Home > Calculus > Integration by Substitution

How do you integrate #int cosxsqrtsinx# using substitution?

Answer 1

William Abdalla

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

#=int(cosxsqrt(u))/cosx du#

#=sqrt(u) du#

#= u^(1/2) du#

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

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

#=2/3sqrt(sin^3x) + C#

#=2/3sinxsqrt(sinx) + C#

Hopefully this helps!

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Luke Alvarez

To integrate ( \int \cos(x) \sqrt{\sin(x)} ) using substitution:

Let ( u = \sin(x) ), then ( du = \cos(x) dx ).
Rewrite the integral in terms of ( u ): [ \int \sqrt{u} , du ]
Integrate ( \sqrt{u} ) with respect to ( u ): [ \frac{2}{3} u^{3/2} + C ]
Substitute back for ( u ): [ \frac{2}{3} \sin^{3/2}(x) + C ]

By signing up, you agree to our Terms of Service and Privacy Policy

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.