Posted by W on Monday, January 2, 2012 at 8:15pm.
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!
- Calculus - Damon, Monday, January 2, 2012 at 8:28pm
when all else fails
go to wolfram alpha (Just type that in your browser)
in the box type:
after you see the answer, click on "show steps" in the box on the upper right.
- Calculus - Damon, Monday, January 2, 2012 at 8:30pm
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
- Calculus - Damon, Monday, January 2, 2012 at 8:31pm
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, Monday, January 2, 2012 at 8:33pm
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
check: take the derivative...
-1cossqrtx/2sqrx+ -sqrtx(-sinsqrtx)1/2sqrx +cossqurx/2sqrtx
1/2 sinSqrtx so it checks.
- Calculus - W, Monday, January 2, 2012 at 8:45pm
very cool. Thanks, both of you! Does wolfram alpha give steps for all integration problems?
- Calculus - Damon, Monday, January 2, 2012 at 8:51pm
Pretty much so
your other one would be
integrate 1/(x ln x)
- Calculus - W, Monday, January 2, 2012 at 9:13pm
awesome! Thanks Damon ;D
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