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- #1

- Thread starter pape
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- Thread starter
- #1

- Feb 21, 2013

- 739

Integral belongs to calculus...Can anyone please help to solve this following integral?

∫(sin(x)-sin^2 (x)/√(sin^2 (x)+c)) dx

c is a constant

is this what you want to integrate?

\(\displaystyle \int\frac{\sin(x)-\sin^2(x)}{\sqrt{\sin^2(x)}}+c\)

Regards,

\(\displaystyle |\pi\rangle\)

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- #3

- Aug 30, 2012

- 1,279

"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

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- #5

- Aug 18, 2013

- 76

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.

- Feb 21, 2013

- 739

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

Regards,

\(\displaystyle |\pi\rangle\)

- Thread starter
- #8

Hello,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.

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

- Sep 16, 2013

- 337

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...