Asked by W

Integrate (1/2)sin(x^(1/2))dx.

I've tried using u-substitution, 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!

Answers

Answered by 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.
Answered by 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) du-u 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
Answered by Damon
Obviously it does not copy and paste very well so go to the original. Your substitution approach was in fact what was used.
Answered by 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.

Answered by W
very cool. Thanks, both of you! Does wolfram alpha give steps for all integration problems?
Answered by Damon
Pretty much so
your other one would be
integrate 1/(x ln x)
Answered by W
awesome! Thanks Damon ;D
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