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How do you integrate #(sin(2x))/(5-sin(x))^(1/2) dx#?

Answer 1

Cameron Alexander

First rewrite : #2int (cos(x)sin(x))/sqrt((5-sin(x))) dx #

Then #u = sin(x)# so #du = cos(x) dx#

Now you have #2int u/sqrt(5-u) du#

now substitute again #t = 5-u # so #dt = -du#

and # u = 5-t#

Now it's more easy : #-2int (5-t)/sqrt(t) dt#

#= -2int 5/sqrt(t)-sqrt(t) dt #

#= 2intsqrt(t) dt - 10int1/sqrt(t) dt#

#= 4/3*t^(3/2) - 20sqrt(t) + C#

Substitute back for #t = 5 - u#

#=4/3(5-u)^(3/2)-20sqrt(5-u) + C#

Again for #u = sin(x)#

#=4/3(5-sin(x))^(3/2)-20sqrt(5-sin(x)) + C#

If you so choose, you can factorize.

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Answer 2

Cameron Alexander

To integrate ( \frac{\sin(2x)}{\sqrt{5 - \sin(x)}} , dx ), you can use the substitution method. Let ( u = 5 - \sin(x) ). Then ( du = -\cos(x) , dx ).

Rewrite the integral in terms of ( u ):

[ \begin{aligned} \int \frac{\sin(2x)}{\sqrt{5 - \sin(x)}} , dx &= -2 \int \frac{1}{\sqrt{u}} , du \ &= -4 \sqrt{u} + C \end{aligned} ]

Substitute back for ( u ):

[ -4 \sqrt{5 - \sin(x)} + C ]

Where ( C ) is the constant of integration.

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Answer from HIX Tutor

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.