Integrals Of Trigonometric Functions Trivia Quiz

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  • 1/20 Questions

    āˆ«cos(u)du

    • Sin(u)+C
    • Ln|sin(u)| + C
    • -cot(u) + C
    • (1/a)arcsec(|u|/a) + C
    • -csc(u) + C
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About This Quiz

How well do you remember the integrals of the trigonometric functions? For this quiz, you must know how to do integrations using various trigonometric identities, solving some questions for us. This quiz is best to strengthen your basics and prepare for an upcoming exam. All the Best!
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Integrals Of Trigonometric Functions Trivia Quiz - Quiz

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  • 2. 

    Sin^2(x) =

    • (1 - cos (2x)) / 2

    • (1 + cos(2x)) / 2

    Correct Answer
    A. (1 - cos (2x)) / 2
    Explanation
    The given equation sin^2(x) = (1 - cos (2x)) / 2 is the correct answer. This can be derived using the trigonometric identity sin^2(x) = 1 - cos^2(x). By substituting cos^2(x) with (1 - cos(2x))/2, we get the given equation.

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  • 3. 

    āˆ«sin(u)du

    • Ln|secu + tanu)| + C

    • Tan(u)+C

    • Sec(u)+C

    • -cos(u) + C

    • Cos(u)+C

    Correct Answer
    A. -cos(u) + C
    Explanation
    The integral of sin(u) with respect to u is equal to -cos(u) + C, where C is the constant of integration. This can be found using the integral rules for trigonometric functions. The antiderivative of sin(u) is -cos(u), and adding the constant of integration C gives the final result.

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  • 4. 

    If the powers of both the sine and cosine are even and positive,

    • Save a sine factor

    • Save a cosine factor

    • Use the sine^2 and cosine^2 identities

    • Convert the remaining factors into sines

    • Convert the remaining factors into cosines

    Correct Answer
    A. Use the sine^2 and cosine^2 identities
    Explanation
    When the powers of both the sine and cosine are even and positive, we can use the sine^2 and cosine^2 identities. These identities state that sine squared plus cosine squared equals 1. By using these identities, we can simplify the expression and eliminate the need for separate sine and cosine factors.

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  • 5. 

    āˆ«cot(u)du

    • Ln|sin(u)| + C

    • (1/a)arctan(u/a) + C

    • -cot(u) + C

    • Sin(u)+C

    • Ln|secu + tanu)| + C

    Correct Answer
    A. Ln|sin(u)| + C
    Explanation
    The integral of cot(u) can be found by using the formula for the integral of the cotangent function, which is ln|sin(u)| + C. This formula is derived from the derivative of the natural logarithm function and the chain rule. Therefore, the correct answer is ln|sin(u)| + C.

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  • 6. 

    Cos^2(x) = 

    • (1 + cos(2x)) / 2

    • (1 - cos (2x)) / 2

    Correct Answer
    A. (1 + cos(2x)) / 2
    Explanation
    A mnemonic device to remember these: It is the solution with the POSITIVE cos(2x) because we started with cos^2(x), not sin^2(x)--did we start with a sine function? Nope, sine^2(x) gives a NEGATIVE on the cosine(2x) because sine is not cosine. Remember, this is not a mathematical proof, just a way to relate things in your head to remember them.

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  • 7. 

    If there are no secant factors and the power of the tangent is even and positive,

    • Convert a tangent-squared factor into a secant-squared factor

    • Save a secant-tangent factor

    • Save a secant squared factor

    Correct Answer
    A. Convert a tangent-squared factor into a secant-squared factor
    Explanation
    When there are no secant factors and the power of the tangent is even and positive, we can convert a tangent-squared factor into a secant-squared factor. This means that if we have an expression with a term like tangent squared, we can rewrite it as secant squared. This conversion allows us to simplify the expression and potentially make it easier to work with or solve.

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  • 8. 

    āˆ«tan(u)du

    • -ln|cos(u)| + C

    • (1/a)arctan(u/a) + C

    • -sec^2(u) + C

    • Csc(u)*cot(u)

    • -cot(u) + C

    Correct Answer
    A. -ln|cos(u)| + C
  • 9. 

    āˆ«csc(u)du

    • Ln|csc(u) + cot(u)| + C

    • Tan^2(u) + C

    • -ln|csc(u) + cot(u)| + C

    Correct Answer
    A. -ln|csc(u) + cot(u)| + C
    Explanation
    The integral of csc(u)du can be found by using the substitution method. Let's substitute v = csc(u) + cot(u). Then, dv = (-csc(u)cot(u) - csc^2(u))du. Rearranging the terms, we have -dv = (csc(u) + cot(u))du. Substituting back into the integral, we get āˆ«-dv = -āˆ«(csc(u) + cot(u))du. Integrating both sides, we have -v = -ln|csc(u) + cot(u)| + C. Therefore, the correct answer is -ln|csc(u) + cot(u)| + C.

