The integral can be expanded by distributing the exponent.

\(\displaystyle \int3(x\sec(x^3))^2dx=\int3x^2\sec^2(x^3)dx\)

We will make the following substitution:

\(\displaystyle u=x^3\).

Differentiating both sides yields

\(\displaystyle du=3x^2dx\).

We can then substitute the left hand side of each equation into our integral and evaluate it now.

\(\displaystyle \int3x^2\sec^2(x^3)dx=\int\sec^2(u)du=\tan(u)+C.\)

Finally, we substitute the original variable back into the expression:

\(\displaystyle \tan(u)+C=\tan(x^3)+C\).

Use substitution:

\(\displaystyle u = sin(x)\), \(\displaystyle du = cos(x)dx\)

Plug the \(\displaystyle u\) and \(\displaystyle du\) into the regular equation, but no need to worry about the bounds yet:

\(\displaystyle \int udu\) \(\displaystyle = \frac{1}{2}u^{2}\)

Plug \(\displaystyle u\) back into the integrated equation from above and evaluate from \(\displaystyle 0\) to \(\displaystyle \pi\).

\(\displaystyle = \frac{1}{2} (sin(x))^2\)

\(\displaystyle \frac{1}{2} (sin(\pi)^2 - sin(0)^2) = 0\)

Use substitution integration:

\(\displaystyle u = ln(x)\)

\(\displaystyle du=\frac{1}{x}dx\)

\(\displaystyle \int u^2 du=\frac{1}{3}u^3+C=\frac{1}{3}(ln(x))^3+C\)

Use substitution:

\(\displaystyle u = x^2\)

\(\displaystyle du = 2xdx\) 

\(\displaystyle \frac{du}{2} = xdx\)

\(\displaystyle \int \frac{e^{u}}{2}du\)   \(\displaystyle = \frac{e^{u}}{2}\) \(\displaystyle = \frac{e^{x^2}}{2}\)

Substitute \(\displaystyle u\) back in.

To evaluate the integral, we must first perform the following substitution:

\(\displaystyle u=2x, du=2dx\)

Now, rewrite the integral and integrate:

\(\displaystyle \frac{1}{2}\int \frac{du}{\sqrt{16-u^2}}=\frac{1}{2}\arcsin(\frac{u}{4})+C\)

The integration was performed using the following rule:

\(\displaystyle \int \frac{dx}{\sqrt{a^2-x^2}}=\arcsin(\frac{x}{a})+C\)

Finally, replace u with the original term:

\(\displaystyle \frac{1}{2}\arcsin(\frac{x}{2})+C\)

To evaluate the integral, we must make the following substitution:

\(\displaystyle u=\sqrt{x+14}, du=\frac{1}{2}(x+14)^{-\frac{1}{2}}dx\)

Now, rewrite the integral and integrate:

\(\displaystyle \int2du=2u+C\)

The integral was performed using the following rule:
\(\displaystyle \int dx=x+C\)

Finally, replace u with our original term:

\(\displaystyle 2\sqrt{x+14}+C\)

We proceed as follows,

\(\displaystyle \int 10e^{-x}dx\). Start

\(\displaystyle 10\int e^{-x}dx\). Factor out the 10.

Use u-substitution with \(\displaystyle u = -x\), then taking derivates of both sides gives\(\displaystyle du = -1 dx\).

\(\displaystyle 10\int -1e^u du\). Substitute values

\(\displaystyle -10\int e^u du\). Factor out the negative.

\(\displaystyle -10e^u+C\). The antiderivative of \(\displaystyle e^u\) is \(\displaystyle e^u\). Don't forget \(\displaystyle +C\).

\(\displaystyle -10e^{-x}+C\). Substitute \(\displaystyle x\) back.

We evaluate the integral as follows,

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

Use u-substitution, let \(\displaystyle u=x^2\), then taking derivatives of both sides gives \(\displaystyle du = 2xdx\). Divide both sides of this equation by \(\displaystyle 2\), giving \(\displaystyle \frac{1}{2}du = xdx\). Now we can substitute out \(\displaystyle xdx, x^2\), and get

\(\displaystyle \int e^{u} \frac{1}{2}du\)

\(\displaystyle \frac{1}{2}\int e^{u} du\). Factor out the \(\displaystyle 1/2\).

\(\displaystyle \frac{1}{2}e^u+C\). Integrate \(\displaystyle e^u\) and add \(\displaystyle C\).

\(\displaystyle \frac{1}{2}e^{x^2}+C\). Substitute back

We use u-substitution to evaluate this integral.

Let \(\displaystyle u = 1+x\). Subtracting \(\displaystyle 1\) gives \(\displaystyle u-1=x\), and taking derivatives gives\(\displaystyle du =dx\) (We subtract \(\displaystyle 1\) from both sides in order to make the expression under the square root as simple as possible). Then we have

\(\displaystyle \int^{15}_{8}x\sqrt{1+x}dx\). Start

\(\displaystyle \int^{16}_{9}(u-1)\sqrt{u}du\). Make our substitutions. (Make sure you change the bounds of integration too, by plugging \(\displaystyle 8\) and \(\displaystyle 15\) into \(\displaystyle u=1+x\) for \(\displaystyle x\)).

\(\displaystyle \int^{16}_{9}(u-1)u^{\frac{1}{2}}du\).

\(\displaystyle \int^{16}_{9} u^{\frac{3}{2}}-u^{\frac{1}{2}}du\)

\(\displaystyle [\frac{2}{5}u^{\frac{5}{2}}-\frac{2}{3}u^{\frac{3}{2}}]^{16}_9\)

\(\displaystyle [\frac{2}{5}u^{\frac{5}{2}}-\frac{2}{3}u^{\frac{3}{2}}]^{16}_9\)

\(\displaystyle (\frac{2}{5}(16)^{\frac{5}{2}}-\frac{2}{3}(16)^{\frac{3}{2}})-(\frac{2}{5}(9)^{\frac{5}{2}}-\frac{2}{3}(9)^{\frac{3}{2}})\)

\(\displaystyle \frac{2}{5}(1024)-\frac{2}{3}(64)-\frac{2}{5}(243)+\frac{2}{3}(27)\)

\(\displaystyle \frac{2}{5}(781)-\frac{2}{3}(37)\)

\(\displaystyle \frac{1562}{5}-\frac{74}{3}\)

\(\displaystyle \frac{4316}{15}\)

The correct answer is \(\displaystyle \frac{1}{2}(\tan^{-1}(\frac{\pi^2}{4}))^2\).

We proceed as follows-

\(\displaystyle \int^{\pi/2}_{0}\frac{2x\tan^{-1}(x^2)}{1+x^4}dx\). Start

Evaluating this integral relies on the fact \(\displaystyle \frac{d}{dx}\tan^{-1}(x) = \frac{1}{1+x^2}\), and the Chain Rule for derivatives.

Use u-substitution \(\displaystyle u = \tan^{-1}(x^2)\), then we obtain \(\displaystyle du = \frac{1}{1+(x^2)^2} \times 2x dx\)

Our integral then becomes

\(\displaystyle \int^{\tan^{-1}(\frac{\pi^2}{4})}_{0} u du\) after substitution. (The new upper bound on the integral cannot be simplified well, so we should leave it as is).

We then integrate to get

\(\displaystyle [\frac{1}{2}u^2]^{\tan^{-1}(\frac{\pi^2}{4})}_0\)

\(\displaystyle \frac{1}{2}(\tan^{-1}(\frac{\pi^2}{4}))^2\)