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Calculus 2 : Indefinite Integrals

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Example Questions

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Example Question #311 : Finding Integrals

Calculate the following integral: $y=\int \frac{4}{16x^{2}+1}dx$

Possible Answers:

$y=ln\left |4x+1 \right |+C$

$y=\frac{4x}{16x^{2}+1}+C$

$y=tan^{-1}(4x)+C$

$y=4ln\left |16x^{2}+1 \right |+C$

Correct answer:

$y=tan^{-1}(4x)+C$

Explanation:

At first glance, it may look like we need to use partial fraction decomposition to solve this integral. However, this integral is much simpler than that. recall that

$\int \frac{1}{1+x^{2}}dx=tan^{-1}x+C$ . The integral we need to solve, is just the derivative of $tan^{-1}x$ , scaled by a factor of four. So, the solution to our integral is:

$y=tan^{-1}(4x)+C$

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Example Question #312 : Finding Integrals

$y=\int sinxcosx dx$

Possible Answers:

$y=sin^{2}x+C$

$y=sin^{2}xcos^{2}x+C$

$y=\frac{1}{2}(sin^{2}x)+C$

$y=-sin^{2}x+C$

Correct answer:

$y=\frac{1}{2}(sin^{2}x)+C$

Explanation:

We can solve this integral with u substitution. let u=sinx , so, du=cosxdx

Making this substitution, our integral looks like this:

$y=\intudu$ $\int udu$

So, $y=\frac{1}{2}u^{2}+C$

$y=\frac{1}{2}sin^{2}x+C$

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Example Question #41 : Indefinite Integrals

Calculate the following integral:

$\int 3cos^{2}xsinxdx$

Possible Answers:

$y=\frac{1}{3}cos^{3}x+C$

$y=-\frac{1}{3}cos^{3}x+C$

$y=cos^{3}x+C$

$y=-cos^{3}x+C$

Correct answer:

$y=-cos^{3}x+C$

Explanation:

We can solve this integral via u substitution:

$y=\int 3cos^{2}xsinxdx$

let u=cosx and du=-sinxdx Thus, our integral becomes:

$y=\int -3u^{2}du$ which equals: $y=-3(\frac{1}{3})u^{3}+C$

Re-substituting our value for u back in, we get our answer:

$y=-cos^{3}x+C$

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Example Question #42 : Indefinite Integrals

Evaluate the following indefinite integral:

$\int(9e^2^x+5cosx) dx$

Possible Answers:

18e^2^x+5sinx+C

18e^2^x-5sinx+C

4.5e^2^x+5sinx+C

4.5e^2x-5sinx+C

Correct answer:

4.5e^2^x+5sinx+C

Explanation:

The antiderivative of a sum is the sum of the antiderivatives. The first term in the integrand can be solved by u-substitution, as follows:

$\int9e^2^x dx = 9\int e^2^x dx = 9\int (e^u)(\frac{du}{2}) = 4.5e^u +C = 4.5e^2^x + C$

The second term in the integrand is a straightforward integration of a trigonometric function:

$\int5cosx dx = 5 \int cosx dx = 5sinx +C$

Hence, the antiderivative of the original function is

$\int(9e^2^x+5cosx)dx = (4.5e^2^x+C) +(5sinx + C)$

=4.5e^2^x+5sinx+C

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Example Question #43 : Indefinite Integrals

Evaluate.

$\int 2x + \frac{1}{x} \ dx$

Possible Answers:

$F(x) = 2x^2 + \frac{1}{x}$

$F(x) = x^2 + \ln(x)$

$F(x) = x^2 + \frac{2}{x}$

Answer not listed

$F(x) =\frac{1}{2} x^2 + \ln(x)$

Correct answer:

$F(x) = x^2 + \ln(x)$

Explanation:

$\int_{a}^{b} f(x) dx$

In order to evaluate an integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = 2x + \frac{1}{x}$ .

The antiderivative is $F(x) = x^2 + \ln(x)$ .

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Example Question #44 : Indefinite Integrals

Evaluate.

$\int \cos (2x-3) \ dx$

Possible Answers:

Answer not listed

$F(x) = - \frac{\cos(2x-3)}{\sin(2x-3)}$

$F(x) =- \sin(2x-3)$

$F(x) = \frac{\sin(2x-3)}{2}$

$F(x) = \frac{2}{\sin(2x-3)}$

Correct answer:

$F(x) = \frac{\sin(2x-3)}{2}$

Explanation:

$\int_{a}^{b} f(x) dx$

In order to evaluate an integral, first find the antiderivative of f(x).

