Finding a closed form of an integral: $\int_0^k\ln(a\sin^2(x)+(a+b)\cos^2(x))dx$

Question

I am trying to find a closed form for the following integral:

$$\int_0^k\ln(a\sin^2(x)+(a+b)\cos^2(x))dx$$

And I know that $a>0$, $b\ge0$ and $k=(\pi(1+n))/2$ where $n$ is a natural number.

How can I approach this problem?

Answer 1

Assignment:

Find a closed form for the following integral:

$$\mathcal{I}_\text{k}\left(\alpha,\beta\right):=\int_0^\text{k}\ln\left(\alpha\sin^2\left(x\right)+\left(\alpha+\beta\right)\cos^2\left(x\right)\right)\space\text{d}x$$

Where $\text{k}:=\frac{\pi\left(1+\text{n}\right)}{2}$ for $\text{n}\in\mathbb{N}$ and $\alpha\space\wedge\space\beta\in\mathbb{R}_{>0}$.

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Solution:

First, let's recall that:

$$\alpha\sin^2\left(x\right)+\left(\alpha+\beta\right)\cos^2\left(x\right)=\alpha\sin^2\left(x\right)+\alpha\cos^2\left(x\right)+\beta\cos^2\left(x\right)=$$ $$\alpha\left(\underbrace{\sin^2\left(x\right)+\cos^2\left(x\right)}_{=\space1}\right)+\beta\cos^2\left(x\right)=\alpha+\beta\cos^2\left(x\right)\tag1$$

So, we have:

$$\mathcal{I}_\text{k}\left(\alpha,\beta\right)=\int_0^\text{k}\ln\left(\alpha+\beta\cos^2\left(x\right)\right)\space\text{d}x\tag2$$

Now, let's find:

$$\frac{\partial\mathcal{I}_\text{k}\left(\alpha,\beta\right)}{\partial\beta}=\frac{\partial}{\partial\beta}\left\{\int_0^\text{k}\ln\left(\alpha+\beta\cos^2\left(x\right)\right)\space\text{d}x\right\}=$$ $$\int_0^\text{k}\frac{\partial}{\partial\beta}\left(\ln\left(\alpha+\beta\cos^2\left(x\right)\right)\right)\space\text{d}x=\int_0^\text{k}\frac{\cos^2\left(x\right)}{\alpha+\beta\cos^2\left(x\right)}\space\text{d}x\tag3$$

Now, we write:

$$\cos^2\left(x\right)=\frac{\alpha+\beta\cos^2\left(x\right)}{\beta}-\frac{\alpha}{\beta}\tag4$$

Using the linearity of the integral we can split it up, so:

$$\frac{\partial\mathcal{I}_\text{k}\left(\alpha,\beta\right)}{\partial\beta}=\frac{1}{\beta}\int_0^\text{k}1\space\text{d}x-\frac{\alpha}{\beta}\int_0^\text{k}\frac{1}{\alpha+\beta\cos^2\left(x\right)}\space\text{d}x=$$ $$\frac{1}{\beta}\cdot\left[x\right]_0^\text{k}-\frac{\alpha}{\beta}\int_0^\text{k}\frac{1}{\alpha+\beta\cos^2\left(x\right)}\space\text{d}x=\frac{\text{k}}{\beta}-\frac{\alpha}{\beta}\int_0^\text{k}\frac{1}{\alpha+\beta\cos^2\left(x\right)}\space\text{d}x\tag5$$

Let $\text{u}:=\tan\left(x\right)$, so $\text{d}x=\frac{1}{\sec^2\left(x\right)}\space\text{du}$. The lower bound gives $\text{u}=\tan\left(0\right)=0$ and notice for the upper bound that when $\text{k}=\frac{\pi\left(1+\text{n}\right)}{2}$ for $\text{n}\in\mathbb{N}$ we have that $\text{u}=\tan\left(\text{k}\right)\to\infty$. So:

$$\frac{\partial\mathcal{I}_\text{k}\left(\alpha,\beta\right)}{\partial\beta}=\frac{\text{k}}{\beta}-\frac{\alpha}{\beta}\underbrace{\int_0^\infty\frac{1}{\alpha\left(\text{u}^2+1\right)+\beta}\space\text{du}}_{=\space\text{I}_1}\tag6$$

Let $\text{s}:=\sqrt{\frac{\alpha}{\alpha+\beta}}\cdot\text{u}$, so:

$$\text{I}_1=\frac{1}{\sqrt{\alpha\left(\alpha+\beta\right)}}\int_0^\infty\frac{1}{\text{s}^2+1}\space\text{ds}=\frac{1}{\sqrt{\alpha\left(\alpha+\beta\right)}}\cdot\lim_{\text{n}\to\infty}\left[\arctan\left(\text{s}\right)\right]_0^\text{n}=$$ $$\frac{1}{\sqrt{\alpha\left(\alpha+\beta\right)}}\cdot\left(\lim_{\text{n}\to\infty}\arctan\left(\text{n}\right)-\arctan\left(0\right)\right)=\frac{1}{\sqrt{\alpha\left(\alpha+\beta\right)}}\cdot\frac{\pi}{2}\tag7$$

So, we have:

$$\frac{\partial\mathcal{I}_\text{k}\left(\alpha,\beta\right)}{\partial\beta}=\frac{\text{k}}{\beta}-\frac{\alpha}{\beta}\cdot\frac{1}{\sqrt{\alpha\left(\alpha+\beta\right)}}\cdot\frac{\pi}{2}=\frac{\text{k}}{\beta}-\frac{\pi}{2}\cdot\frac{\sqrt{\alpha}}{\beta}\cdot\frac{1}{\sqrt{\alpha+\beta}}\tag8$$

Now, we must find:

$$\mathcal{I}_\text{k}\left(\alpha,\beta\right)=\int\frac{\partial\mathcal{I}_\text{k}\left(\alpha,\beta\right)}{\partial\beta}\space\text{d}\beta\tag9$$