Posted by W on .
Integrate (1/2)sin(x^(1/2))dx.
I've tried using usubstitution, with u=sin(x^(1/2)). du would then = ((1/2)x^(1/2))(cos(x^(1/2))). As you can see, this only make the problem more complicated. I don't get what to do. Thank you in advance!

Calculus 
Damon,
when all else fails
go to wolfram alpha (Just type that in your browser)
in the box type:
Integrate (1/2)sin(x^(1/2))
after you see the answer, click on "show steps" in the box on the upper right. 
Calculus 
Damon,
integral (sin(sqrt(x)))\/2 dx = sin(sqrt(x))sqrt(x) cos(sqrt(x))+constant
Possible intermediate steps:\n integral (sin(sqrt(x)))\/2 dx\nFactor out constants:\n = 1\/2 integral sin(sqrt(x)) dx\nFor the integrand sin(sqrt(x)), substitute u = sqrt(x) and du = 1\/(2 sqrt(x)) dx:\n = integral u sin(u) du\nFor the integrand u sin(u), integrate by parts, integral f dg = f g integral g df, where \n f = u, dg = sin(u) du,\n df = du, g = cos(u):\n = integral cos(u) duu cos(u)\nThe integral of cos(u) is sin(u):\n = sin(u)u cos(u)+constant\nSubstitute back for u = sqrt(x):\n = sin(sqrt(x))sqrt(x) cos(sqrt(x))+constant 
Calculus 
Damon,
Obviously it does not copy and paste very well so go to the original. Your substitution approach was in fact what was used.

Calculus 
bobpursley,
let u= sqrtx
du= 1/2*1/sqrtx dx
dx= 2u du
INT 1/2 sin (sqrtx) dx
INT 1/2 sin u 2u du
int u sin u du=u cos u + sin u
then change it back to x
sqrtx cosSqrtx+sinsqrtx
check: take the derivative...
1cossqrtx/2sqrx+ sqrtx(sinsqrtx)1/2sqrx +cossqurx/2sqrtx
1/2(sinSqrtx)
1/2 sinSqrtx so it checks. 
Calculus 
W,
very cool. Thanks, both of you! Does wolfram alpha give steps for all integration problems?

Calculus 
Damon,
Pretty much so
your other one would be
integrate 1/(x ln x) 
Calculus 
W,
awesome! Thanks Damon ;D