The derivative of $\displaystyle e^x=e^x$. Therefore, the antiderivative of $\displaystyle e^x$ is equal to itself. 

$\displaystyle f(x) = x^{9} + tan(x) - x^{3}cos(x)$ and

$\displaystyle f(-x) = (-x)^{9} + tan(-x) - (-x)^{3}cos(-x) =$

$\displaystyle -x^{9} - tan(x) + x^{3}(cos(x)) =$ 

$\displaystyle -[x^{9} + tan(x) - x^{3}(cos(x))] = -f(x)$

Recall that $\displaystyle tan(x)$ is an odd function and $\displaystyle cos(x)$ is an even function.

Thus, since $\displaystyle f(x)$ is an odd function, the integral of this function from $\displaystyle -4$ to $\displaystyle 4$ will be zero.

To approach this problem, first rewrite the integral expression as shown below:

$\displaystyle \int sin^{4}(x)cos^{2}(x)cos(x)dx$.

Then, recognize that $\displaystyle cos^2(x) = 1 - sin^2(x)$, and substitute this into the integral expression:

$\displaystyle \int sin^{4}(x)(1-sin^{2}x)cos(x)dx$

Use substitution, letting $\displaystyle u = sin(x)$ and $\displaystyle du = cos(x)dx$.  The integral can then be rewritten as

$\displaystyle \int u^{4}(1-u^{2})du = \int (u^{4} - u^{6})du.$  

Evaluating this integral gives

$\displaystyle \frac{u^{5}}{5} - \frac{u^{7}}{7} + C$.

Finally, substituting $\displaystyle u = sin(x)$ back into this expression gives the final answer:

$\displaystyle \frac{sin^{5}(x)}{5} - \frac{sin^{7}(x)}{7} + C$

(As this is an indefinite integral, $\displaystyle C$ must be included).

$\displaystyle \int_{1}^{e^{100}} \frac{dx}{x} = \ln e^{100} - \ln 1 = 100-0 = 100$ 

This is most easily solved by recognizing that $\displaystyle \sin^{2}x+\cos^{2}x=1$.  

Remember the fundamental theorem of calculus!

$\displaystyle \int_{a}^{b}f(x){\mathrm{d} x}=\int f(b){\mathrm{d} x}-\int f(a){\mathrm{d} x}$

Since our $\displaystyle f(x)=\sin(2x)$, we can't use the power rule. Instead we end up with: 

$\displaystyle \int f(x){\mathrm{d} x}=\int \sin(2x){\mathrm{d} x}=\frac{-\cos(2x)}{2}+c$

Remember to include the $\displaystyle + c$ for any anti-derivative or integral taken!

Now we can plug that equation into our FToC equation:

$\displaystyle \int_{a}^{b}f(x){\mathrm{d} x}=(\frac{-\cos(2b)}{2}+c)-(\frac{-\cos(2a)}{2}+c)$

Notice that the c's cancel out. Plug in the given values for a and b and solve:

$\displaystyle \int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(\frac{-\cos(4\pi)}{2})-(\frac{-\cos(2\pi)}{2})$

$\displaystyle \int_{\pi}^{2\pi}f(x){\mathrm{d} x}=(\frac{-1}{2})-(\frac{-1}{2})$

$\displaystyle \int_{\pi}^{2\pi}f(x){\mathrm{d} x}=0$

The integral of $\displaystyle \frac{1}{x}$ is $\displaystyle \ln \left | x \right |+C$.  The constant 3 is simply multiplied by the integral.  

To integrate $\displaystyle \cos(x)\sin(x)$, we need to get the two equations in terms of each other. We are going to use "u-substitution" to create a new variable, $\displaystyle u$, which will equal $\displaystyle \cos(x)$.

Now, if $\displaystyle u=\cos(x)$, then 

$\displaystyle \frac{\mathrm{d}u }{\mathrm{d} x}=\frac{\mathrm{d} }{\mathrm{d} x}\cos(x)=-sin(x)$ 

Multiply both sides by $\displaystyle \mathrm d {x}$ to get the more familiar: 

$\displaystyle {\mathrm{d} u}=-\sin(x)\mathrm{d}x$

Note that our $\displaystyle \mathrm d{u}=-\sin(x)$, and our original equation was asking for a positive $\displaystyle \sin(x)$.

