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 \(\displaystyle u=x^3+1\):

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