Question

Evaluate \(\int \cos \sqrt{x} d x\) using a substitution followed by integration by parts.

Short answer

Question: Evaluate the integral: \(\int x\cos(\sqrt{x}) dx\) Answer: \(-\frac{1}{24}x^2 \sin(\sqrt{x}) - \frac{1}{48}x^{3/2} \cos(\sqrt{x}) + C\)

Step-by-step solution

Choose a substitution

Let \(u = \sqrt{x}\). Then, \(u^2 = x\), and \(d u = \frac{1}{2} x^{-1/2} d x\). We substitute these expressions into the integral to make it simpler.

Perform the substitution

Now, we substitute \(u\), \(x\), and \(d x\) into the integral: $$\int \cos (u) (\frac{1}{2}u^2) d u$$

Choose functions for integration by parts

To apply integration by parts, we need to choose two functions so that one of them is differentiated and the other is integrated. Let \(v = \cos(u)\) and \(w = \frac{1}{2}u^2\). Then, we differentiate \(v\) and integrate \(w\): \(v' = -\sin(u)\) \(w = \int \frac{1}{2}u^2 d u = \frac{1}{6}u^3\)

Apply integration by parts

Now, we will apply the integration by parts formula: $$\int v'w du = vw - \int vw' du $$ We can plug in our expressions for \(v\), \(v'\), and \(w\): $$-\int \sin(u) \frac{1}{6}u^3 d u = \frac{1}{6}u^3 \cos(u) - \frac{1}{6} \int u^3 (-\cos(u)) d u$$

Integrate the remaining integral

We can now integrate the remaining integral inside the expression: $$-\frac{1}{6} \int u^3 \cos(u) d u = -\frac{1}{24}u^4 \sin(u) - \frac{1}{48}u^3 \cos(u) + C$$

Substitute back for x

Finally, we can substitute back for \(x\): $$-\frac{1}{24}\sqrt{x}^4 \sin(\sqrt{x}) - \frac{1}{48}\sqrt{x}^3 \cos(\sqrt{x}) + C$$ Simplifying, we get the final answer: $$-\frac{1}{24}x^2 \sin(\sqrt{x}) - \frac{1}{48}x^{3/2} \cos(\sqrt{x}) + C$$