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  • 10. 

    āˆ«csc(u)cot(u)du

    • Csc(u) + C

    • -csc^2(u) + C

    • Sec(u)tan(u) + C

    • Tan(u) + C

    • -csc(u) + C

    Correct Answer
    A. -csc(u) + C
    Explanation
    The given integral is āˆ«csc(u)cot(u)du. By using the derivative of csc(u) which is -csc(u)cot(u), we can see that the correct answer is -csc(u) + C, where C represents the constant of integration.

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  • 11. 
    āˆ«du/(a^2 - u^2) - ProProfs

    āˆ«du/(a^2 - u^2)

    • Arctan(u/a) + C

    • Arcsin(u/a) + C

    • (1/a) arcsin (u/a) + C

    • (1/u) arcsin (|u|/a) + C

    • (1/a) arcsec (|u|/a) + C

    Correct Answer
    A. Arcsin(u/a) + C
    Explanation
    The integral of du/(a^2 - u^2) can be solved using the substitution method. Let u = a*sin(theta), then du = a*cos(theta)d(theta). Substituting these values into the integral, we get āˆ«(a*cos(theta)d(theta))/(a^2 - a^2*sin^2(theta)). Simplifying further, we have āˆ«(a*cos(theta)d(theta))/(a^2*cos^2(theta)). Canceling out the common terms, we are left with āˆ«(1/a)d(theta). Integrating this, we get (1/a)theta + C. Replacing theta with arcsin(u/a), we get arcsin(u/a) + C as the final answer.

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  • 12. 
    āˆ«du/(a^2 + u^2) - ProProfs

    āˆ«du/(a^2 + u^2)

    • (1/u) arcsin (|u|/a) + C

    • Arctan(u/a) + C

    • Arcsin(u/a) + C

    • (1/a) arctan (u/a) + C

    • (1/a) arctan (|u|/a) + C

    Correct Answer
    A. (1/a) arctan (u/a) + C
    Explanation
    The given integral can be solved using the substitution method. Let u = a tanĪø. Then, du = a sec^2Īø dĪø. Substituting these values into the integral, we get āˆ«(a sec^2Īø)/(a^2 + a^2 tan^2Īø) dĪø. Simplifying this expression gives us āˆ«(a sec^2Īø)/(a^2 sec^2Īø) dĪø. The sec^2Īø terms cancel out, leaving us with āˆ«(1/a) dĪø. Integrating this expression gives us (1/a)Īø + C. Substituting back Īø = arctan(u/a), we get (1/a) arctan (u/a) + C as the final answer.

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  • 13. 

    If the power of the cosine is odd and positive,

    • Save a sine factor

    • Save a cosine factor

    • Convert the remaining factors into sines

    • Convert the remaining factors into cosines

    Correct Answer(s)
    A. Save a cosine factor
    A. Convert the remaining factors into sines
    Explanation
    When the power of the cosine is odd and positive, we save a cosine factor because it cannot be converted into a sine. The remaining factors, which are cosines, can be converted into sines using the identity cos^2(x) = 1 - sin^2(x). Therefore, we convert the remaining factors into sines.

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  • 14. 

    For integrals involving √(a^2 + u^2)

    • U = asinĪ˜

    • U = atanĪ˜

    • U=asecĪ˜

    • āˆš(a^2 + u^2) = acosĪ˜

    • āˆš(u^2 + a^2) = asecĪ˜

    Correct Answer(s)
    A. U = atanĪ˜
    A. āˆš(u^2 + a^2) = asecĪ˜
  • 15. 

    For integrals involving √(u^2 - a^2)

    • U = asinĪ˜

    • U = atanĪ˜

    • U=asecĪ˜

    • āˆš(u^2 - a^2) = acosĪ˜

    • āˆš(u^2 - a^2) = atanĪ˜

    Correct Answer(s)
    A. U=asecĪ˜
    A. āˆš(u^2 - a^2) = atanĪ˜
  • 16. 
    āˆ«du/((u(u^2 - a^2))^.5) - ProProfs

    āˆ«du/((u(u^2 - a^2))^.5)

    • (1/a) arctan (u/a) + C

    • Arcsin(u/a) + C

    • (1/a)arcsec(u/a) + C

    • (1/a) arctan (|u|/a) + C

    • (1/a)arcsec(|u|/a) + C

    Correct Answer
    A. (1/a)arcsec(|u|/a) + C
    Explanation
    The integral given is in the form of āˆ«du/((u(u^2 - a^2))^.5). To evaluate this integral, we can use the substitution method. Let's substitute u = a secĪø, du = a secĪø tanĪø dĪø.

