1. **Problem Statement:** Evaluate the integrals from 11 to 15 using integration techniques. 2. **Integral 11: \(\int \cos \sqrt{x} \, dx\)** - Use substitution: let \(u = \sqrt{x} = x^{1/2}\), then \(x = u^2\) and \(dx = 2u \, du\). - Rewrite integral: \(\int \cos u \cdot 2u \, du = 2 \int u \cos u \, du\). - Use integration by parts with \(v = u\), \(dw = \cos u \, du\). - Then \(dv = du\), \(w = \sin u\). - So, \(2 (u \sin u - \int \sin u \, du) = 2 (u \sin u + \cos u) + C\). - Substitute back \(u = \sqrt{x}\): $$\int \cos \sqrt{x} \, dx = 2 \sqrt{x} \sin \sqrt{x} + 2 \cos \sqrt{x} + C$$ 3. **Integral 12: \(\int (1 + 2x)^x e^x \, dx\)** - This integral is complex and does not have a standard elementary antiderivative. - The provided answer is \(x e^{x^2} + C\), which suggests a possible typo or approximation. - Without further context, this integral cannot be solved by elementary methods here. 4. **Integral 13: \(\int \sin^3 x \cos^2 x \, dx\)** - Use power-reduction and substitution: - Write \(\sin^3 x = \sin x (\sin^2 x) = \sin x (1 - \cos^2 x)\). - So integral becomes \(\int \sin x (1 - \cos^2 x) \cos^2 x \, dx = \int \sin x (\cos^2 x - \cos^4 x) \, dx\). - Substitute \(u = \cos x\), \(du = -\sin x \, dx\). - Integral becomes \(-\int (u^2 - u^4) \, du = -\left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C\). - Substitute back \(u = \cos x\): $$\int \sin^3 x \cos^2 x \, dx = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$$ 5. **Integral 14: \(\int \sin^2 x \cos x \, dx\)** - Use substitution: let \(u = \sin x\), then \(du = \cos x \, dx\). - Integral becomes \(\int u^2 \, du = \frac{u^3}{3} + C = \frac{\sin^3 x}{3} + C\). 6. **Integral 15: \(\int \sqrt{1 + \cos 4x} \, dx\)** - Use identity: \(1 + \cos 4x = 2 \cos^2 2x\). - So integral becomes \(\int \sqrt{2} |\cos 2x| \, dx\). - Assuming \(\cos 2x \geq 0\) in the interval, integral is \(\sqrt{2} \int \cos 2x \, dx = \sqrt{2} \cdot \frac{\sin 2x}{2} + C = \frac{\sqrt{2}}{2} \sin 2x + C\). **Final answers:** $$\int \cos \sqrt{x} \, dx = 2 \sqrt{x} \sin \sqrt{x} + 2 \cos \sqrt{x} + C$$ $$\int \sin^3 x \cos^2 x \, dx = -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$$ $$\int \sin^2 x \cos x \, dx = \frac{\sin^3 x}{3} + C$$ $$\int \sqrt{1 + \cos 4x} \, dx = \frac{\sqrt{2}}{2} \sin 2x + C$$ Integral 12 is not solved here due to complexity and possible typo.