For ∫ sin(√x) dx
let y = √x = x1/2
→ dy = 1/2 x-1/2 dx
→ 2x1/2 dy = dx
→ 2y dy = dx
→ ∫ sin(x1/2) dx = ∫(sin y) 2y dy
Now:
∫ uv dx = u∫v dx - ∫(u'∫v dx) dx
→ ∫(sin y) 2y dy = ∫2y sin y dy
= 2y ∫sin y dy - ∫(2 ∫sin y dy) dy
= -2y cos y + 2 sin y + C
= 2 sin y - 2y cos y + C
→ ∫ sin(√x) dx = 2 sin(√x) - 2(√x) cos(√x) + C
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The integral of root(sin(x)) is -2 time the elliptic integral of the second order of .25(pi-2x) at 2. For this and other integrals, go to http://integrals.wolfram.com/index.jsp?expr=sqrt(sin(x))&random=false For more information on the elliptic integral functions, go to http://en.wikipedia.org/wiki/Elliptic_integral Hope this helps!
x/sqrt(x)=sqrt(x) integral sqrt(x)=2/3x3/2
The square root of the fifth root of x is the tenth root of x.
Square root of x = x^0.5
Integral[sin(x)cos(x)sin2(x)cos3(x)] dxgather termsintegral[sin3(x) cos4(x)] dxpull one sin(x) as sin is oddintegral[sin2(x) cos4(x) sin(x)] dxusing trig identitiesintegral[(1 - cos2(X)) cos4(x) sin(X)] dxu substitutionu = cos(x)du = - sin(x) dxsointegral[(1 - u2)) (u4) - du] dx- integral[(1- u2)) u4 du] dx= u - 1/3u3 + 1/5u5 du= cos(x) - 1/3cos3(x) + 1/5cos5(x) + C============================