NIMCET 2014 Question Paper with solutions

We have $\underset{0}{\overset{\mathit{\pi }∕4}{\int }}$ log (1 tanx) dx
let l = $\underset{0}{\overset{\mathit{\pi }∕4}{\int }}$ log (1 + tanx) dx
= $\underset{0}{\overset{\mathit{\pi }∕4}{\int }}$ log $\left[1+\mathit{t}\mathit{a}\mathit{n}\left(\frac{\mathit{\pi }}{4}-\mathit{x}\right)\right]$ dx
= $\underset{0}{\overset{\mathit{\pi }∕4}{\int }}$ log $\left(1+\frac{\mathit{t}\mathit{a}\mathit{n}\mathit{\pi }∕4-\mathit{t}\mathit{a}\mathit{n}\mathit{x}}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{\pi }∕4\mathit{t}\mathit{a}\mathit{n}\mathit{x}}\right)$ dx
= $\underset{0}{\overset{\mathit{\pi }∕4}{\int }}$ log $\left(1+\frac{1-\mathit{t}\mathit{a}\mathit{n}\mathit{x}}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{x}}\right)$ dx
= $\underset{0}{\overset{\mathit{\pi }∕4}{\int }}$ log $\left(\frac{1+\mathit{t}\mathit{a}\mathit{n}\mathit{x}+1-\mathit{t}\mathit{a}\mathit{n}\mathit{x}}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{x}}\right)$ dx
= $\underset{0}{\overset{\mathit{\pi }∕4}{\int }}$ log $\left(\frac{2}{1+\mathit{t}\mathit{a}\mathit{n}\mathit{x}}\right)$ dx
= $\underset{0}{\overset{\mathit{\pi }∕4}{\int }}$ log 2 dx - $\underset{0}{\overset{\mathit{\pi }∕4}{\int }}$ log (1 + tanx) dx
⇒ 2l = log ${\left[\mathit{x}\right]}_0^{\mathit{\pi }∕4}$
⇒ l = log 2 [π/4]
∴ l = $\frac{\mathit{\pi }}{8}$ log 2