Techniques of antidifferentiation

Help Questions

AP Calculus AB › Techniques of antidifferentiation

Questions 1 - 10

Integrate:

$\int 7x^4 \sin(3x^5)dx$

$-\frac{7}{15}\cos(3x^5)+C$

$\frac{7}{15}\cos(3x^5)+C$

$-\frac{7}{15}\cos(3x^5)$

$-\cos(3x^5)+C$

Explanation

To integrate, the following substitution was made:

Now, we rewrite the integral in terms of u and integrate:

$\frac{7}{15}\int\sin(u)du=-\frac{7}{15}\cos(u)+C$

The following rule was used for integration:

$\int \sin(x)dx=-\cos(x)+C$

Finally, rewrite the final answer in terms of our original x term:

$-\frac{7}{15}\sin(3x^5)+C$

Determine the value of $\int_{0}^{2}\sqrt{9x^2+12x+4} dx$ .

Explanation

We can factor the equation inside the square root:

$\\int_0^2\sqrt{9x^2+12x+4}dx \\\int_0^2\sqrt{(3x+2)^2}dx \\\int_0^2 (3x+2)dx$

From here, increase each term's exponent by one and divide the term by the new exponent.

$\frac{3x^2}{2}+2x\big|_0^2$

Now, substitute in the upper bound into the function and subtract the lower bound function value from it.

$\\frac{3(2)^2}{2}+2(2)-\left(\frac{3(0)^2}{2}+2(0)\right) \\\frac{3(4)}{2}+4 \\\frac{12}{2}+4 \\6+4 \\10$

Therefore,

$\int_{0}^{2}\sqrt{(3x+2)^2}dx=\int_{0}^{2}(3x+2)dx=10$

Integrate,

$\int[x^2 +\sin(x)-3]dx$

$\frac{1}{3}x^3 - \cos(x) - 3x +C$

$\frac{1}{2}x^2 + \cos(x) - 3x +C$

$x -x^2+ \cos(x) +C$

$\frac{1}{2}x - \sin(x) - 3x^2 +C$

$\frac{1}{2}x^3 -\cos(x) - \frac{3}{2}x^2 +C$

Explanation

Integrate

$\int[x^2 +\sin(x)-3]dx$

1) Apply the sum rule for integration,

$\int x^2 dx +\int \sin(x)dx -\int 3dx$

2) Integrate each individual term and include a constant of integration,

$\frac{1}{3}x^3 - \cos(x) - 3x +C$

Further Discussion

Since indefinite integration is essentially a reverse process of differentiation, check your result by computing its' derivative.

$\frac{d}{dx}\left(\frac{1}{3}x^3-\cos(x)-3x+C \right )=x^2+\sin(x) -3$

This is the same function we integrated, which confirms our result. Also, because the derivative of a constant is always zero, we must include "C" in our result since any constant added to any function will produce the same derivative.

Given , find the general form for the antiderivative .

$\frac{4}{3}x^3 +10x^2 +25x+C$

None of the other answers

Explanation

To answer this, we will need to FOIL our function first.

Now can find the antiderivatives of each of these three summands using the power rule.

$f(x) = \frac{4}{3}x^3 +10x^2 +25x+C$ (Don't forget )!

Find the antiderivative of the following.

$\sec^2(x)$

$\tan (x)$

$2\sec (x)$

$\frac{\tan (x)}{2}$

$2\tan (x)$

Explanation

$\sec^2(x)$ is the derivative of $\tan (x)$ . Thus, the antiderivative of $\sec^2(x)$ is $\tan (x)$ .

$\begin{align*}&\text{Evaluate the integral: }\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx\end{align*}$

Explanation

$\begin{align*}&\text{In performing an integral, familiarity with derivative rules}\&\text{can prove useful. After all, an integral is essentially the}\&\text{opposite operation of a derivative. Consider how in a derivative}\&\text{we’d use the chain rule to multiply the derivative of the “inside”}\&\text{of a function by the derivative of the outside? Integrals, in}\&\text{similar fashion, entail a division.}\&\text{Utilizing integral rules:}\&\int[cos(ax)]=\frac{sin(ax)}{a}\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=\frac{(27sin(x + 1))}{8}|_{8}^{11.5}\&\text{Now we just subtract the function value at the lower bound from that of the upper bound:}\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=\frac{(27sin((11.5) + 1))}{8}-(\frac{(27sin((8) + 1))}{8})\&\int_{8}^{11.5}(\frac{(27cos(x + 1))}{8})dx=-1.61\end{align*}$

