# Solving a Trigonometric Integral

#### pape

##### New member
∫(sin(x)-sin^2 (x)/√(sin^2 (x)+c)) dx
c is a constant

#### Petrus

##### Well-known member
Re: 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?
$$\displaystyle \int\frac{\sin(x)-\sin^2(x)}{\sqrt{\sin^2(x)}}+c$$

Regards,
$$\displaystyle |\pi\rangle$$

#### pape

##### New member
Re: integral

Please check the JPG file for the correct expression of
the function.

Best...

Last edited:

#### topsquark

##### Well-known member
MHB Math Helper
Re: integral

View attachment 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?

-Dan

#### pape

##### New member
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?

-Dan

I hope that I have been more specific this time.

Regards

#### eddybob123

##### Active member
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

##### Well-known member
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 $$\displaystyle u= \sin(x)$$ Then $$\displaystyle du=\cos(x)\,dx$$ which is not same as that function he wants to integrate?

Regards,
$$\displaystyle |\pi\rangle$$

#### pape

##### New member
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

##### Well-known member
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 $$\displaystyle c=b^2\,$$, so that $$\displaystyle b=\sqrt{c}\,$$, you can take the constant outside of the square root...