    After substitution, the integral becomes āˆ«(a secĪø tanĪø)/(a secĪø(a sec^2Īø - a^2))^.5 dĪø. Simplifying this expression, we get āˆ«(tanĪø)/(secĪø(a^2 sec^2Īø - a^2))^.5 dĪø.

    Further simplifying, we have āˆ«(tanĪø)/(a secĪø)^.5 dĪø. Using the trigonometric identity tanĪø = sinĪø/cosĪø and secĪø = 1/cosĪø, we can rewrite the integral as āˆ«(sinĪø)/(a cosĪø)^1.5 dĪø.

    Now, we can use the substitution method again, letting t = cosĪø, dt = -sinĪø dĪø. The integral becomes āˆ«(-1)/(a t^1.5) dt.

    Integrating this expression, we get (-1/a)āˆ«t^(-1.5) dt = (-1/a)(-2t^(-0.5)) = (2/aāˆšt) + C.

    Substituting back t = cosĪø, we have (2/aāˆšcosĪø) + C. Finally, using the identity secĪø = 1/cosĪø, we can rewrite the answer as (1/a)arcsec(|u|/a) + C.

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  • 17. 

    If the power of the tangent is odd and positive,

    • Save a secant squared factor

    • Save a secant-tangent factor

    • Convert the remaining factors into tangents

    • Convert the remaining factors into secants

    • Expand and integrate

    Correct Answer(s)
    A. Save a secant-tangent factor
    A. Convert the remaining factors into secants
    A. Expand and integrate
    Explanation
    When the power of the tangent is odd and positive, we can save a secant-tangent factor and convert the remaining factors into secants. This is because the derivative of tangent is secant squared and the derivative of secant is secant-tangent. By saving a secant-tangent factor, we can use it later to simplify the integral. Converting the remaining factors into secants allows us to use the derivative of secant to simplify the integral further. Finally, we expand and integrate the simplified expression to find the solution.

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  • 18. 

    For integrals involving √(a^2 - u^2),

    • U = asinĪ˜

    • U = tanĪ˜

    • U=secĪ˜

    • āˆš(a^2 - u^2) = acosĪ˜

    • āˆš(u^2 - a^2) = atanĪ˜

    Correct Answer(s)
    A. U = asinĪ˜
    A. āˆš(a^2 - u^2) = acosĪ˜
    Explanation
    The given answer, u = asinĪ˜ and āˆš(a^2 - u^2) = acosĪ˜, is correct because when we substitute u = asinĪ˜ into āˆš(a^2 - u^2), we get āˆš(a^2 - (asinĪ˜)^2) = āˆš(a^2 - a^2sin^2Ī˜) = āˆš(a^2(1 - sin^2Ī˜)) = āˆš(a^2cos^2Ī˜) = acosĪ˜. Therefore, the answer is valid and provides the correct relationship between u and āˆš(a^2 - u^2).

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  • 19. 

    If the power of the cosine is odd and positive...

    • Save a sine factor

    • Save a cosine factor

    • Convert the remaining factors into sines

    • Convert the remaining factors into cosines

    Correct Answer(s)
    A. Save a sine factor
    A. Convert the remaining factors into cosines
    Explanation
    If the power of the cosine is odd and positive, it means that there is an odd number of cosine factors in the expression. In this case, we save a sine factor because the sine and cosine are complementary functions. Then, we convert the remaining factors into cosines because an odd number of cosines multiplied together will result in a positive value.

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  • 20. 

    If the power of the secant is even and positive,

    • Save a secant squared factor

    • Convert the remaining factors into tangents

    • Save a secant-tangent factor

    • Convert the remaining factors into secants

    • Expand and integrate

    Correct Answer(s)
    A. Save a secant squared factor
    A. Convert the remaining factors into tangents
    A. Expand and integrate
    Explanation
    When the power of the secant is even and positive, we can save a secant squared factor. Then, we can convert the remaining factors into tangents. After that, we save a secant-tangent factor and convert the remaining factors into secants. Finally, we expand and integrate the expression.

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  • Mar 22, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Apr 20, 2010
    Quiz Created by
    Kirakiwibug
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