Finding a closed form of an integral: $\int_0^k\ln(a\sin^2(x)+(a+b)\cos^2(x))dx$ 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?
 A: 
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 $\alpha\space\wedge\space\beta\in\mathbb{R}_{>0}$.


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