# [SOLVED]Trigonometric integrals:

##### Member
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
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

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

Kind Regards,
Sudharaka.

##### Member

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

Kind Regards,
Sudharaka.

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

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.

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, $$\cos(2x)=1-2\sin^{2}(x)$$.
$$\displaystyle \frac{cos(2x)-7}{3\sqrt{sin(x)}}+C$$