No calculation is necessary here. 

If $\displaystyle f (x) = \sin\left ( x^{3} \right )$, then

$\displaystyle f (-x) = \sin\left [ \left (- x \right ) ^{3} \right ] = \sin \left (- x ^{3} \right )$.

Since a sine is an odd function, 

$\displaystyle f (-x) = \sin \left (- x ^{3} \right ) = - \sin \left ( x ^{3} \right ) = - f (x)$.

This makes $\displaystyle f (x) = \sin\left ( x^{3} \right )$ an odd function, and for any $\displaystyle A$,

$\displaystyle \int_{-A}^{A} \sin\left ( x^{3} \right ) \; \mathrm{d} x = 0$.

This integral meets the criterion.

$\displaystyle \int x ^{n} \;\mathrm{d} x = \frac{ x ^{n+ 1 }}{n+1}$,

so

$\displaystyle \int_{-1}^{1} x ^{555}\; \mathrm{d} x$

$\displaystyle =\frac{ x ^{555+ 1 } }{555+ 1} \left| \begin{matrix} 1 \\ -1 \end{matrix}\right.$

$\displaystyle =\frac{ x ^{556 } }{556} \left| \begin{matrix} 1 \\ -1 \end{matrix}\right.$

$\displaystyle =\frac{ 1 ^{556 } }{556} - \frac{ (-1) ^{556 } }{556} = \frac{ 1 }{556} - \frac{ 1 }{556} =0$

Alternatively, any monomial of the form $\displaystyle f(x) = x ^ {2N+ 1}$ for an integer $\displaystyle N$ is an odd function. Since the two bounds of integration are each other's opposite, the definite integral is 0.

If $\displaystyle f(x)$ is an odd function - that is, if $\displaystyle f(-c) = -f(c)$ for all real $\displaystyle c$ - then $\displaystyle \int_{-A}^{A} f(x) \; \mathrm{d}x = 0$ for all real $\displaystyle A$, including $\displaystyle A = 1$.

If $\displaystyle f(x)$ is positive on the interval $\displaystyle [-1,1]$, then $\displaystyle \int_{-1}^{1} f(x) \; \mathrm{d}x > 0$.

We can sleuth out the correct answer by demonstrating that four of these functions are odd, and the fifth is always nonnegative and sometimes positive on the interval.

$\displaystyle f(x) = x^{2} \tan {x} \cos{2x}$

$\displaystyle f(-x) =\left ( -x \right ) ^{2} \tan {\left ( -x \right )} \cos{2\left ( -x \right )}$

$\displaystyle =\left ( -x \right ) ^{2} \tan {\left ( -x \right )} \cos{\left ( -2x \right )}$

$\displaystyle =x^{2} \left (-\tan {x} \right ) \cos{2x}$

$\displaystyle =-x^{2} \tan {x} \cos{2x} = -f(x)$

making this function odd.

$\displaystyle f(x) = x^{3} \sin^{2} {x}$

$\displaystyle f(-x) = (-x) ^{3} \sin^{2} {(-x) }$

$\displaystyle = (-x) ^{3}\left ( -\sin {x } \right )^{2}$

$\displaystyle = -x^{3} \sin ^{2} {x }= -f(x)$

making this function odd.

$\displaystyle f(x) = x^{2} \left ( \sin {x} + \tan {x} \right )$

$\displaystyle f(-x) = (-x)^{2} \left ( \sin {(-x)} + \tan {(-x)} \right )$

$\displaystyle =x^{2} \left ( -\sin {x} - \tan {x} \right )$

$\displaystyle =- x^{2} \left ( \sin {x} + \tan {x} \right ) = - f(x)$

making this function odd.

$\displaystyle f(x)=e^{3x}-e^{-3x}$

$\displaystyle f(-x)=e^{3(-x)}-e^{-3(-x)}$

$\displaystyle =e^{-3x}-e^{3x}$

$\displaystyle =-\left ( e^{3x} - e^{-3x} \right ) = - f(x)$

making this function odd.

