Compute the following integral $$\int_{0}^{\frac{\pi}{2}} \frac{\mathrm{~d}x}{(\sqrt{\sin x} + \sqrt{\cos x})^{4}}.$$

A continuous function $f ( x )$ defined on $x > 0$ satisfies $$2 f ( x ) + \frac { 1 } { x ^ { 2 } } f \left( \frac { 1 } { x } \right) = \frac { 1 } { x } + \frac { 1 } { x ^ { 2 } }$$ for all positive $x$. What is the value of $\int _ { \frac { 1 } { 2 } } ^ { 2 } f ( x ) d x$? [4 points]
(1) $\frac { \ln 2 } { 3 } + \frac { 1 } { 2 }$
(2) $\frac { 2 \ln 2 } { 3 } + \frac { 1 } { 2 }$
(3) $\frac { \ln 2 } { 3 } + 1$
(4) $\frac { 2 \ln 2 } { 3 } + 1$
(5) $\frac { 2 \ln 2 } { 3 } + \frac { 3 } { 2 }$

Evaluate $\displaystyle I = \int_{1/2014}^{2014} \frac{\tan^{-1} x}{x}\, dx$.
(A) $\dfrac{\pi}{2} \log(2014)$ (B) $\pi \log(2014)$ (C) $2\pi \log(2014)$ (D) $\dfrac{\pi}{4} \log(2014)$

The value of the integral $$\int _ { \pi / 2 } ^ { 5 \pi / 2 } \frac { e ^ { \tan ^ { - 1 } ( \sin x ) } } { e ^ { \tan ^ { - 1 } ( \sin x ) } + e ^ { \tan ^ { - 1 } ( \cos x ) } } d x$$ equals
(A) 1
(B) $\pi$
(C) $e$
(D) none of these

The value of the integral $$\int _ { \pi / 2 } ^ { 5 \pi / 2 } \frac { e ^ { \tan ^ { - 1 } ( \sin x ) } } { e ^ { \tan ^ { - 1 } ( \sin x ) } + e ^ { \tan ^ { - 1 } ( \cos x ) } } d x$$ equals
(A) 1
(B) $\pi$
(C) $e$
(D) none of these

If
$$\alpha = \int _ { \frac { 1 } { 2 } } ^ { 2 } \frac { \tan ^ { - 1 } x } { 2 x ^ { 2 } - 3 x + 2 } d x$$
then the value of $\sqrt { 7 } \tan \left( \frac { 2 \alpha \sqrt { 7 } } { \pi } \right)$ is $\_\_\_\_$. (Here, the inverse trigonometric function $\tan ^ { - 1 } x$ assumes values in $\left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$.)

If $\int _ { 0 } ^ { 1 } \frac { 1 } { \left( 5 + 2 x - 2 x ^ { 2 } \right) \left( 1 + e ^ { ( 2 - 4 x ) } \right) } d x = \frac { 1 } { \alpha } \log _ { e } \left( \frac { \alpha + 1 } { \beta } \right) , \alpha , \beta > 0$, then $\alpha ^ { 4 } - \beta ^ { 4 }$ is equal to
(1) 19
(2) $- 21$
(3) 0
(4) 21

If $( a , b )$ be the orthocentre of the triangle whose vertices are $( 1,2 ) , ( 2,3 )$ and $( 3,1 )$, and $I _ { 1 } = \int _ { \mathrm { a } } ^ { \mathrm { b } } \mathrm { x } \sin \left( 4 \mathrm { x } - \mathrm { x } ^ { 2 } \right) \mathrm { dx } , \mathrm { I } _ { 2 } = \int _ { \mathrm { a } } ^ { \mathrm { b } } \sin \left( 4 \mathrm { x } - \mathrm { x } ^ { 2 } \right) \mathrm { dx }$, then $36 \frac { I _ { 1 } } { I _ { 2 } }$ is equal to:
(1) 72
(2) 88
(3) 80
(4) 66