MHB Solving a Trigonometric Integral

[yanic]

Can anyone please help to solve this following integral?
∫(sin(x)-sin^2 (x)/√(sin^2 (x)+c)) dx
c is a constant

 

[Petrus]

Re: integral

Can anyone please help to solve this following integral?
∫(sin(x)-sin^2 (x)/√(sin^2 (x)+c)) dx
c is a constant

Integral belongs to calculus...
is this what you want to integrate?
$$\int\frac{\sin(x)-\sin^2(x)}{\sqrt{\sin^2(x)}}+c$$

Regards,
$$|\pi\rangle$$

 

[yanic]

Re: integral

View attachment 1302
Please check the JPG file for the correct expression of
the function.

Best...

 

[topsquark]

Re: integral

https://www.physicsforums.com/attachments/1302
Please check the JPG file for the correct expression of
the function.

Best...

"And what I have found so far is" What is "what you have found"? How did you get this line? What does it mean?

How does the derivative relate to the given problem?

Please be more specific.

-Dan

 

[yanic]

Re: integral

"And what I have found so far is" What is "what you have found"? How did you get this line? What does it mean?

How does the derivative relate to the given problem?

Please be more specific.

-Dan

View attachment 1303

Please check the attached file.
I hope that I have been more specific this time.

Regards

 

[eddybob123]

Re: integral

You can start by substituting $u=\sin(x)$ to obtain
$$\int u-\frac{u^2}{\sqrt{u^2+c}}\, du$$
From there you can try integration by parts.

 

[Petrus]

Re: integral

You can start by substituting $u=\sin(x)$ to obtain
$$\int u-\frac{u^2}{\sqrt{u^2+c}}\, du$$
From there you can try integration by parts.

If $$u= \sin(x)$$ Then $$du=\cos(x)\,dx$$ which is not same as that function he wants to integrate?

Regards,
$$|\pi\rangle$$

 

[yanic]

Re: integral

You can start by substituting $u=\sin(x)$ to obtain
$$\int u-\frac{u^2}{\sqrt{u^2+c}}\, du$$
From there you can try integration by parts.

Hello,
Integration by part leads to a more complicated expression at the end for the finding of the primitive.
Do you have any other suggestion please?

Regards

 

[DreamWeaver]

Re: integral

Hello,
Integration by part leads to a more complicated expression at the end for the finding of the primitive.
Do you have any other suggestion please?

Regards

Unfortunately, that's because the primitive is quite complex...

Incidentally, if you apply Eddybob's method, but set $$c=b^2\,$$, so that $$b=\sqrt{c}\,$$, you can take the constant outside of the square root... ;)