Question

Two of these three integrals are elementary; evaluate them (A) \(\int \sin ^{2} x d x\) (B) \(\int \sin \sqrt{x} d x\)\text { (C) } \int \sin x^{2} d x

Short answer

Question: Identify the two elementary integrals and provide their solutions. Answer: The two elementary integrals are (A) \(\int \sin^2{x} dx\) and (B) \(\int \sin{(\sqrt{x})} dx\). Their solutions are: (A) \(\int \sin^2{x} dx = \frac{1}{2}(x - \frac{1}{2}\sin{2x}) + C\). (B) \(\int \sin{(\sqrt{x})} dx = -2x\cos{(\sqrt{x})} + 2\sin{(\sqrt{x})} + C\).

Step-by-step solution

Use the double-angle identity for cosine

We can rewrite \(\sin^2{x}\) using the double-angle identity for cosine: \(\sin^2{x} = \frac{1}{2}(1 - \cos{2x})\). Now, the integral becomes: \(\int \sin^2{x} dx = \int \frac{1}{2}(1 - \cos{2x}) dx\).

Integrate the function

Integrate the function term by term. \(\int \frac{1}{2}(1 - \cos{2x}) dx = \frac{1}{2} \int (1 - \cos{2x}) dx = \frac{1}{2} \int 1 dx - \frac{1}{2} \int \cos{2x} dx\).

Find the antiderivatives

Now, we need to find the antiderivatives for each integral. The antiderivative of \(1\) with respect to \(x\) is just \(x\), and the antiderivative of \(\cos{2x}\) with respect to \(x\) is \(\frac{1}{2}\sin{2x}\). So the overall antiderivative is: \(\frac{1}{2}(x - \frac{1}{2}\sin{2x}) + C\), where C is the constant of integration. Therefore, \(\int \sin^2{x} dx = \frac{1}{2}(x - \frac{1}{2}\sin{2x}) + C\). Integral (B): \(\int \sin{(\sqrt{x})} dx\)

Use substitution

We can use substitution to solve this integral. Let \(u = \sqrt{x}\), then \(x = u^2\), and \(dx = 2u du\). Now, we substitute these values in the integral, which becomes: \(\int \sin{(\sqrt{x})} dx = \int \sin{u} \cdot 2u du\).

Integrate using integration by parts

To integrate this, we can use integration by parts. Let \(dv = \sin{u} du\), and \(v = -\cos{u}\). Let \(u = u\), and \(du = du\). Therefore, by integration by parts, we have: \(\int u\sin{u} du = -u\cos{u} - \int(-\cos{u}) du = -u\cos{u} + \int \cos{u} du\) Now, the antiderivative of \(\cos{u}\) is \(\sin{u}\). So, the integral becomes: \(\int u\sin{u} du = -u\cos{u} + \sin{u} + C_1\)

Substitute back the original variable

Now, we need to substitute back the original variable \(x\). We know that \(u = \sqrt{x}\). So, we have: \(-u\cos{u} + \sin{u} + C_1 = -\sqrt{x}\cos{(\sqrt{x})} + \sin{(\sqrt{x})} + C_1\) Multiplying by the factor we got when substituting for \(x\), we obtain the final answer: \(\int \sin{(\sqrt{x})} dx = -2x\cos{(\sqrt{x})} + 2\sin{(\sqrt{x})} + C\). So, to summarize the solutions: Integral (A): \(\int \sin^2{x} dx = \frac{1}{2}(x - \frac{1}{2}\sin{2x}) + C\). Integral (B): \(\int \sin{(\sqrt{x})} dx = -2x\cos{(\sqrt{x})} + 2\sin{(\sqrt{x})} + C\).