Integrate:

$\int \frac{dx}{\sqrt{25-6x^2}}$

$\frac{1}{\sqrt{6}}\arcsin(\frac{\sqrt{6}x}{5})+C$

$\arcsin(\frac{\sqrt{6}x}{5})+C$

$\frac{1}{\sqrt{6}}\arcsin(\frac{x}{5})+C$

$\frac{1}{\sqrt{6}}\arcsin(\frac{\sqrt{6}x}{25})+C$

$\frac{1}{\sqrt{6}}\arccos(\frac{\sqrt{6}x}{5})+C$

Explanation

To integrate, the following substitution must be made:

$u=\sqrt{6} x, du = \sqrt{6} dx$

Now, we rewrite the integral in terms of u and integrate:

$\frac{1}{\sqrt{6}} \int \frac{du}{\sqrt{25-u^2}}= \frac{1}{\sqrt{6}}\arcsin(\frac{u}{5})+C$

The following integration rule was used:

$\int \frac{du}{\sqrt{a^2-u^2}}=\arcsin(\frac{u}{a})+C$

To finish, we replace u with our original x term:

$\frac{1}{\sqrt{6}}\arcsin(\frac{\sqrt{6}x}{5})+C$

Evaluate the integral $\int \frac{2x[\ln(x^2-4)]^5}{x^2-4}dx$

$\frac{[\ln(x^2-4)]^6}{6}+C$

$5[\ln(x^2-4)]^4+C$

$[\ln(x^2-4)]^7+C$

Explanation

Evaluating integrals requires knowledge of the basic integral forms. In this problem, there is a group raised to an exponent, which is the $[\ln(x^2-4)]^5$ .

This arrangement typically follows the basic integral form $\int u^n du = \frac{1}{n+1}u^{n+1}+ C$ , where is a variable expression, is some constant number and is the constant of integration, which stays unknown. In this case, we can use u-substitution to match this form.

The exponent is 5, so the in the basic integral form will be5.

Since the inside of the exponent group is $\ln(x^2-4)$ , set $u = \ln(x^2-4)$ .

Now we differentiate to find . Recall that the derivative of $\ln(v)= \frac{dv}{v}$ .

$du = \frac{2x}{x^2-4}dx$

Notice that the parts of our match up with the integral we are evaluating, and there are no variables that aren't accounted for. This is clearer if we write out and simplify what our substitution says.

$\int u^n du = \int [\ln(x^2-4)]^5\frac{2x}{x^2-4}dx$

$\int \frac{2x[\ln(x^2-4)]^5}{x^2-4}dx$

This is exactly what we are asked to integrate.

This means we can evaluate the integral by using basic integral form directly. Plugging in our and into the right side of $\int u^n du = \frac{1}{n+1}u^{n+1}+C$ , we get

$\frac{1}{5+1}[\ln(x^2-4)]^{5+1}$

$\frac{1}{6}[\ln(x^2-4)]^6+C$

This is equivalent to the answer $\frac{[\ln(x^2-4)]^6}{6}+C$ , which is the correct answer.

Evaluate the integral $\int \frac{-2 \sin(1/x^2)e^{-\cos(1/x^2)}}{x}dx$

$e^{-\cos(1/x^2)}+C$

$e^{2\cos(1/x^2)}+C$

$e^{\sin(1/x^2)}+C$

$e^{-\sin(-1/x)}+C$

Not integrable

Explanation

There are a lot of pieces inside this integral. There are trigonometry functions, exponential functions, and a rational arrangement. Lots of possibilities.

This is where u-substitution is best. Try making u represent various parts and see if du gets all the other parts. After doing this enough times, you will see that we should make u be the exponent of the e.

$u = -\cos(1/x^2)$

Let's write the inner fraction as an x with a negative exponent

$u = -\cos(x^{-2})$

Now we differentiate and see what we get for du. This will require the chain rule. The outer structure is the $-\cos$ , which requires trig functions integration rules, and the chained inner structure is a power rule arrangement