What is the integral of #sqrt(sinx)cosx dx#?

Answer 1

Luke Alvarez

The integral equals #2/3(sinx)^(3/2) + C#

Let #u =sinx#. Then #du= cosxdx# and #dx= (du)/(cosx)#.

#I = int sqrt(u) cosx * (du)/cosx#

#I = int sqrt(u) du#

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

#I = 2/3u^(3/2) + C#

#I = 2/3(sinx)^(3/2) + C#

Hopefully this helps!

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Answer 2

Cameron Alexander

The integral of √(sin(x))cos(x) dx can be solved using trigonometric substitution. By letting u = sin(x), the integral transforms into ∫ √(u) du. This integral can then be evaluated using standard integral rules. After finding the antiderivative, the substitution is reversed to find the final result in terms of x.

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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.