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Calculus 2 : Finding Integrals

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Example Questions

Example Question #421 : Finding Integrals

$\int \frac{4x^3}{x^4+3}dx$

Possible Answers:

$ln\left | 4x^3 \right |+C$

$ln\left | x^4-3 \right |+C$

$ln\left | x^4+3 \right |$

$2ln\left | x^4+3 \right |+C$

$ln\left | x^4+3 \right |+C$

Correct answer:

$ln\left | x^4+3 \right |+C$

Explanation:

First, you must use u substitution to integrate this expression:

u=x^4+3; du=4x^3dx

Now, plug back in so you can integrate:

$\int \frac{1}{u}du$

Now integrate:

$ln\left | u \right |$

Now substitute back in your initial expression and add C because it is an indefinite integral:

$ln\left | x^4+3 \right |+C$

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Example Question #141 : Indefinite Integrals

$\int t^3-2t+4dt$

Possible Answers:

$\frac{t^4}{4}-t^2+4t$

$\frac{t^4}{4}-2t^2+4t+C$

$\frac{t^3}{4}-t^2+4t+C$

$\frac{t^4}{4}-t^2+8t+C$

$\frac{t^4}{4}-t^2+4t+C$

Correct answer:

$\frac{t^4}{4}-t^2+4t+C$

Explanation:

To integrate this expression, remember to add one to the exponent and then put that result on the denominator:

$\frac{t^4}{4}-\frac{2t^2}{2}+4t$

Simplify and add C because it is an indefinite integral:
$\frac{t^4}{4}-t^2+4t+C$

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Example Question #144 : Indefinite Integrals

Evaluate.

$\int \cos(2x-9) \ dx$

Possible Answers:

$F(x) =2\sin(2x-9) +C$

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

$F(x) = \frac{\sin(2x-9)}{2} +C$

$F(x) = \frac{2}{\sin(2x-9)} +C$

Answer not listed.

Correct answer:

$F(x) = \frac{\sin(2x-9)}{2} +C$

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, $f(x) = \cos(2x-9)$ .

The antiderivative is $F(x) = \frac{\sin(2x-9)}{2} +C$ .

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Example Question #145 : Indefinite Integrals

Evaluate.

$\int -\frac{\sin(2x)}{2} \ dx$

Possible Answers:

$F(x) = -\frac{\cos(x)}{2x} +C$

$F(x) = -\frac{4}{\cos(2x)} +C$

$F(x) = \frac{\cos(2x)}{4} +C$

$F(x) = -4\cos(2x)} +C$

Answer not listed.

Correct answer:

$F(x) = \frac{\cos(2x)}{4} +C$

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, $f(x) = -\frac{\sin(2x)}{2}$ .

The antiderivative is $F(x) = \frac{\cos(2x)}{4} +C$ .

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Example Question #146 : Indefinite Integrals

Evaluate.

$\int \sec^2 (12x-1)\ dx$

Possible Answers:

Answer not listed.

$F(x) = 12x\tan(12x-1) +C$

$F(x) = \frac{\tan(12x-1)}{12} +C$

$F(x) = -\frac{\tan(x)}{2} +C$

$F(x) = \frac{x}{\tan(12x-1)} +C$

Correct answer:

$F(x) = \frac{\tan(12x-1)}{12} +C$

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, $f(x) =\sec^2 (12x-1)$ .

The antiderivative is $F(x) = \frac{\tan(12x-1)}{12} +C$ .

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Example Question #772 : Integrals

$\int x^3-2x^2+xdx$

Possible Answers:

$\frac{x^5}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+C$

$\frac{x^4}{4}-\frac{8x^3}{3}+\frac{x^2}{2}+C$

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

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

$\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}$

Correct answer:

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

Explanation:

Remember, that when integrating, raise the exponent by 1 and then also put that result on the denominator:

$\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}$

Now, remember to add C because it is an indefinite integral:
$\frac{x^4}{4}-\frac{2x^3}{3}+\frac{x^2}{2}+C$

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Example Question #421 : Finding Integrals

Integrate

$\int{\frac{2x}{x^2+6x+13}}$

Possible Answers:

$\ln|x^2+6x+13|-3\arctan(x+3)+C$

$\frac{x^2}{\frac{1}{3}x^3+3x^2+13x}$

$\ln|x^2+6x+13|-3\arctan \left(\frac{x+3}{2}\right)+C$

$\ln|x^2+6x+13|-3\arccos \left(\frac{x+3}{2}\right)+C$

Correct answer:

$\ln|x^2+6x+13|-3\arctan \left(\frac{x+3}{2}\right)+C$

Explanation:

Untitled

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Example Question #777 : Integrals

Rewrite the integral in terms of the three basic trigonometric ratios

$\int{\frac{\sec^3 x}{\tan x\csc x}}dx$

Possible Answers:

$\int{\frac{1}{\sin^2x}}dx$

$\int{\frac{1}{\arccos x}}dx$

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

$\int{\frac{1}{\cos^2x}}dx$

$\int{\frac{1}{\tan^2 x-1}}dx$

Correct answer:

$\int{\frac{1}{\cos^2x}}dx$

Explanation:

$\int{\frac{\sec^3 x}{\tan x\csc x}}dx$

Step 1: Rewrite secant and cosecant in terms of cosine and sine:

$\frac{\sec^3 x}{\tan x\csc x}=\frac{\frac{1}{\cos^3 x}}{\tan x\times\frac{1}{\sin x}}=\frac{\frac{1}{\cos^3 x}}{\frac{\sin x}{\cos x}\times\frac{1}{\sin x}}\\$

Step 2: Rewrite and cancel out any common terms:

$\rightarrow \frac{\frac{1}{\cos^3 x}}{\frac{1}{\cos x}}=\frac{1}{\cos^3 x}\times\frac{\cos x}{1}=\frac{1}{\cos^2 x}$

Step 3: Make an equivalence relation between the answer and the original function.

