\right) \, du \\ &= \int_1^\phi \left(\frac13 \frac{d\left(u^4-2u^3+2u^2+2u+1\right)}{u^4-2u^3+2u^2+2u+1} - \frac16 \frac{d\left(u^2+2u-1\right)}{u^2+2u-1} - \frac12 \frac{d\left(u^2+1\right)}{u^2+1} + \frac{du}{u^2+1}\right) \\ &= \frac16 \ln\frac{\left(\phi^4-2\phi^3+2\phi^2+2\phi+1\right)^2}{\left(\phi^2+2\phi-1\right) \left(\phi^2+1\right)^3} + \arctan \phi - \frac\pi4 \\ &= \bbox[1pt, pink]{\frac16 \ln \frac{\left(\cos\theta\sin\theta-1\right)^2}{\cos\theta+\sin\theta} + \arctan\left(\sec\theta+\tan\theta\right)} - \frac\pi4 \\ &= \frac16 \ln \frac{\left(\cos\theta\sin\theta-1\right)^2}{\cos\theta+\sin\theta} + \frac\theta2 \end{align*}$$

which indeed yields $\displaystyle\lim_{\theta\to\tfrac\pi2^-}I(\theta) = \boxed{\dfrac\pi4}$.

user170231
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