However, 

$\displaystyle f(x) = x ^{2} (2^x + 2^{-x})$ is nonnegative for all real values of $\displaystyle x$:

$\displaystyle x^{2}, 2^x, 2^ {-x }\geq 0$, so  $\displaystyle f(x) = x ^{2} (2^x + 2^{-x}) \geq 0$ on $\displaystyle [-1,1]$. Also, since $\displaystyle f(x) = x ^{2} (2^x + 2^{-x}) > 0$ for at least one real $\displaystyle c \in [-1,1]$:

$\displaystyle f(1) = 1 ^{2} (2^1 + 2^{-1}) = 1\left ( 2 + \frac{1}{2} \right )= 2\frac{1}{2}$

Then $\displaystyle \int_{-1}^{1} f(x) \; \mathrm{d}x > 0$.

$\displaystyle f(x) = x ^{2} (2^x + 2^{-x})$ is the correct choice.

$\displaystyle x ^2- 3 > 0$ when $\displaystyle x > \sqrt{3}$ or $\displaystyle x < - \sqrt{3}$ and $\displaystyle x ^2- 3 < 0$ when $\displaystyle - \sqrt{3}< x < \sqrt{3}$.

The function

$\displaystyle f(x)=|x ^2- 3 |$

can then be rewritten as 

$\displaystyle f(x)= \left\{\begin{matrix} x^2 - 3, & x > \sqrt{3} \\ -x^2 + 3, &-\sqrt{3} \leq x \leq \sqrt{3} \\x^2 - 3, & x < -\sqrt{3} \end{matrix}\right.$.

Therefore,

$\displaystyle \int_{0}^{4} |x ^2- 3| \; \mathrm{d}x$

$\displaystyle = \int_{0}^{4}f(x) \; \mathrm{d}x$

$\displaystyle = \int_{0}^{\sqrt{3}}f(x) \; \mathrm{d}x + \int_{\sqrt{3}}^{4}f(x) \; \mathrm{d}x$

$\displaystyle = \int_{0}^{\sqrt{3}}( -x^2+3) \; \mathrm{d}x + \int_{\sqrt{3}}^{4}( x^2-3)\; \mathrm{d}x$

$\displaystyle =\left.\begin{matrix} -\frac{x^3}{3}+ 3x \\ \\ \end{matrix}\right|\begin{matrix} \sqrt{3}\\ \\ 0 \end{matrix}+ \left.\begin{matrix} \frac{x^3}{3}- 3x \\ \\ \end{matrix}\right|\begin{matrix} 4\\ \\ \sqrt{3} \end{matrix}$

$\displaystyle =\left ( -\frac{(\sqrt{3}) ^3}{3}+ 3 (\sqrt{3}) \right ) - \left ( -\frac{0^3}{3}+ 3\cdot 0 \right ) + \left ( \frac{4^3}{3}- 3 \cdot 4 \right ) - \left ( \frac{(\sqrt{3})^3}{3}- 3(\sqrt{3}) \right )$

$\displaystyle =\left ( - \sqrt{3} + 3 \sqrt{3} \right ) - 0 + \left ( \frac{64}{3}- 12 \right ) - \left ( \sqrt{3} - 3 \sqrt{3}\right )$

$\displaystyle = 2 \sqrt{3} + \frac{28}{3} - \left ( -2 \sqrt{3} \right )$

$\displaystyle = 4 \sqrt{3} + \frac{28}{3} = \frac{28 + 12\sqrt{3}}{3}$

The definite integral $\displaystyle \int_{\pi}^{5\pi} sin(\theta)-cos(\theta) \:d\theta$ can be integrated as is:

$\displaystyle [-cos(\theta)-sin(\theta)]_{\pi}^{5\pi}$

Evaluate the bounds.  Be aware of the sign changes.

$\displaystyle =[-cos(5\pi)-sin(5\pi)]-[-cos(\pi)-sin(\pi)]$

$\displaystyle =[-(-1)-0]-[-(-1)-0]=[1]-[1]=0$

Be careful, we are integrating with the respect to y, not x.  This means that the integrand itself is treated as a constant.

