To integrate, we must make the following subsitution:

\(\displaystyle u=(x+5)^{\frac{1}{2}}, du=\frac{1}{2}(x+5)^{-\frac{1}{2}}dx\)

Now, rewrite the integral, and integrate:

\(\displaystyle 2\int \frac{du}{u}=2\ln \left | u\right |+C\)

The following rule was used for the integration:

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

Finally, replace u with our original term:

\(\displaystyle 2\ln(\sqrt{x+5})+C\)

Note that the absolute value sign went away, because the square root is always positive.

To integrate, we must split the integral in two:

\(\displaystyle \int e^xdx+\int (x\cos(x^2+3))dx\)

The first integral is:

\(\displaystyle \int e^xdx=e^x+C\)

and is identical to the rule used to integrate.

The second integral is performed using the following substitution:

\(\displaystyle u=x^2+3, du=2xdx\)

Now, rewrite the integral and integrate:

\(\displaystyle \frac{1}{2}\int \cos(u)du=\frac{1}{2}\sin(u)+C\)

We used the following rule to integrate:

\(\displaystyle \int \cos(x)dx=\sin(x)+C\)

Finally, replace u with the original term:

\(\displaystyle e^x+\frac{1}{2}\sin(x^2+3)+C\)

The integral is equal to
\(\displaystyle 15\ln\left | \sec(x)+\tan(x)\right |+\frac{x^3}{3}+C\)

and was found using the following rules:

\(\displaystyle \int \sec(x)=\ln\left | \sec(x)+\tan(x)\right |+C\), 

\(\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C\).

The integral is equal to

\(\displaystyle \frac{x^3}{3}+2\sin(x)+C\)

and we used the following rules for integration:

\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\), 

\(\displaystyle \int \cos(x)dx=\sin(x)+C\)

The integral was performed using the following rules:

\(\displaystyle \int x^n dx=\frac{x^{n+1}}{n+1}+C\), 

\(\displaystyle \int \sec^2(x)dx=\tan(x)+C\), 

\(\displaystyle \int \frac{dx}{x^2+1}=\arctan(x)+C\)

Applying the above rules we get the following.

\(\displaystyle \frac{8}{5}x^5+\tan(x)+\arctan(x)+C\)

To integrate, we must make the following substitution:

\(\displaystyle u=3x^2, du=6xdx\)

Now, rewrite the integral and integrate:

\(\displaystyle \frac{1}{6}\int e^udu=\frac{1}{6}e^u+C\)

The following rule was used to integrate:

\(\displaystyle \int e^xdx=e^x+C\)

Finally, replace u wiht our original term:

\(\displaystyle \frac{1}{6}e^{3x^2}+C\)

The integral can be split into two seperate ones:

\(\displaystyle \int e^xdx+\int x\cos(x^2)dx\)

The first integral is simple:

\(\displaystyle \int e^xdx=e^x+C\) 

and is identical to the rule used.

The second integral can be performed using the following substitution:

\(\displaystyle u=x^2, du=2xdx\)

Rewrite and integrate:

\(\displaystyle \frac{1}{2} \int \cos(u)du=\frac{1}{2}\sin(u)+C\)

The following rule was used to integrate:

\(\displaystyle \int \cos(x)dx=\sin(x)+C\)

Finally, replace u with the original term and add the integrals together:

\(\displaystyle e^x+\frac{1}{2}\sin(x^2)+C\)

Since we have a piecewise function we will need to set up an integral with two parts. One will \(\displaystyle f(x)=cos(x)\) from \(\displaystyle -\frac{\pi}{2}\) to zero and the other, \(\displaystyle f(x)=e^{2x}\) from zero to two.

Setting up the integral and plugging in the bounds looks like,

\(\displaystyle \\ \int_{-\frac{\pi }{2}}^{2}f(x)dx=\int_{-\frac{\pi }{2}}^{0}f(x)dx+\int_{0}^{2}f(x)dx\\ \\=\int_{-\frac{\pi }{2}}^{0}cos(x)dx+\int_{0}^{2}e^{2x}dx\\ \\=sin(x)]\begin{matrix}-\frac{\pi }{2}, \end{matrix}0+\frac{e^{2x}}{2}]\begin{matrix}0, \end{matrix}2\\ \\=[sin(0)-sin(-\frac{\pi }{2})]+[\frac{e^{4}}{2}-\frac{e^0}{2}]\\ \\=[0-(-1)]+[\frac{e^4}{2}-\frac{1}{2}]\\ \\=1-\frac{e^4-1}{2}\\ \\=\frac{2+e^4-1}{2}\\ \\=\frac{1+e^4}{2}\).

We have a product of two easy functions, so to get our answer we need to use integration by parts. The general formula for this is:

\(\displaystyle \int u dv = uv- \int v du\).

Choose \(\displaystyle u=x^2\) and \(\displaystyle dv=e^x dx\). Then \(\displaystyle du=2xdx\) and \(\displaystyle v=e^x.\) 

So integration by parts tells us that our antiderivative equals

 \(\displaystyle {} x^2e^x-\int 2xe^x dx.\)

Now we must apply integration by parts again, because we still have an integral of a product of two functions. Choose \(\displaystyle u=2x\) and \(\displaystyle dv=e^x dx\) again, so \(\displaystyle du=2dx\) and \(\displaystyle v=e^x\).

Then we have 

\(\displaystyle {} \int 2xe^x dx = 2xe^x-\int2e^x dx.\)

But of course, \(\displaystyle 2e^x\) is its own antiderivative, because we can pull out constants and \(\displaystyle e^x\) is its own anti-derivative.

So combining these two results we get our final answer:

\(\displaystyle {} \int x^2e^x dx = x^2e^x-2xe^x+2e^x+C.\)

The indefinite integral of acceleration is velocity. The indefinite integral of velocity is position. We can write this as: 

\(\displaystyle v(t)=\int a(t)dt\), where \(\displaystyle a(t)\) is acceleration and \(\displaystyle v(t)\) is velocity

\(\displaystyle p(t)=\int v(t)dt\), where \(\displaystyle p(t)\) is position

\(\displaystyle p(t)=\int \left[\int a(t)dt\right]dt\)