Calculus II - Trigonometric Integrals - Evaluate Integral tan(x)^5*sec(x)^4 dx

In summary: i see the problem∫ [tan9(x) + 2 tan7(x)+tan5(x)] dx= tan8(x)/8 - tan6(x)/6 + tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|) + tan6(x)/3 - tan4(x)/2 + tan2(x) - 2 ln(|sec(x)|) + tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|)= tan8(x)/8 - tan6(x)/6 + tan6(x)/3 + tan4(x)/4 - tan4(x)/2 + tan4(x)/4 - tan2(x)/2 + tan2(x) -
  • #1
GreenPrint
1,196
0

Homework Statement


Hi,
I'm trying to solve this problem and guess I'm doing something wrong.

Evaluate Integral tan(x)^5*sec(x)^4 dx

Homework Equations



integral tan(x) dx = ln(|sec(x)|)

integral tan(x)^n dx = tan(x)^(n-1)/(n-1) - integral tan(x)^(n-1) dx

tan(x)^2+1=sec(x)^2

The Attempt at a Solution



My Answer
Integral tan(x)^5*sec(x)^4 dx = 1/8*tan(x)^8 + 1/6*tan(x)^6 + 1/2*tan(x)^2 + ln(|sec(x)|)+c

I don't see what I'm doing wrong...
You can see my work attached. Thanks in advance for any assistance you can provide! Don't forget you can click on the window that pops up after clicking attachments to open the image in a new tab to view it at a larger scale.
 

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  • #2
∫ tan5(x) sec4(x) dx
= ∫ tan5(x) (tan2(x)+1) sec2(x) dx
= ∫ [tan7(x) + tan5(x)] sec2(x) dx
 
  • #3
∫ tan5(x) sec4(x) dx
= ∫ tan5(x) (tan2(x)+1) sec2(x) dx
= ∫ [tan7(x) + tan5(x)] sec2(x) dx
= ∫ [tan7(x) + tan5(x)] (tan2(x)+1) dx
= ∫ [tan9(x) + tan7(x)+tan7(x)+tan5(x)] dx
= ∫ [tan9(x) + 2tan7(x)+tan5(x)] dx
is this not correct?
 
  • #4
Why would you do that when d/dx tan(x) = sec2(x)?
 
  • #5
I didn't think of making the substitution at the time <_< it would of made the problem much easier @_@ I should still be able to obtain the correct answer though, yes? I find it odd that I don't get the right answer

as you can see in my work
∫ tan5(x) sec4(x) dx
= ∫ tan5(x) (tan2(x)+1) sec2(x) dx
= ∫ [tan7(x) + tan5(x)] sec2(x) dx
= ∫ [tan7(x) + tan5(x)] (tan2(x)+1) dx
= ∫ [tan9(x) + tan7(x)+tan7(x)+tan5(x)] dx
= ∫ [tan9(x) + 2tan7(x)+tan5(x)] dx

one i get here i evaluated each integral separately using
integral tan(x)^n dx = tan(x)^(n-1)/(n-1) - integral tan(x)^(n-1) dx
I should be able to get the correct answer and i don't think my answer
1/8*tan(x)^8 + 1/6*tan(x)^6 + 1/2*tan(x)^2 + ln(|sec(x)|)+c
is correct for whatever strange reason when i can't find a error in my work at all
 
  • #6
GreenPrint said:
I should be able to get the correct answer and i don't think my answer
1/8*tan(x)^8 + 1/6*tan(x)^6 + 1/2*tan(x)^2 + ln(|sec(x)|)+c
is correct

It's not correct. The bolded terms are extraneous. Here's your work:

∫ tan5(x) sec4(x) dx
= ∫ tan5(x) (tan2(x)+1)2 dx
= ∫ tan5(x) (tan2(x)+1)2 dx
= ∫ tan5(x) (tan4(x)+2 tan2(x) +1) dx
= ∫ [tan9(x) + 2 tan7(x)+tan5(x)] dx

Let's evaluate each individually, using your formula. I'm not even sure whether or not your formula is true, but let's assume that it is.

∫ tann(x) dx = tann-1(x)/(n-1) - ∫ tann-1(x) dx
∫ [tan9(x) dx = tan8(x)/8 - ∫ tan8(x) dx
∫ [tan8(x) dx = tan7(x)/7 - ∫ tan7(x) dx
∫ [tan7(x) dx = tan6(x)/6 - ∫ tan6(x) dx
∫ [tan6(x) dx = tan5(x)/5 - ∫ tan5(x) dx

The problem here is that you used your own formula incorrectly. Using the formula correctly:

