# How do you integrate #int (sin3x)/sqrt(5+cos3x)# using substitution?

Generally, when we have integrals where two trigonometric functions are being divided, we should tend to substitute them such that the numerator or denominator cancels out.

We have:

Thus, our integral is equal to:

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To integrate ( \int \frac{\sin(3x)}{\sqrt{5+\cos(3x)}} ) using substitution, let:

[ u = 5 + \cos(3x) ]

Then, differentiate ( u ) with respect to ( x ):

[ \frac{du}{dx} = -3\sin(3x) ]

Solve for ( dx ):

[ dx = -\frac{1}{3\sin(3x)} , du ]

Now, substitute ( u ) and ( dx ) into the integral:

[ \int \frac{\sin(3x)}{\sqrt{5+\cos(3x)}} , dx = \int \frac{\sin(3x)}{\sqrt{u}} \left(-\frac{1}{3\sin(3x)} \right) , du ]

[ = -\frac{1}{3} \int \frac{1}{\sqrt{u}} , du ]

This integral can be easily integrated:

[ = -\frac{2}{3} \sqrt{u} + C ]

Finally, substitute ( u ) back in terms of ( x ):

[ = -\frac{2}{3} \sqrt{5 + \cos(3x)} + C ]

So, ( \int \frac{\sin(3x)}{\sqrt{5+\cos(3x)}} , dx = -\frac{2}{3} \sqrt{5 + \cos(3x)} + C ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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