How do you find the indefinite integral of #int cosx/sqrtsinx#?

Answer 1

William Abdalla

Let #u = sinx#.

#du = cosx(dx)#

#dx = (du)/cosx#

#=int(cosx/u^(1/2))dx#

#=int(cosx/u^(1/2))((du)/cosx)#

#=int(u^(-1/2))du#

#=2u^(1/2) + C#

#=2(sinx)^(1/2) + C#

#=2sqrt(sinx) + C#

You can check your result by differentiating. You will get the initial function

Hopefully this helps!

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Answer 2

Isabella Ackerman

To find the indefinite integral of ∫cos(x)/√sin(x), you can use the substitution method. Let ( u = \sqrt{\sin(x)} ), then ( du = \frac{1}{2\sqrt{\sin(x)}}\cos(x)dx ). Now, rearrange to solve for ( dx ). Substituting ( du = \frac{1}{2\sqrt{\sin(x)}}\cos(x)dx ) into the original integral, we get ( 2\int du ). This simplifies to ( 2u + C ), where ( C ) is the constant of integration. Finally, replace ( u ) with ( \sqrt{\sin(x)} ) to get the final answer: ( 2\sqrt{\sin(x)} + C ).

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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.