To evaluate the integral, we can first use the fact that cosine and secant are inverses of each other, so they cancel:

$\displaystyle \int 2te^{3t^2}dt$

Now, we must make the following substitution:

$\displaystyle u=3t^2, du=6tdt$

Rewriting the integral in terms of u and integrating, we get

$\displaystyle \frac{1}{3}\int e^udu=\frac{1}{3}e^u+C$

We used the following rule to integrate:

$\displaystyle \int e^xdx=e^x+C$

Finally, replace u with our original x term:

$\displaystyle \frac{1}{3}e^{3t^2}+C$

To integrate this problem, you have to use "u" substitution. Assign $\displaystyle x^2-2=u$. Then, find du, which is 2x. That works out since we can then replace the other x in the original problem. We will have to offset the 2 though: $\displaystyle du=2x; \frac{1}{2}du=x$. Now plug in all the parts: $\displaystyle \frac{1}{2}\int u^{\frac{1}{3}}du$. Now, integrate as normal, remembering to raise the exponent by 1 and then also putting that result on the bottom: $\displaystyle \frac{1}{2}(\frac{3}{4})(u^{\frac{4}{3}})$. Simplify, add a C because it is an indefinite integral, and substitute your original expression back in: $\displaystyle \frac{3}{2}(x^2-2)^{\frac{4}{3}}+C$.

To integrate this problem, use "u" substitution. Assign $\displaystyle u=x^3-1$, $\displaystyle du=3x^2; \frac{1}{3}du=x^2$. Substitute everything in so you can integrate: $\displaystyle \frac{1}{3}\int \frac{1}{u}du$. Recall that when there is a single variable on the denominator, the integral is ln of that term. Therefore, after integrating, you get $\displaystyle \frac{1}{3}ln\left | u \right |$. Sub back in your original expression and add C because it is an indefinite integral: $\displaystyle \frac{1}{3}ln\left | x^3-1 \right |+C$.

$\displaystyle \int^1_0 x^2(x^3+4)^3dx$

Make the substitution:

$\displaystyle u=x^3+4$

$\displaystyle du=3x^2dx$

$\displaystyle \frac{du}{3}=x^2dx$

$\displaystyle \int^1_0 \frac{u^3}{3}du=\frac{1}{12}u^4\left.\begin{matrix} \\ \end{matrix}\right|^1_0=\frac{1}{12}(x^3+4)^4 \left.\begin{matrix} \\ \end{matrix}\right|^1_0=\frac{123}{4}$

Substitute $\displaystyle u=10x, du = 10 dx$:

$\displaystyle \int e^{u}du$, 

which is equal to

$\displaystyle e^{u}+ C$.

Replace u with 10x:

$\displaystyle e^{10x}+C$.

Substitute $\displaystyle u=cosx, du=-sinxdx$:

$\displaystyle \int- u^3du=-\frac{u^4}{4} + C$.

Replace $\displaystyle u=cos x$:

$\displaystyle \frac{-cos^4x}{4} + C$.

Substitute $\displaystyle u=sin x, du=cosx$:

$\displaystyle \int u^3 du=\frac{u^4}{4} +C$.

Replace $\displaystyle u=sin x$:

$\displaystyle \frac{sin^4x}{4} +C$.

Substitue $\displaystyle u=5x, du=5$.

$\displaystyle \int e^udu=e^u+C$.

Replace $\displaystyle u=5x$:

$\displaystyle e^{5x} + C$.

Substitute $\displaystyle u=x+2, du=dx$:

$\displaystyle \int \frac{1}{u^4}du=-\frac{u^{-3}}{3}$.

Replace $\displaystyle u=x+2$:

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

Substitute $\displaystyle u=x^3 + 1, du=3x^2 dx$.

$\displaystyle \int \frac{1}{3}u^{-1}du=\frac{1}{3}ln\left | u \right |+C$.

Replace $\displaystyle u=x^3+1$:

$\displaystyle \frac{1}{3}ln\left | x^3+1\right | + C$.