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I need to evaluate two kind of similar integrals

The first one: $$\lim _{n\to \infty }\int _0^{\frac{\pi }{3}}\:\frac{\sin ^n\left(x\right)}{\sin ^n\left(x\right)+\cos ^n\left(x\right)}dx$$

The second one: $$\int _0^{2\pi }\:\frac{x\sin ^{100}\left(x\right)}{\sin ^{100}\left(x\right)+\cos ^{100}\left(x\right)}dx$$

For the first one, I got to this form $$\lim _{n\to \infty }\int _0^{\frac{\pi }{3}}\:\frac{\sin ^n\left(x\right)}{\sin ^n\left(x\right)+\cos ^n\left(x\right)}dx = \frac{\pi }{3}-\int _0^{\frac{\pi }{3}}\:\frac{\cos ^n\left(x\right)dx}{\sin \:^n\left(x\right)+\cos \:^n\left(x\right)}$$

Now, I would make a substitution $u = \frac{\pi}{3}-x$ but it seems it doesn't help in this situation.

I have put the second integral here because I think it should have the same solving concept.

  • 1
    Both these integrals are easily solved using $\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ – 2017-02-25
  • 1
    Second integral evaluates to $\pi^{2}$. – 2017-02-25
  • 0
    @ParamanandSingh What can I do then with $\cos ^n\left(\frac{\pi }{3}-x\right)$ – 2017-02-25

2 Answers 2

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For the first try substituting $x=\frac{2}{3}u$ and divide through by $\sin^n(\frac{2}{3}x)$ top and bottom.

So now we have,

$$\frac{2}{3} \lim_{n \to \infty} \int_{0}^{\frac{\pi}{2}} \frac{1}{1+\cot^n (\frac{2}{3}x)} dx $$

Now for $\frac{2}{3}x \in (0,\frac{\pi}{4})$ we have $\cot \frac{2}{3} x>1$ so there the integrand tends to $0$. This is when $x \in (0,\frac{3\pi}{8})$. Whereas for $x \in (\frac{3\pi}{8},\frac{3\pi}{4})$ we have $\cot \frac{2}{3} x \in (0,1)$ so there the integrand tends to $1$ and thus we have,

$$=\frac{2}{3} \int_{\frac{3}{8}\pi}^{\frac{\pi}{2}} 1 dx$$

$$=\color{red}{\frac{\pi}{12}}$$

For the second integral with $x=u+\pi$ we have,

$$I=\int_{-\pi}^{\pi} (x+\pi)\frac{\sin^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx$$

Notice that $x\frac{\sin^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)}$ is odd.

$$\color{blue}{I}=\color{blue}{\pi \int_{-\pi}^{\pi} \frac{\sin^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx}$$

$$=\pi \int_{-\pi}^{\pi} \frac{\sin^{100} (x)+\cos^2(x)-\cos^2(x)}{\sin^{100} (x)+\cos^{100} (x)}$$

$$=\pi(2\pi-\int_{-\pi}^{\pi}\frac{\cos^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx)$$

$$=2\pi^2-\color{blue}{\pi\int_{-\pi}^{\pi}\frac{\cos^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)}} dx$$

But,

$$\color{blue}{I}=\pi \int_{-\pi}^{\pi}\frac{\cos^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx$$

$$=\pi \int_{-\pi}^{\pi} \frac{\sin^{100} (x)}{\sin^{100} (x)+\cos^{100} (x)} dx$$

Follows from $x=\frac{\pi}{2}-u$ then $u-\frac{\pi}{2}=v$.

So,

$$I=2\pi^2-I$$

$$I=\color{red}{\pi^2}$$

In fact everything I wrote for this will work for $n$ an even positive integer instead of $100$.

  • 0
    Doesn't get me anywhere... – 2017-02-25
  • 0
    Sorry @Liviu please check edits – 2017-02-26
  • 1
    Ok I think everything is good now @Liviu – 2017-02-26
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The first limit can be tackled by the dominated convergence theorem. As $n\to +\infty$, $f_n(x)=\frac{\sin(x)^n}{\sin(x)^n+\cos(x)^n}$ is pointwise convergent to a function that equals $0$ on the interval $\left(0,\frac{\pi}{4}\right)$ and $1$ on the interval $\left(\frac{\pi}{4},\frac{\pi}{3}\right)$. It follows that the value of the limit is $\color{red}{\large\frac{\pi}{12}}.$ The second integral is screaming for symmetry.