Evaluate $\int_{0}^{\frac{\pi}{2}} \ln(1+\sin^3 x)\text{d}x$

Question

Here's the integral that I would like to solve. Purely for recreational purposes:

$$I=\int_{0}^{\frac{\pi}{2}} \ln(1+\sin^3x)\text{d}x$$

Here's my shot at it. I would like to stick to this method if possible. Let $I(\alpha)$ be defined as follows:

$$I(\alpha) = \int_{0}^{\frac{\pi}{2}} \ln(1+\alpha\sin^3 x)\text{d}x$$

$$\implies I'(\alpha) = \int_{0}^{\frac{\pi}{2}} \frac{\sin^3x}{1+\alpha \sin^3x}\text{d}x$$

Let $t = \tan(\frac{x}{2})$, and we have

$$I'(\alpha) = \int_{0}^{1}\frac{\frac{8t^3}{(1+t^2)^3}}{1 + \alpha\frac{8t^3}{(1+t^2)^3}}\cdot\frac{2}{1+t^2}\text{d}t$$

I will now get rid of the fraction

$$I'(\alpha) = \int_{0}^{1}\frac{\frac{8t^3}{(1+t^2)^3}}{1 + \alpha\frac{8t^3}{(1+t^2)^3}}\cdot\frac{2}{1+t^2}\cdot\frac{(1+t^2)^3}{(1+t^2)^3}\text{d}t = \int_{0}^{1} \frac{16t^3}{(1+t^2)(t^6 + 3t^4 + 8\alpha t^3 + 3t^2 + 1)}\text{d}t$$

Perform partial fractions (which was so time consuming):

$$I'(\alpha) = \frac{-2}{\alpha}\int_{0}^{1} \left(\frac{(t^2+1)^2}{t^6+3t^4+8\alpha t^3+3t^2+1}-\frac{1}{1+t^2}\right)\text{d}t$$

The second integral is inverse tangent. How would I go about doing the first integral? It's a $4^{th}$-degree polynomial over a $6^{th}$ degree polynomial, but parts of the polynomials look kind of simple.

Answer 1

Not a finished answer, just the beginning of a variation of the $I(\alpha)$ approach from the OP.

Using $$I(\alpha)=\int_{0}^{\pi/2} \log(1+\alpha^3\sin^3x)\,dx$$

Then use $$1+z^3=(1+z)\left(1+\omega z\right) \left(1+\omega^2z\right)$$ where $\omega=e^{2\pi i/3}.$

Then, for $k=0,1,2,$ let$$I_k(\alpha)=\int\log(1+\omega^k\alpha\sin x)\,dx.$$

Then $$\begin{align}I_k’(\alpha)&=\omega^k\int_{0}^{\pi/2}\frac{\sin x}{1+\omega^k\alpha \sin x}\,dx\\ &=\omega^k\int_0^2 \dfrac{\frac{2t}{1+t^2}}{1+\omega^k\alpha\frac{2t}{1+t^2}}\cdot\dfrac2{1+t^2}\,dt\\ &=-\frac{2}{\alpha}\int_0^1\left(\frac{1}{t^2+2\omega^k\alpha t+1}-\frac{1}{1+t^2}\right)\,dt \end{align}$$

Now:

$$\int_0^1\frac{1}{1+t^2}\,dt =\arctan(t)\Bigg\vert_0^1=\frac{\pi}4.$$

So now you need to compute:

$$I_k’(\alpha)-\frac\pi{2\alpha}=-\frac2{\alpha}\int_{0}^1\frac{1} {t^2+2\omega^k\alpha t+1}\,dt$$

The roots of the denominator are $$r_k^{\pm}=-\omega^k\alpha \pm\sqrt{\omega^{2k}\alpha^2-1}$$

And we get the partial fractions for this integrand:

$$2\sqrt{\omega^{2k}\alpha^2-1}\left(\frac1{t-r_k^+}-\frac1{t-r_k^-}\right)$$

So

$$\int_0^1\frac{dt}{1+2\omega^k\alpha t+t^2}\\=2 \sqrt{\omega^{2k}\alpha^2-1} \left(\log(t-r_k^+)-\log(t-r_k^-)\right)\Bigg\vert_0^1\\ =2 \sqrt{\omega^{2k}\alpha^2-1} \left(\log(1-r_k^- )-\log(1-r_k^+)\right) $$

A cursory look indicates it might be tough to take the antiderivative of the resulting expression. Remember, the $r_k^\pm$ are functions of $\alpha.$

When $k=0$ and $\alpha=\cos\theta$ then $r_k^\pm=e^{\pm i\theta}$ and $$-\frac{2}{\alpha}\int_0^1\frac{dt}{1+2\alpha t+t^2} =4i\theta\tan(\theta)$$

But there isn’t any nice way that I can see to get $k=1,2,$ and not seeing much nice cancellation when summing, either.