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[SOLVED] Trigonometric integrals:

MacLaddy

Member
Jan 29, 2012
52
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
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.
 

MacLaddy

Member
Jan 29, 2012
52
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
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
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.
 

MacLaddy

Member
Jan 29, 2012
52
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