∫ [tan9(x) + 2 tan7(x)+tan5(x)] dx
= tan8(x)/8 - ∫ tan8(x) dx + ∫ [2 tan7(x)+tan5(x)] dx
= tan8(x)/8 - tan7(x)/7 + ∫ tan7(x) dx + ∫ [2 tan7(x)+tan5(x)] dx
= tan8(x)/8 - tan7(x)/7 + ∫ [3 tan7(x)+tan5(x)] dx
= tan8(x)/8 - tan7(x)/7 + 3 [tan6(x)/6 - ∫ tan6(x) dx] + ∫ tan5(x) dx
= tan8(x)/8 - tan7(x)/7 + 3 [tan6(x)/6 - tan5(x)/5 + ∫ tan5(x) dx] + ∫ tan5(x) dx
= tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 ∫ tan5(x) dx
= tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 ∫ tan4(x) dx
= tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 tan3(x)/3 + 4 ∫ tan3(x) dx
= tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 tan3(x)/3 + 4 tan2(x)/2 - 4 ∫ tan2(x) dx
= tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 tan3(x)/3 + 4 tan2(x)/2 - 4 tan(x) - 4 ∫ tan(x) dx
= tan8(x)/8 - tan7(x)/7 + 3 tan6(x)/6 - 3 tan5(x)/5 + 4 tan4(x)/4 - 4 tan3(x)/3 + 4 tan2(x)/2 - 4 tan(x) + 4 log|cos(x)| +C

It seems like something is seriously wrong with your formula.
 
Last edited:
  • #7
I thought the formula was correct, see attachment
ah yes the correct formula is
∫ tann(x) dx = tann-1(x)/(n-1) - ∫ tann-2(x) dx
which although i wrote the wrong formula in my posts i still used this one in my work and the correct

I don't see how you get rid of terms... isn't this correct

∫ tan9(x) dx = tan8(x)/8 - ∫ tan7(x) dx
= tan8(x)/8 - tan6(x)/6 +∫ tan5(x) dx
= tan8(x)/8 - tan6(x)/6 + tan4(x)/4 - ∫ tan3(x) dx
= tan8(x)/8 - tan6(x)/6 + tan4(x)/4 - tan2(x)/2 + ∫ tan(x) dx
= tan8(x)/8 - tan6(x)/6 + tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|)

2∫ tan7(x) = 2*tan6(x)/6 - 2 ∫ tan5(x) dx
= tan6(x)/3 - 2*tan4(x)/4 + 2 ∫ tan3(x) dx
= tan6(x)/3 - tan4(x)/2 + 2*tan2(x)/2 - 2 ∫ tan(x) dx
= tan6(x)/3 - tan4(x)/2 + tan2(x) - 2 ln(|sec(x)|)

∫ tan5(x) dx = tan4(x)/4 - ∫ tan3(x) dx
= tan4(x)/4 - tan2(x)/2 + ∫ tan(x) dx
= tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|)

∫ [tan9(x) + 2 tan7(x)+tan5(x)] dx

= tan8(x)/8 - tan6(x)/6 + tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|) + tan6(x)/3 - tan4(x)/2 + tan2(x) - 2 ln(|sec(x)|) + tan4(x)/4 - tan2(x)/2 + ln(|sec(x)|)

= tan8(x)/8 - tan6(x)/6 + tan6(x)/3 + tan4(x)/4 - tan4(x)/2 + tan4(x)/4 - tan2(x)/2 + tan2(x) - tan2(x)/2 + ln(|sec(x)|) - 2 ln(|sec(x)|) + ln(|sec(x)|)


= tan8(x)/8 + tan6(x)/6

hmm interesting
 

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Related to Calculus II - Trigonometric Integrals - Evaluate Integral tan(x)^5*sec(x)^4 dx

1. What is the purpose of Calculus II?

Calculus II is a continuation of Calculus I and focuses on more advanced topics such as integration, techniques of integration, and applications of integration. It is an important subject for students majoring in mathematics, science, engineering, and other related fields.

2. What are trigonometric integrals?

Trigonometric integrals involve integrating functions that contain trigonometric functions such as sine, cosine, tangent, and their inverses. These integrals are useful in solving real-world problems involving periodic phenomena, such as motion and waves.

3. What is the general process for evaluating integrals?

The general process for evaluating integrals involves using various techniques, such as substitution, integration by parts, and trigonometric identities, to transform the integral into a simpler form that can be easily evaluated. It also involves understanding the fundamental properties of integrals and applying them appropriately.

4. How do you approach evaluating the integral tan(x)^5*sec(x)^4 dx?

To evaluate this integral, we can use the substitution method by letting u = tan(x). This will transform the integral into a simpler form that can be evaluated using the power rule for integration and the identity sec^2(x) = 1 + tan^2(x). After integrating, we can then back-substitute u = tan(x) to get the final answer.

5. What are some common mistakes to avoid when evaluating trigonometric integrals?

Some common mistakes to avoid when evaluating trigonometric integrals include forgetting to use trigonometric identities, not properly applying the substitution method, and making errors in algebraic simplification. It is important to double-check your work and be familiar with the properties and rules of integration to avoid these mistakes.

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