That means if we want $\displaystyle \int\cos(x)\sin(x)$ in terms of $\displaystyle u$, it looks like this:

$\displaystyle \int u\cdot (-\mathrm d{u}})$

Bring the negative sign to the outside:

$\displaystyle -\int u\mathrm d{u}$.

We can use the power rule to find the integral of $\displaystyle u$:

$\displaystyle -(\frac{1}{2}u^2+c)$

Since we said that $\displaystyle u=\cos(x)$, we can plug that back into the equation to get our answer:

$\displaystyle -(\frac{1}{2}\cos^2(x)+c)$

In this case we have a rational function as $\displaystyle \frac{N(x)}{D(x)}$, where

$\displaystyle N(x)=1$

and

 $\displaystyle D(x)=\frac{1}{x^2-16}$

$\displaystyle D(x)$ can be written as a product of linear factors:

$\displaystyle \frac{1}{x^2-16}=\frac{1}{(x-4)(x+4)}\equiv \frac{A}{x-4}+\frac{B}{x+4}$

It is assumed that A and B are certain constants to be evaluated. Denominators can be cleared by multiplying both sides by (x - 4)(x + 4). So we get:

$\displaystyle 1=A(x+4)+B(x-4)$

First we substitute x = -4 into the produced equation:

$\displaystyle 1=-8B\Rightarrow B=-\frac{1}{8}$

Then we substitute x = 4 into the equation:

$\displaystyle 1=8A\Rightarrow A=\frac{1}{8}$

Thus:

$\displaystyle \frac{1}{(x-4)(x+4)}=\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4}$

Hence:

$\displaystyle \int \frac{dx}{x^2-16}=\int(\frac{1}{8}\frac{1}{x-4}-\frac{1}{8}\frac{1}{x+4})dx$

$\displaystyle =\frac{1}{8}ln\left | x-4 \right |-\frac{1}{8}ln\left |x+4 \right |+c$

$\displaystyle =\frac{1}{8}(ln\left | x-4 \right |-ln\left |x+4 \right |)+c$

$\displaystyle =\frac{1}{8}ln\left | \frac{x-4}{x+4}\right |+c$

$\displaystyle \log_{10} x = \frac{\ln x}{\ln 10}$, so this can be rewritten as 

$\displaystyle \dpi{120}\int\frac{\sqrt{\log_{10} x}}{x}\; \; dx$

$\displaystyle \dpi{120}=\int\frac{\sqrt{\ln x}}{x \sqrt{\ln 10} }\; \; dx$

$\displaystyle \dpi{120}= \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

Set $\displaystyle u = \ln x$. Then

$\displaystyle \frac{\mathrm{d} u}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} \ln x = \frac{1}{x}$

and 

$\displaystyle du = \frac{dx}{x}$

Substitute:

$\displaystyle \dpi{120} \frac{1}{\sqrt{\ln 10}}\int\frac{\sqrt{\ln x}}{x }\; \; dx$

$\displaystyle \dpi{120} =\frac{1}{\sqrt{\ln 10}} \int\sqrt{u} \; du$

$\displaystyle \dpi{120} =\frac{1}{\sqrt{\ln 10}} \int u ^{\frac{1}{2}} \; du$

$\displaystyle \dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{u^{\frac{3}{2}}}{{\frac{3}{2}}} + C \right )$

$\displaystyle \dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2u \sqrt{u}} {3} + C \right )$

The outer factor can be absorbed into the constant, and we can substitute back:

$\displaystyle \dpi{120} = \frac{1}{\sqrt{\ln 10}} \left (\frac{ 2 \ln x \cdot \sqrt{ \ln x}} {3} \right )+ C$

$\displaystyle = \frac{ 2 \sqrt{ \ln x} \cdot \ln x} {3 \cdot \sqrt{\ln 10}} + C$

$\displaystyle = \frac{ 2 \sqrt{ \log_{10} x} \cdot \ln x} {3 } + C$