# Prove $\int_0^1 (1+a+a^2\pi^2x^{2a})\sin(\pi x^a)dx = a\pi$ using previous results

If $$a>0$$ prove,
$$\int_0^1 \sin(\pi x^a)dx+a\pi\int_0^1 \ x^a\cos(\pi x^a)dx = 0 \tag1$$ $$\int_0^1 \cos(\pi x^a)dx-a\pi\int_0^1 \ x^a\sin(\pi x^a)dx = -1 \tag2$$ $$\int_0^1 (1+a+a^2\pi^2x^{2a})\sin(\pi x^a)dx = a\pi\tag3$$ Parts (1) and (2) are ok, using integration by parts on $$\int_0^1 \sin(\pi x^a)dx$$, and $$\int_0^1 \cos(\pi x^a)dx$$, respectively.

To illustrate, in (1)., let $$u = \sin(\pi x^a) \Rightarrow du = a \pi x^{a-1}cos(\pi x^a)dx$$, then with $$dv = 1dx \Rightarrow v = x$$
On integrating by parts, $$\int_0^1 \sin(\pi x^a)dx = [xsin(\pi x^a)]_0^1 -a\pi\int_0^1 x^acos(\pi x^a)dx$$, which gives
$$\int_0^1 \sin(\pi x^a)dx = -a\pi\int_0^1 \ x^acos(\pi x^a)dx$$
and therefore, $$\int_0^1 \sin(\pi x^a)dx +a\pi\int_0^1 \ x^acos(\pi x^a)dx = 0$$\

Same procedure for part (2)., except let $$u = \cos(\pi x^a) \Rightarrow du = -a \pi x^{a-1}sin(\pi x^a)dx$$, then with $$dv = 1dx \Rightarrow v = x$$ Integrating by parts this time leads to the desired result, $$\int_0^1 \cos(\pi x^a)dx$$ - $$a\pi\int_0^1 \ x^asin(\pi x^a)dx = -1$$

To prove (3), I believe the previous results are to be used somehow. I have re written (3) as, $$\int_0^1 (1+a+a^2\pi^2x^{2a})sin(\pi x^a)dx = (1+a)\int_0^1 \sin(\pi x^a)dx + a^2 \pi^2 \int_0^1 x^{2a}\sin(\pi x^a)dx$$, and trying to resolve the integral $$\int_0^1 x^{2a}\sin(\pi x^a)dx$$, with the awkward factor $$x^{2a}$$.

The first thing I tried was let $$u=x^{2a}\Rightarrow du=2ax^{2a-1}dx$$

Then $$dv= sin(\pi x^a)dx \Rightarrow v=\int sin(\pi x^a)dx$$, but because $$v$$ is an integral, and remains to be if using the result from part (1), I can't integrate this by parts, because I get:

$$\int_0^1 x^{2a}\sin(\pi x^a)dx = [x^{2a} \int sin(\pi x^a)dx]_0^1 - \int_0^1 (\int sin(\pi x^a)(2ax^{2a-1})dx)dx$$

I have also tried writing the integral as, $$\int_0^1 x^{2a}\sin(\pi x^a)dx = \int x^a(x^a \sin(\pi x^a))dx$$, then differentiating $$x^a \sin(\pi x^a)$$, using product rule to give, $$du = a \pi x^{2a-1} \cos(\pi x^a)+ax^{a-1} \sin(\pi x^a)$$
Then with $$dv=x^adx \Rightarrow v=\frac {x^{a+1}}{a+1}$$

Integrating by parts gives the result, $$(2a+1)\int x^a(x^a \sin(\pi x^a))dx = -a \pi \int_0^1 x^{3a} \cos(\pi x^a)dx$$

Integrating $$\int_0^1 x^{3a} \cos(\pi x^a)dx$$ by parts by differentiating the product $$x^{2a} \cos(\pi x^a)$$ and integrating $$x^a$$ then brings in a term $$\int_0^1 x^{4a} \sin(\pi x^a) dx$$, which looks to be going the wrong way!

Any fantastic tips to nudge me in the right direction will be greatly appreciated. Thank you.

$$\int_0^1 \sin(\pi x^a)dx=-a\pi\int_0^1 \ x^a\cos(\pi x^a)dx \tag 1$$ and integrate the RHS by parts $$\int_0^1 \sin(\pi x^a)dx=a\pi\left( 1+\int_0^1 a x^a \cos(\pi x^a)dx - a\pi \int_0^1 x^{2a}\sin(\pi x^2)dx\right)\tag2$$ Then, substitute (1) into (2) to arrive at the third result $$\int_0^1 (1+a+a^2\pi^2x^{2a})\sin(\pi x^a)dx = a\pi$$