Question

In Exercises \(43-46\) , evaluate the integral by using a substitution prior to integration by parts. $$\int \sin \sqrt{x} d x$$

Short answer

The integral of \(\sin\sqrt{x} dx\) equals to \(-\sqrt{x}\cos(\sqrt{x}) + \sin(\sqrt{x}) + C\).

Step-by-step solution

Substitution

Firstly, we start with a substitution to simplify the function. Let \( u = \sqrt{x} \). Therefore, \( x = u^2 \) and \( dx = 2u du \).

Re-write the Integral

Re-write the integral in terms of \( u \). Now the integral becomes: \[ \int \sin(u) \cdot 2u \cdot du \]

Apply Integration by Parts

Now the integral becomes a product of two functions \( u \) and \( \sin(u) \) and can be solved by integration by parts. The formula is given by: \[ \int u dv = uv - \int v du \] Here, let \( dv = \sin(u) du \) and \( u = u \). Then \( v = -\cos(u) \) and \( du = du \). Substituting back to equation: \[ uv - \int v du = u \cdot -\cos(u) - \int -\cos(u) du \] Simplifying, we get: \[ -u \cos(u) + \int \cos(u) du \]

Evaluate the Integral and Substitute Back

\[ -u \cos(u) + \int \cos(u) du \] becomes \[ -u \cos(u) + \sin(u) \]. Now we substitute back \( u = \sqrt{x} \). Therefore: \[-\sqrt{x} \cos(\sqrt{x}) + \sin(\sqrt{x}) + C\]. \( C \) is the constant of integration.