How do you integrate #int sin(sqrtx)# by integration by parts method?

Question

Isaiah Allen · Accepted Answer

# 2sin(sqrt(x))-2 sqrt(x) cos(sqrt(x))+C#

Making #x = y^2# in #int sin(sqrtx)dx#

after #dx = 2 y dy# we have

#int sin(sqrtx)dx equiv 2inty sin y dy#

but #d/(dy)(y cos y) = cosy -y sin y# so

#2inty sin y dy=2int cos y dy -2y cos y = 2sin y -2y cos y + C#

Finally

#int sin(sqrtx)dx = 2sin(sqrt(x))-2 sqrt(x) cos(sqrt(x))+C#

Scarlett Allison · Answer

undefined To integrate $ \int \sin(\sqrt{x}) $ by integration by parts, let's designate $ u = \sin(\sqrt{x}) $ and $ dv = dx $. Then, differentiate $ u $ and integrate $ dv $ to find $ du $ and $ v $, respectively.

$ u = \sin(\sqrt{x}) $  
$ du = \frac{1}{2\sqrt{x}}\cos(\sqrt{x})dx $

$ dv = dx $  
$ v = x $

Now, we apply the integration by parts formula:

$ \int u \, dv = uv - \int v \, du $

Substitute the values:

$ \int \sin(\sqrt{x}) \, dx = x \sin(\sqrt{x}) - \int x \left(\frac{1}{2\sqrt{x}}\cos(\sqrt{x})\right) \, dx $

The integral $ \int x \left(\frac{1}{2\sqrt{x}}\cos(\sqrt{x})\right) \, dx $ can be solved by using substitution. Let $ t = \sqrt{x} $, then $ x = t^2 $ and $ dx = 2t \, dt $. Substituting these values, we get:

$ \int x \left(\frac{1}{2\sqrt{x}}\cos(\sqrt{x})\right) \, dx = \int t^2 \cos(t) \, (2t \, dt) $

Simplify and integrate:

$ = 2\int t^3 \cos(t) \, dt $

Now, we can use integration by parts again:

$ u = t^3 $  
$ du = 3t^2 \, dt $

$ dv = \cos(t) \, dt $  
$ v = \sin(t) $

Apply integration by parts:

$ 2\int t^3 \cos(t) \, dt = 2(t^3\sin(t) - \int 3t^2 \sin(t) \, dt) $

Applying integration by parts once more to $ \int 3t^2 \sin(t) \, dt $:

$ u = 3t^2 $  
$ du = 6t \, dt $

$ dv = \sin(t) \, dt $  
$ v = -\cos(t) $

Substitute into the formula:

$ 2(t^3\sin(t) - (3t^2(-\cos(t)) - \int 6t(-\cos(t)) \, dt)) $

Simplify and integrate:

$ = 2(t^3\sin(t) + 3t^2\cos(t) - 6\int t\cos(t) \, dt) $

Applying integration by parts to $ \int t\cos(t) \, dt $:

$ u = t $  
$ du = dt $

$ dv = \cos(t) \, dt $  
$ v = \sin(t) $

Substitute into the formula:

$ = 2(t^3\sin(t) + 3t^2\cos(t) - 6(t\sin(t) - \int \sin(t) \, dt)) $

$ = 2(t^3\sin(t) + 3t^2\cos(t) - 6(t\sin(t) + \cos(t))) + C $

Finally, substituting back $ t = \sqrt{x} $, we get the integral:

$ \int \sin(\sqrt{x}) \, dx = x \sin(\sqrt{x}) + 3x\sqrt{x}\cos(\sqrt{x}) - 6\sqrt{x}\sin(\sqrt{x}) - 6\cos(\sqrt{x}) + C $