$\displaystyle \int_{1}^{6}(3x+4)\: dy$

Since we just have a constant, to integrate means to increase the value of the variable y. Thus we get the following,

$\displaystyle (3x+4)\int_1^6dy=(3x+4)\frac{y^{0+1}}{0+1}=(3x+4)y$

$\displaystyle = (3x+4)y|_{1}^{6}$.

From here we plug in our bounds. We take the upper bound function value and subtract lower bound function value.

$\displaystyle = (3x+4)(6)-(3x+4)(1)$

$\displaystyle =(3x+4)(5)$

$\displaystyle =15x+20$

To help us evalute the integral, we can split up the expression into 3 parts:

$\displaystyle y(x)= \int_{0}^{1}(e^{2x}+x+2) dx = \int_{0}^{1}e^{2x} dx + \int_{0}^{1}x dx+\int_{0}^{1}2 dx$.

This allows us to evaluate the integral of each of the three parts, sum them up, and then evaluate the summed up parts from 0 to 1.

The first integral is $\displaystyle \int_{0}^{1}e^{2x} dx = \frac{1}{2}e^{2x}$.

The second integral is $\displaystyle \int_{0}^{1}xdx=\frac{1}{2}x^2$.

The third integral is $\displaystyle \int_{0}^{1}2 dx=2x$.

Sum up all these terms in evaluate between 0 and 1.

$\displaystyle \left(\frac{1}{2}e^{2x}+\frac{1}{2}x^2+2x\right)|_0^1$

$\displaystyle \left(\frac{1}{2}e^{2}+\frac{1}{2}+2\right)-\left(\frac{1}2}\right)= \frac{1}{2}e^2 +2$

The general expression for the integral of a sine function is:

$\displaystyle \int sin(ax)=-\frac{1}{a}cos(ax)$

Using this, the integral above now becomes,

$\displaystyle -\frac{3}{2}cos(2x)+x |_{-\frac{\pi}{2}}^{\frac{\pi}{2}}$

$\displaystyle \left(-\frac{3}{2}cos\left(2\left(\frac{\pi}{2}\right)\right)+\frac{\pi}{2} \right)-\left(-\frac{3}{2}cos\left(2\left(-\frac{\pi}{2}\right)\right)-\frac{\pi}{2} \right)$, which simplifies to

$\displaystyle \left(\frac{\pi}{2} \right)-\left(-\frac{\pi}{2} \right)=\pi$

To integrate, we must first make the following subsitution:

$\displaystyle u=(8x+9)^{\frac{1}{2}}, du=4(8x+9)^{-\frac{1}{2}}$

The derivative was found using the following rules:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$, $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Now, rewrite the given integral, change the bounds in terms of $\displaystyle u$ (by plugging in the upper and lower bounds into the equation for $\displaystyle u$ in terms of $\displaystyle x$), and integrate:

$\displaystyle 4 \int_{3}^{5} \frac{du}{u}=4(\ln\left | 5\right |-\ln\left | 3\right |)=4\ln(\frac{5}{3})$

The integral was performed using the following rule:

$\displaystyle \int \frac{dx}{x} = \ln\left | x\right |+C$

To perform the definite integration, simply plug in the upper limit of integration and subtract from the result of plugging in the lower limit of integration, as shown above. Simplifying the result, we get 

$\displaystyle 4\ln(\frac{5}{3})$

In order to calculate the definite integral, we apply the inverse power rule which states

$\displaystyle \int x^n\,dx = \frac{x^{n+1}}{n+1}$

Applying this to the problem in this question term by term we get

$\displaystyle \int_{0}^{9}x+5\,dx = \Big(\frac{x^2}{2}+5x\Big)\Big|_{0}^{9}$

And by the corollary of the Fundamental Theorem of Calculus the definite integral becomes

$\displaystyle =\frac{9^2}{2}+5(9)-\Big(\frac{0^2}{2}+5(0)\Big)$

And so

$\displaystyle \int_{0}^{9}x+5\,dx = 85.5$