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How to Integrate Sin(Sqrt(x)) and Cos(Sqrt(x))

Updated on April 23, 2015
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TR Smith is a product designer and former teacher who uses math in her work every day.

Trigonometric functions of the form sin(x^p) and cos(x^p) are non-integrable unless the exponent p is the reciprocal of a positive integer. That is, if p = 1, 1/2, 1/3, 1/4, etc., then the antiderivative can be expressed in terms of elementary functions, otherwise not. The case when p = 1/2 gives you the functions sin(sqrt(x)) and cos(sqrt(x)). These functions are the primitive solutions to the differential equation

y'' + y'/(2x) + y/(4x) = 0

The general solution of this differential equation is y = A*sin(sqrt(x)) + B*cos(sqrt(x)), where A and B are arbitrary constants. To integrate the functions sin(√x) and cos(√x), you must use a combination of u-substitution and integration by parts. Here we show how to find the antiderivatives of sin(√x) and cos(√x) with some examples.

See also How to Integrate Sqrt(Sin(x)) and Sqrt(Cos(x)).

Graphs of y = sin(sqrt(x)) in red and y = cos(sqrt(x)) in blue.
Graphs of y = sin(sqrt(x)) in red and y = cos(sqrt(x)) in blue.

Antiderivative of Sin(Sqrt(x))

The first step is to make the substitution of variables x = u^2 and dx = 2u du. This transforms our integral into

∫ sin(sqrt(x)) dx
= ∫ 2u*sin(u) du

Next, recall the integration by parts formula:

∫ f * dg = fg - ∫ g * df

If we take f = 2u and dg = sin(u) du, we get df = 2 du and g = -cos(u). This gives us

∫ 2u*sin(u) du
= -2u*cos(u) + ∫ 2*cos(u) du
= -2u*cos(u) + 2*sin(u) + c

And now we make the reverse substitution u = sqrt(x), which gives us the final expression for the antiderivative of sin(sqrt(x)):

∫ sin(sqrt(x)) dx
= -2*sqrt(x)*cos(sqrt(x)) + 2*sin(sqrt(x)) + c
= -2√x * cos(√x) + 2 * sin(√x) + c


Antiderivative of Cos(Sqrt(x))

As before, the first step in integrating cos(sqrt(x)) is t0 make the substitution x = u^2 and dx = 2u du. This transforms our integral into

∫ cos(sqrt(x)) dx
= ∫ 2u*cos(u) du

And again, we use the integration by parts formula

∫ f * dg = fg - ∫ g * df

this time using the assignment f = 2u and dg = cos(u) du. This in turn gives us df = 2 du and g = sin(u). Now we have

∫ 2u*cos(u) du
= 2u*sin(u) - ∫ 2*sin(u) du
= 2u*sin(u) + 2*cos(u) + c

And finally we make the reverse substitution u = sqrt(x), which gives us the equation for the antiderivative of cos(sqrt(x)):

∫ cos(sqrt(x)) dx
= 2*sqrt(x)*sin(sqrt(x)) + 2*cos(sqrt(x)) + c
= 2√x * sin(√x) + 2 * cos(√x) + c

Example 1

Find the area under the curve y = sin(sqrt(x)) from x = 0 to x = π^2. A graph of the function y = sin(sqrt(x)) is shown below with the region in question shaded in pink.

Since we know the antiderivative of sin(sqrt(x) is

-2*sqrt(x)*cos(sqrt(x)) + 2*sin(sqrt(x))

we simply need to plug in the limits of integration x = π^2 and x = 0 and subtract. This gives us

[-2*sqrt(π^2)*cos(sqrt(π^2)) + 2*sin(sqrt(π^2))] - [-2*0*cos(0) + 2*sin(0)]
= -2π(-1) + 2(0)
= 2π


Example 2

Find the area under the curve y = cos(2*sqrt(x)) from x = 1 to x = 3. This region is shaded in the graph below.

If we use the substitution x = w/4 and dx = (1/4) dw, we get the integral equation

∫ cos(2*sqrt(x)) dx
= (1/4) * ∫ cos(sqrt(w)) dw
= (1/2)*sqrt(w)*sin(sqrt(w)) + (1/2)*cos(sqrt(w))
= (1/2)*sqrt(4x)*sin(sqrt(4x)) + (1/2)*cos(sqrt(4x))
= sqrt(x)*sin(2*sqrt(x)) + (1/2)*cos(2*sqrt(x))

Plugging in the endpoints x = 3 and x = 1 and subtracting gives us

sqrt(3)*sin(2*sqrt(3)) + (1/2)*cos(2*sqrt(3)) - sin(2) - (1/2)*cos(2)

≈ -1.7244


Alternative Integration Method: Easier or Harder?

Here's an alternative method of integration that uses integration by parts first, and then substitution. We rewrite the integral in an equivalent form as

∫ sin(sqrt(x)) dx = ∫ [ sqrt(x) ] [ sin(sqrt(x)) / sqrt(x) ] dx

Now we let f = sqrt(x) and dg = sin(sqrt(x))/sqrt(x) dx. The choice of dg is because the expression sin(sqrt(x))/sqrt(x) is recognizable as the derivative of -2*cos(sqrt(x)) if you mentally apply substitution. So we have df = 0.5*sqrt(x) dx, and g = -2*cos(sqrt(x)). Applying the integration by parts formula gives us

∫ sin(sqrt(x)) dx = ∫ f * dg = f * g - ∫ g * df
= -2*sqrt(x)*cos(sqrt(x)) + ∫ cos(sqrt(x))/sqrt(x) dx

And now applying substitution once more gives us

-2*sqrt(x)*cos(sqrt(x)) + ∫ cos(sqrt(x))/sqrt(x) dx
= -2*sqrt(x)*cos(sqrt(x)) + 2*sin(sqrt(x)) + C.

We get the same answer as before, but using a different method. Whether this alternative integration method is easier or harder depends on the individual.


Integrating sin(x^(1/n)) and cos(x^(1/n))

To integrate sin(x^(1/n)) and cos(x^(1/n)), that is the sine and cosine of the nth root of x, you make the substitution

x = u^n, dx = n*u^(n-1),

which gives you the integral transformations

∫ sin(x^(1/n)) dx = n * ∫ sin(u) * u^(n-1) du

∫ cos(x^(1/n)) dx = n * ∫ cos(u) * u^(n-1) du

which can then be reduced with n-1 iterations of integration by parts. For example, applying this method to integrate sin(x^(1/3)) = sin(cbrt(x)), i.e., sine of the cube root of x, gives you the antiderivative formula

∫ sin(x^(1/3)) dx = 6*sin(x^(1/3))*x^(1/3) - 3*[x^(2/3) - 2]*cos(x^(1/3)) + c

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    • pstraubie48 profile image

      Patricia Scott 3 years ago from sunny Florida

      I do so admire you for understanding and sharing this. I am challenged in this area. About 30 years ago a friend of mine talked me into taking a class on this topic one summer. Keep in mind I had no math background beyond algebra. I got a C in the class and felt as if it equaled any A ever got. Bless you.

      Angels are on the way to you this evening.

    • calculus-geometry profile image
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      TR Smith 3 years ago

      Thanks pstraubie!

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      c.andrew 3 years ago

      it's helpful to see the alternative method as well. almost all the problems on our practice midterm have to be solved with integration by parts and substitution combined. :/

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