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

7. Integrate
(A) $\int \frac { x ^ { 3 } + 3 x + 2 } { \left( x ^ { 2 } + 1 \right) ^ { 2 } ( x + 1 ) } \mathrm { dx }$.
(B) $\quad \int _ { 0 } ^ { \pi } \frac { e ^ { \cos x } } { e ^ { \cos x } + e ^ { - \cos x } } \mathrm { dx }$. 8. Let $\mathrm { f } ( \mathrm { x } )$ be a continuous function given by
$$f ( x ) = \left\{ \begin{array} { c c } 2 x , & | x | \leq 1 \\ x ^ { 2 } + a x + b , & | x | > 1 \end{array} \right.$$
Find the area of the region in the third quadrant bounded by the curves $x = - 2 y 2$ and $y = \mathrm { f } ( \mathrm { x } )$ lying on the left on the line $8 \mathrm { x } + 1 = 0$. 9. Find the co-ordinates of all the P on the ellipse $\mathrm { x } 2 / \mathrm { a } 2 + \mathrm { y } 2 / \mathrm { b } 2 = 1$, for which the area of the triangle PON is maximum, where O denotes the origin and N , the foot of the perpendicular from O to the tangent at P . 10. A curve passing through the point $( 1,1 )$ has the property that the perpendicular distance of the origin from the normal at any point P of the curve is equal to the distance of P from the x -axis. Determine the equation of the curve. 11. Eight players $\mathrm { P } 1 , \mathrm { P } 2 , \ldots \ldots . \mathrm { P } 8$ play a knock-out tournament. It is known that whenever the players Pi and Pj play, the play Pi will win if $\mathrm { i } < \mathrm { j }$. Assuming that the players are paired at random in each round, what is the probability that the player P4 reaches the final? 12. Let $\vec { u }$ and $\vec { v }$ be unit vectors. If $\vec { w }$ is a vector such that $\vec { w } + ( \vec { w } \times \vec { u } ) = \vec { v }$, then prove that $| ( \vec { u } \times \vec { v } ) \cdot \vec { w } | \leq \frac { 1 } { 2 }$ and that the equality holds if and only if $\vec { u }$ is perpendicular to $\vec { v }$.

14. Evaluate $\int _ { - \pi / 3 } ^ { \pi / 3 } \frac { \pi + 4 x ^ { 3 } } { 2 - \cos \left( | x | + \frac { \pi } { 3 } \right) } d x$.
Sol. $I = \int _ { - \pi / 3 } ^ { \pi / 3 } \frac { \left( \pi + 4 x ^ { 3 } \right) d x } { 2 - \cos \left( | x | + \frac { \pi } { 3 } \right) }$ $2 \mathrm { I } = \int _ { - \pi / 3 } ^ { \pi / 3 } \frac { 2 \pi \mathrm { dx } } { 2 - \cos \left( | \mathrm { x } | + \frac { \pi } { 3 } \right) } = \int _ { 0 } ^ { \pi / 3 } \frac { 2 \pi \mathrm { dx } } { 2 - \cos \left( \mathrm { x } + \frac { \pi } { 3 } \right) }$ $\mathrm { I } = \int _ { \pi / 3 } ^ { 2 \pi / 3 } \frac { 2 \pi \mathrm { dt } } { 2 - \cos \mathrm { t } } \Rightarrow \mathrm { I } = 2 \pi \int _ { \pi / 3 } ^ { 2 \pi / 3 } \frac { \sec ^ { 2 } \frac { \mathrm { t } } { 2 } \mathrm { dt } } { 1 + 3 \tan ^ { 2 } \frac { \mathrm { t } } { 2 } } = 2 \pi \int _ { 1 / \sqrt { 3 } } ^ { \sqrt { 3 } } \frac { 2 \mathrm { dt } } { 1 + 3 \mathrm { t } ^ { 2 } } = \frac { 4 \pi } { 3 } \int _ { 1 / \sqrt { 3 } } ^ { \sqrt { 3 } } \frac { \mathrm { dt } } { \left( \frac { 1 } { \sqrt { 3 } } \right) ^ { 2 } + \mathrm { t } ^ { 2 } }$ $\mathrm { I } = \frac { 4 \pi } { 3 } \sqrt { 3 } \left[ \tan ^ { - 1 } \sqrt { 3 } \mathrm { t } \right] _ { 1 / \sqrt { 3 } } ^ { \sqrt { 3 } } = \frac { 4 \pi } { \sqrt { 3 } } \left[ \tan ^ { - 1 } 3 - \frac { \pi } { 4 } \right] = \frac { 4 \pi } { \sqrt { 3 } } \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$.

14. Evaluate $\int _ { - \pi / 3 } ^ { \pi / 3 } \frac { \pi + 4 x ^ { 3 } } { 2 - \cos \left( | x | + \frac { \pi } { 3 } \right) } d x$.
Sol. $I = \int _ { - \pi / 3 } ^ { \pi / 3 } \frac { \left( \pi + 4 x ^ { 3 } \right) d x } { 2 - \cos \left( | x | + \frac { \pi } { 3 } \right) }$ $2 \mathrm { I } = \int _ { - \pi / 3 } ^ { \pi / 3 } \frac { 2 \pi \mathrm { dx } } { 2 - \cos \left( | \mathrm { x } | + \frac { \pi } { 3 } \right) } = \int _ { 0 } ^ { \pi / 3 } \frac { 2 \pi \mathrm { dx } } { 2 - \cos \left( \mathrm { x } + \frac { \pi } { 3 } \right) }$ $\mathrm { I } = \int _ { \pi / 3 } ^ { 2 \pi / 3 } \frac { 2 \pi \mathrm { dt } } { 2 - \cos \mathrm { t } } \Rightarrow \mathrm { I } = 2 \pi \int _ { \pi / 3 } ^ { 2 \pi / 3 } \frac { \sec ^ { 2 } \frac { \mathrm { t } } { 2 } \mathrm { dt } } { 1 + 3 \tan ^ { 2 } \frac { \mathrm { t } } { 2 } } = 2 \pi \int _ { 1 / \sqrt { 3 } } ^ { \sqrt { 3 } } \frac { 2 \mathrm { dt } } { 1 + 3 \mathrm { t } ^ { 2 } } = \frac { 4 \pi } { 3 } \int _ { 1 / \sqrt { 3 } } ^ { \sqrt { 3 } } \frac { \mathrm { dt } } { \left( \frac { 1 } { \sqrt { 3 } } \right) ^ { 2 } + \mathrm { t } ^ { 2 } }$ $\mathrm { I } = \frac { 4 \pi } { 3 } \sqrt { 3 } \left[ \tan ^ { - 1 } \sqrt { 3 } \mathrm { t } \right] _ { 1 / \sqrt { 3 } } ^ { \sqrt { 3 } } = \frac { 4 \pi } { \sqrt { 3 } } \left[ \tan ^ { - 1 } 3 - \frac { \pi } { 4 } \right] = \frac { 4 \pi } { \sqrt { 3 } } \tan ^ { - 1 } \left( \frac { 1 } { 2 } \right)$.

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

The value of the integral $\int _ { 1 / 2 } ^ { 2 } \frac { \tan ^ { - 1 } x } { x } d x$ is equal to (1) $\pi \log _ { e } 2$ (2) $\frac { 1 } { 2 } \log _ { e } 2$ (3) $\frac { \pi } { 4 } \log _ { \mathrm { e } } 2$ (4) $\frac { \pi } { 2 } \log _ { \mathrm { e } } 2$

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