Can You Help Me Solve This Trigonometric Integral?

In summary: Everything you have done is correct. The answer that Wolfram gives you can be obtained from your answer. Try to simplify your solution, by taking a common denominator. You will need to use the double angle formula, \(\cos(2x)=1-2\sin^{2}(x)\).Kind Regards,Sudharaka.Everything you have done is correct. The answer that Wolfram gives you can be obtained from your answer. Try to simplify your solution, by taking a common denominator. You will need to use the double angle formula, \(\cos(2x)=1-2\sin^{2}(x)\).Kind Regards,Sudharaka.
  • #1
MacLaddy1
52
0
Hello again. I am hoping that someone can assist in checking my work regarding a trigonometric integral.

The problem and my attempt to solve is as follows.

\(\displaystyle \int\sin^{\frac{-3}{2}}(x)*cos^3(x) dx\)

\(\displaystyle \int\sin^{\frac{-3}{2}}(x)*cos^2(x)*cos(x) dx\)

Using a Pythagorean identity,

\(\displaystyle \int\sin^{\frac{-3}{2}}(x)*(1-sin^2(x))*cos(x) dx\)

Distributing

\(\displaystyle \int[\sin^{\frac{-3}{2}}(x)-sin^{-3}(x)]*cos(x)dx\)

Substituting

U=sin(x) du=cos(x)dx

\(\displaystyle \int [u^{\frac{-3}{2}}-u^{-3}]du\)

Integrating

\(\displaystyle -2u^{\frac{-1}{2}}+\frac{1}{2}u^{-2}\)

And finally, and simplified as I can see,

\(\displaystyle -\frac{2}{\sqrt{\sin(x)}}+\frac{1}{2\sin^2(x)}+C\)

I've looked at this a few times and couldn't find any errors, but Wolfram is coming up with an answer that doesn't Jive with what I have. That being the case, I would really appreciate it if someone could take a look at this problem and let me know where I went wrong.

Thanks much,
Mac
 
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  • #2
MacLaddy said:
Hello again. I am hoping that someone can assist in checking my work regarding a trigonometric integral.

The problem and my attempt to solve is as follows.

\(\displaystyle \int\sin^{\frac{-3}{2}}(x)*cos^3(x) dx\)

\(\displaystyle \int\sin^{\frac{-3}{2}}(x)*cos^2(x)*cos(x) dx\)

Using a Pythagorean identity,

\(\displaystyle \int\sin^{\frac{-3}{2}}(x)*(1-sin^2(x))*cos(x) dx\)

Distributing

\(\displaystyle \color{red}{\int[\sin^{\frac{-3}{2}}(x)-sin^{-3}(x)]*cos(x)dx}\)

Substituting

U=sin(x) du=cos(x)dx

\(\displaystyle \int [u^{\frac{-3}{2}}-u^{-3}]du\)

Integrating

\(\displaystyle -2u^{\frac{-1}{2}}+\frac{1}{2}u^{-2}\)

And finally, and simplified as I can see,

\(\displaystyle -\frac{2}{\sqrt{\sin(x)}}+\frac{1}{2\sin^2(x)}+C\)

I've looked at this a few times and couldn't find any errors, but Wolfram is coming up with an answer that doesn't Jive with what I have. That being the case, I would really appreciate it if someone could take a look at this problem and let me know where I went wrong.

Thanks much,
Mac

Hi MacLaddy,

The highlighted part in incorrect. You have multiplied the exponents of \(\sin(x)\) instead of adding them.

Kind Regards,
Sudharaka.
 
  • #3
Sudharaka said:
Hi MacLaddy,

The highlighted part in incorrect. You have multiplied the exponents of \(\sin(x)\) instead of adding them.

Kind Regards,
Sudharaka.

Ah, yes. Good old Algebra getting in the way. Okay, how about this.

Distributing

\(\displaystyle \int[\sin^{\frac{-3}{2}}(x)-sin^{\frac{1}{2}}(x)]*cos(x)dx\)

Substituting

U=sin(x) du=cos(x)dx

\(\displaystyle \int [u^{\frac{-3}{2}}-u^{\frac{1}{2}}]du\)

Integrating

\(\displaystyle -2u^{\frac{-1}{2}}-\frac{2}{3}u^{\frac{3}{2}}\)

And finally, and simplified as I can see,

\(\displaystyle -\frac{2}{\sqrt{\sin(x)}}-\frac{2\sqrt{sin^3(x)}}{3}+C\)

This still doesn't match what Wolfram has, but I'm probably getting something wrong with that.

Let me know if this looks right. My brain isn't functioning at 100% this morning, and most of it's limited power is going into keeping the Latex code straight.

I appreciate the help, Sudharaka

Mac
 
  • #4
MacLaddy said:
Ah, yes. Good old Algebra getting in the way. Okay, how about this.

Distributing

\(\displaystyle \int[\sin^{\frac{-3}{2}}(x)-sin^{\frac{1}{2}}(x)]*cos(x)dx\)

Substituting

U=sin(x) du=cos(x)dx

\(\displaystyle \int [u^{\frac{-3}{2}}-u^{\frac{1}{2}}]du\)

Integrating

\(\displaystyle -2u^{\frac{-1}{2}}-\frac{2}{3}u^{\frac{3}{2}}\)

And finally, and simplified as I can see,

\(\displaystyle -\frac{2}{\sqrt{\sin(x)}}-\frac{2\sqrt{sin^3(x)}}{3}+C\)

This still doesn't match what Wolfram has, but I'm probably getting something wrong with that.

Let me know if this looks right. My brain isn't functioning at 100% this morning, and most of it's limited power is going into keeping the Latex code straight.

I appreciate the help, Sudharaka

Mac

Everything you have done is correct. The answer that Wolfram gives you can be obtained from your answer. Try to simplify your solution, by taking a common denominator. You will need to use the double angle formula, \(\cos(2x)=1-2\sin^{2}(x)\).

Kind Regards,
Sudharaka.
 
  • #5
Sudharaka said:
Everything you have done is correct. The answer that Wolfram gives you can be obtained from your answer. Try to simplify your solution, by taking a common denominator. You will need to use the double angle formula, \(\cos(2x)=1-2\sin^{2}(x)\).

Kind Regards,
Sudharaka.

Okay, I got it now. I wasn't seeing the double angle connection.

\(\displaystyle \frac{cos(2x)-7}{3\sqrt{sin(x)}}+C\)

Thanks again, Sudharaka.

Mac
 

FAQ: Can You Help Me Solve This Trigonometric Integral?

What are trigonometric integrals?

Trigonometric integrals are mathematical expressions that involve both trigonometric functions (such as sine, cosine, and tangent) and variables. They are used to calculate the area under a curve in trigonometric functions.

What is the process for solving trigonometric integrals?

The process for solving trigonometric integrals involves using trigonometric identities and integration techniques, such as substitution, integration by parts, and partial fractions. It also requires knowledge of the properties and rules of trigonometric functions.

Why are trigonometric integrals important?

Trigonometric integrals are important in mathematics, physics, engineering, and other fields as they allow us to solve problems involving trigonometric functions. They are also used in real-world applications, such as calculating the motion of objects in circular or periodic motion.

What are some common trigonometric identities used in solving integrals?

Some common trigonometric identities used in solving integrals include the Pythagorean identities, double angle identities, and power-reducing identities. These identities help simplify the integrals and make them easier to solve.

What are some tips for solving trigonometric integrals?

Some tips for solving trigonometric integrals include being familiar with trigonometric identities and their properties, using substitution and other integration techniques, and practicing regularly to improve problem-solving skills.

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