$\int{\frac{\sec^3 x}{\tan x\csc x}}dx=\int{\frac{1}{\cos^2x}}dx$

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Example Question #778 : Integrals

Evaluate the integral:

$\int x^2\sin2x dx$

Possible Answers:

$\frac{-x^2}{2}\cos2x+\frac{x}{2}\sin2x-\frac{1}{4}\cos2x+C$

$\frac{-x^2}{2}\cos2x+\frac{x}{2}\sin2x+\frac{1}{4}\cos2x$

$\frac{-x^2}{2}\cos2x+\frac{x}{2}\sin2x+\frac{1}{4}\cos2x+C$

$\frac{x^2}{2}\cos2x+\frac{x}{2}\sin2x+\frac{1}{4}\cos2x+C$

Correct answer:

$\frac{-x^2}{2}\cos2x+\frac{x}{2}\sin2x+\frac{1}{4}\cos2x+C$

Explanation:

There are no substitutions that would work in this instance. Therefore, you must integrate by parts. The formula for integrating by parts is:

$\int u dv=uv-\int v du$

TO begin, you must assign values of u and dv based from the two different functions in the original integral. Specifically, $u=x^2, dv=\sin2x dx$ .

From here, you can find values for du and v:

$du=2x dx, v=\frac{-1}{2}\cos2x$

was found by using the following formula:

$\frac{d}{dx}x^n=nx^{n-1}$

V was found by using a u-substitution where u=2x, du=2dx and the fact that:

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

Following the formula, the integral can be rewritten as:

$\frac{-x^2}{2}\cos2x+\frac{1}{2}\int 2x\cos2x dx$

Once again, there is another integral that cannot be evaluated by using a substitution. Therefore, integration by parts must be used again.

$u=2x, du=2dx, dv=\cos2xdx, v=\frac{1}{2}\sin2x$

The integral can be rewritten again as:

$\frac{-x^2}{2}\cos2x+\frac{1}{2}(x\sin2x-\frac{1}{2}\int 2\sin 2x dx)$

The integral that is left is easily solvable by using a u-substitution where u=2x, dv=2dx and the fact that $\int \sin x dx=-\cos x +C$ . This leaves you with:

$\frac{-1}{2}\int2\sin2x dx=\frac{-1}{2}\int \sin u du=\frac{1}{2}\cos u+C=\frac{1}{2}\cos 2x +C$

Now, the final answer can be written as:

$\frac{-x^2}{2}\cos2x+\frac{x}{2}\sin2x+\frac{1}{4}\cos2x+C$

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Example Question #779 : Integrals

Evaluate the integral:

$\int 3x\ln\sqrt[3]{x} dx$

Possible Answers:

$(\ln x)(\frac{x^2}{2})+\frac{1}{4}x^2+C$

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

$(\ln x)(\frac{x^2}{2})-\frac{1}{4}x^2$

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

Correct answer:

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

Explanation:

Despite the intimidating look of the problem, simplifications can be made. Since $\sqrt[3]{x}=x^\frac{1}{3}$ , the integrand can be rewritten as:

$\int 3x\ln x^\frac{1}{3} dx$

Because of the properties of logarithms where $\ln x^a=a\ln x$ , the integrand can be rewritten as:

$\int3x\cdot \frac{1}{3}\ln x dx=\int x\ln x dx$

There are no substitutions that can be done here. Therefore, you must integrate by parts. The formula for integrating by parts is:

$\int u dv=uv-\int v du$

To begin, you must assign values of u and dv based from the two different functions in the original integral. Specifically, $u= \ln x, dv=x dx$ .

From here, you can find values for du and v:

$du=\frac{1}{x}dx, v=\frac{1}{2}x^2$

was found by using the following formula:

$\frac{d}{dx}x^n=nx^{n-1}$

V was found by using the following formula:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$

Following the formula, the integral can be rewritten as:

$(\ln x)(\frac{1}{2}x^2)-\frac{1}{2}\int x^2\cdot \frac{1}{x} dx=(\ln x)(\frac{1}{2}x^2)-\frac{1}{2}\int x dx$

The integral can be taken by using the following formula:

$\int x^ndx=\frac{x^{n+1}}{n+1}+C$

The final answer is:

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

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Calculus 2 : Finding Integrals

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #421 : Finding Integrals

Example Question #141 : Indefinite Integrals

Example Question #144 : Indefinite Integrals

Example Question #145 : Indefinite Integrals

Example Question #146 : Indefinite Integrals

Example Question #772 : Integrals

Example Question #421 : Finding Integrals

Example Question #777 : Integrals

Example Question #778 : Integrals

Example Question #779 : Integrals

All Calculus 2 Resources

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