If then
If then $F(x) = \frac{x^{a+1}}{a+1} + C$
If $f(x) = \frac{1}{x}$ then
If then
If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$
If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$
If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$
Substitution rule $\int f (g(x)) g'(x) \ dx = \int f(u) \ du$ where

In this case, $f(x) = \cos(2x-3)$ where u= 2x-3 .

The antiderivative is $F(x) = \frac{\sin(2x-3)}{2}$ .

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Example Question #45 : Indefinite Integrals

$\int cot(3x+3)dx$

Possible Answers:

3ln|sin(3x+3)|+C

$\frac{1}{3}ln|cos(3x+3)|+C$

ln|sin(3x+3)|+C

$\frac{1}{3}ln|sin(3x+3)|+C$

3ln|cos(3x+3)|+C

Correct answer:

$\frac{1}{3}ln|sin(3x+3)|+C$

Explanation:

This is a u-substitution problem. First, let us rewrite our trigonometric function:

$\int cot(3x+3)dx=\int \frac{cos(3x+3)}{sin(3x+3)}dx.$

We want to replace our variables to make the integral easier to solve.

u=sin(3x+3), du=3cos(3x+3)dx.

Now, we have replaced everything in our integral in terms of our new variable However, we have an extra that was not part of our original function. Therefore, we must divide by so that we match everything exactly. Rewriting, we get:

$\frac{1}{3}\int \frac{du}{u}=\frac{1}{3}ln|u|+C.$ Now, all we have to is substitute our original variable back in.

$\frac{1}{3}ln|sin(3x+3)|+C.$

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Example Question #46 : Indefinite Integrals

Evaluate.

$\int 12x^5 - 3x^2 \ dx$

Possible Answers:

F(x) =6x

F(x) =2x^6 - x^3

F(x) =x^6 - x^3

Answer not listed

F(x) =12x^3 - 3x^4

Correct answer:

F(x) =2x^6 - x^3

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, f(x) = 12x^5 - 3x^2 .

The antiderivative is F(x) =2x^6 - x^3 .

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Example Question #47 : Indefinite Integrals

Evaluate.

$\int \cos (3x) - 1\ dx$

Possible Answers:

$F(x) = \frac{\sin(x)}{3}$

$F(x) = \frac{\sin(3x)}{3}$

$F(x) = - \frac{\sin(3x)}{3x} - x$

Answer not listed

$F(x) = \frac{\sin(3x)}{3} - x$

Correct answer:

$F(x) = \frac{\sin(3x)}{3} - x$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \cos(3x) -1$ where u= 3x .

The antiderivative is $F(x) = \frac{\sin(3x)}{3} - x$ .

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Example Question #48 : Indefinite Integrals

$\int \frac{1}{3x-1} \ dx$

Possible Answers:

$F(x) =\frac{\ln(x)}{3}$

$F(x) =\frac{\ln(3x-1)}{3x^2}$

$F(x) =\frac{\ln(3x-1)}{3}$

$F(x) =-\frac{\ln(3x-1)}{3}$

Answer not listed

Correct answer:

$F(x) =\frac{\ln(3x-1)}{3}$

Explanation:

$\int f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of f(x).

If f(x) = a then F(x) = ax + C

If f(x) = x a then $F(x) = \frac{x^{a+1}}{a+1} + C$

If $f(x) = \frac{1}{x}$ then F(x) = ln(x) + C

If f(x) = e ^x then F(x) = e^x + C

If $f(x) = \cos(x)$ then $F(x) = \sin(x) + C$

If $f(x) = \sin(x)$ then $F(x) = - \cos(x) + C$

If $f(x) = \sec^2 (x)$ then $F(x) = \tan(x) + C$

In this case, $f(x) = \frac{1}{3x-1}$ and u= 3x-1 .

The antiderivative is $F(x) =\frac{\ln(3x-1)}{3}$ .

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Calculus 2 : Indefinite Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #311 : Finding Integrals

Example Question #312 : Finding Integrals

Example Question #41 : Indefinite Integrals

Example Question #42 : Indefinite Integrals

Example Question #43 : Indefinite Integrals

Example Question #44 : Indefinite Integrals

Example Question #45 : Indefinite Integrals

Example Question #46 : Indefinite Integrals

Example Question #47 : Indefinite Integrals

Example Question #48 : Indefinite Integrals

All Calculus 2 Resources

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