A calculus problem by Kishore S. Shenoy

Relevant wiki: Integration Tricks

Take $I(x) = \int_0^{\pi/2} \dfrac{\ln\left(1+x\sin^2\theta\right)}{\sin^2\theta}\, d\theta$

Differentiating w.r.t $x$ , $\begin{aligned}I'(x) &= \int_0^{\pi/2} \dfrac1{\left(1+x\sin^2\theta\right)}\, d\theta &\color{#3D99F6}{\text{Dividing top and bottom by } \cos^2\theta}\\ &=\int_0^{\pi/2} \dfrac{\sec^2\theta}{\sec^2\theta+x\tan^2\theta}\, d\theta &\color{#3D99F6}{t =\tan\theta\Rightarrow dt = \sec^2\theta\, d\theta}\\ &=\int_0^\infty \dfrac1{1+t^2 + xt^2}\,dt\\ &=\dfrac1{\sqrt{1+x}}\left[\tan^{-1} \left(t\sqrt{x+1}\right)\right]_{t=0}^{t\to\infty}\\ &= \dfrac1{\sqrt{1+x}}\dfrac\pi2\end{aligned}$

Since $I(0) = 0$ , $\begin{aligned}I(x) &= \int_0^x I'(x)\, dx\\&=\pi\int_0^x\dfrac1{2\sqrt{1+x}}\, dx \\&=\pi\left(\sqrt{1+x} - 1\right)\end{aligned}$

So, we have $I(x) = \pi\left(\sqrt{1+x} - 1\right)$

Integrating, $\int_0^{100}I(x)\, dx = \pi\int_0^{100} \left(\sqrt{1+x} - 1\right)\, dx = \boxed{\dfrac{2\pi}3\left(101\sqrt{101} - 151\right)}$

So, $A+B+C+D = 2+3+101+151 = 257$

$I = \int_{0}^{\frac{\pi}{2}} \frac{\ln\left(1 + x\sin^2\theta \right)}{\sin^2 \theta} d\theta =$ Integrating by parts $= \left[- \frac{\ln(\left(1 + x\sin^2\theta \right)}{\tan \theta}\right]_0^{\pi/2} + \int_0^{\pi/2} \frac{2x\cos^2\theta}{1 + x\sin^2\theta} d\theta$ $\displaystyle \color{#D61F06}{ \lim_{\theta\to \pi/2} - \frac{\ln(\left(1 + x\sin^2\theta \right)}{\tan \theta} = 0} \text{, remember x } \in [0,100]$

$\displaystyle \color{#D61F06}{ \lim_{\theta\to 0} \frac{\ln\left(1 + x\sin^2\theta \right)}{\tan \theta} = \frac{0}{0} = \displaystyle \lim_{\theta\to 0} \frac{\frac{2x\sin\theta \cos\theta}{1 + x\sin^2\theta}}{1 + \tan^2\theta} = \lim_{\theta\to 0} \frac{2x\sin\theta \cos^3\theta}{1 + x\sin^2\theta} = 0}$ .

Therefore, $I = \int_0^{\pi/2} \frac{2x\cos^2\theta}{1 + x\sin^2\theta} d\theta = \int_0^{\pi/2} \frac{2x}{\frac{1}{\cos^2\theta} + x\tan^2\theta} d\theta = \int_0^{\pi/2} \frac{2x}{1 + (x + 1)\tan^2\theta} d\theta$ Now, we make the variable change $t = \tan\theta \Rightarrow dt = (1 + \tan^2\theta) d\theta \Rightarrow$ $I = \int_0^{\infty} \frac{2x}{1 + (x + 1)t^2}\cdot \frac{1}{1 + t^2} dt = \int _0^{\infty} \frac{2x + 2}{1 + (x + 1)t^2} - \frac{2}{1 + t^2} dt =$ $\left[2\sqrt{x + 1}\text{ arctan }(t\cdot \sqrt{x +1}) - 2\text{ arctan t}\right]_0^{\infty} = \pi \cdot(\sqrt{x + 1} - 1)$

Hence, $\int_0^{100} \int_0^{\pi/2} \frac{\ln\left( 1 + x \sin^2\theta\right)}{\sin^2\theta} d\theta \space dx = \int_0^{100} \pi(\sqrt{x +1} -1) dx = \pi\frac{2}{3}\left(101\sqrt{101} - 151\right)...$

Calculator Wolfram Alpha (BETA)
A=2, B=3, C=101, D=151.
A+B+C+D=257.