Calculus II - Trigonometric Integrals

In summary: integral sin(x)^2/(sin(x)^4*cos(x)^2) dxused the fact that x^m/x^n = x^(m-n)used the fact that x^-m = 1/x^mintegral dx/(sin(x)^2*cos(x)^2)used the fact that sin(x)^2 + cos(x)^2 = 1solved for sin(x)^2sin(x)^2 = 1 - cos(x)^2integral dx/((1-cos(x)^2)*cos(
  • #1
GreenPrint
1,196
0

Homework Statement



Evaluate
integral csc(x)^4/cot(x)^2 dx

Homework Equations



The Attempt at a Solution



Apparently I'm doing something wrong, what I'm not sure, thanks for any help

My Answer: 2*tan(x) - (sec(x)^2*tan(x))/3 + c

integral csc(x)^4/cot(x)^2 dx

used fact that csc(x) = 1/sin(x)
used fact that cot(x) = 1/tan(x)
used fact that tan(x) = sin(x)/cos(x)

integral sin(x)^2/(sin(x)^4*cos(x)^2) dx

used the fact that x^m/x^n = x^(m-n)
used the fact that x^-m = 1/x^m

integral dx/(sin(x)^2*cos(x)^2)

used the fact that sin(x)^2 + cos(x)^2 = 1
solved for sin(x)^2
sin(x)^2 = 1 - cos(x)^2

integral dx/((1-cos(x)^2)*cos(x)^2)

expanded

integral dx/(cos(x)^2 - cos(x)^4)

broke into two separate integrals

integral dx/cos(x)^2 - integral dx/cos(x)^4

used fact that 1/cos(x) = sec(x)

integral sec(x)^2 dx - integral sec(x)^4 dx

begun evaluating first integral using
integral sec(x)^n = ( sec(x)^(n-2)*tan(x) )/( n-1 ) + ( n-2)/(n-1)*integral sec(x)^(n-2) dx, n =/= 1

(sec(x)^(2-2)*tan(x))/(2-1) - (2-2)/(2-1) integral sec(x)^(2-2) dx - integral sec(x)^4 dx

simplified

tan(x) - integral sec(x)^4 dx

begun evaluating second integral

tan(x) - [(sec(x)^(4-2)*tan(x))/(4-1) - integral sec(x)^2 dx]

simplified

tan(x) - (sec(x)^2*tan(x))/3 + integral sec(x)^2 dx

established already that integral sec(x)^2 dx = tan(x)

2*tan(x) - (sec(x)^2*tan(x))/3 + c

Wolfram Alpha Answer: -2 cot(2 x)+constant

MATLAB Answer: tan(x) - 1/tan(x)

Back of the Book Answer: tan(x) - cot(x) + c

My answer is not equivalent
2*tan(5) - (sec(5)^2*tan(5))/3 is about 7.243183523
tan(5) - 1/tan(5) is about -3.084702091
-2 cot(2*5) is about -3.084702091
tan(5) - cot(5) is about -3.084702091
 
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  • #2
The way you broke up the fraction is incorrect, you can't expand a minus sign in the denominator as you did. I suggest you start over, and this time instead expand the numerator with the identity csc2(x)=1+cot2(x). You will then find this problem to be significantly easier.
 
  • #3
GreenPrint said:

Homework Statement



Evaluate
integral csc(x)^4/cot(x)^2 dx

Homework Equations



The Attempt at a Solution



Apparently I'm doing something wrong, what I'm not sure, thanks for any help

My Answer: 2*tan(x) - (sec(x)^2*tan(x))/3 + c

integral csc(x)^4/cot(x)^2 dx

used fact that csc(x) = 1/sin(x)
used fact that cot(x) = 1/tan(x)
used fact that tan(x) = sin(x)/cos(x)

integral sin(x)^2/(sin(x)^4*cos(x)^2) dx

used the fact that x^m/x^n = x^(m-n)
used the fact that x^-m = 1/x^m

integral dx/(sin(x)^2*cos(x)^2)
...

At this point use the double angle formula for sine: sin(2x) = 2sin(x)*cos(x).
 

FAQ: Calculus II - Trigonometric Integrals

1. What are the main concepts covered in Calculus II - Trigonometric Integrals?

Calculus II - Trigonometric Integrals covers topics such as integration by parts, trigonometric substitutions, and partial fraction decomposition.

2. How does integration by parts work for trigonometric integrals?

Integration by parts involves using the product rule of differentiation to simplify the integrand, and then using a trigonometric identity to solve the resulting integral.

3. What is a trigonometric substitution and when should it be used?

A trigonometric substitution involves replacing a variable in an integral with a trigonometric function, such as sine or cosine. It should be used when the integral involves a radical expression or a quadratic expression.

4. How do you use partial fraction decomposition to solve trigonometric integrals?

Partial fraction decomposition is used to break down a rational function into simpler fractions, which can then be integrated. This method is particularly useful for integrals involving trigonometric functions and polynomial expressions.

5. Can you give an example of a real-world application of trigonometric integrals?

Trigonometric integrals can be used in engineering and physics to solve problems related to motion, such as finding the displacement or velocity of an object using its acceleration function.

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