Evaluate the following: 1. ∫√7x−1 dx 2. ∫sin(4πx + 7) dx 3. ∫√sin x cos x dx 4. ∫x cos x dx 5. ∫e^x sin x dx

Here are the evaluations of the integrals:

1. For $\int \sqrt{7x-1}\,dx$: Step 1: Let $u = 7x-1$. Then $du = 7\,dx$, which means $dx = \frac{1}{7}\,du$. Step 2: Substitute $u$ into the integral. $\int \sqrt{u} \cdot \frac{1}{7}\,du = \frac{1}{7} \int u^{1/2}\,du$ Step 3: Integrate $u^{1/2}$. $\frac{1}{7} \cdot \frac{u^{3/2}}{3/2} + C = \frac{1}{7} \cdot \frac{2}{3} u^{3/2} + C = \frac{2}{21} u^{3/2} + C$ Step 4: Substitute back $u = 7x-1$. \frac{2{21} (7x-1)^{3/2} + C}

2. For $\int \sin(4\pi x + 7)\,dx$: Step 1: Let $u = 4\pi x + 7$. Then $du = 4\pi\,dx$, which means $dx = \frac{1}{4\pi}\,du$. Step 2: Substitute $u$ into the integral. $\int \sin(u) \cdot \frac{1}{4\pi}\,du = \frac{1}{4\pi} \int \sin(u)\,du$ Step 3: Integrate $\sin(u)$. $\frac{1}{4\pi} (-\cos(u)) + C = -\frac{1}{4\pi} \cos(u) + C$ Step 4: Substitute back $u = 4\pi x + 7$. -\frac{1{4\pi} \cos(4\pi x + 7) + C}

3. For $\int \sqrt{\sin x} \cos x\,dx$: Step 1: Let $u = \sin x$. Then $du = \cos x\,dx$. Step 2: Substitute $u$ into the integral. $\int \sqrt{u}\,du = \int u^{1/2}\,du$ Step 3: Integrate $u^{1/2}$. $\frac{u^{3/2}}{3/2} + C = \frac{2}{3} u^{3/2} + C$ Step 4: Substitute back $u = \sin x$. \frac{2{3} (\sin x)^{3/2} + C}

4. For $\int x \cos x\,dx$: Step 1: Use integration by parts, $\int u\,dv = uv - \int v\,du$. Let $u = x$ and $dv = \cos x\,dx$. Step 2: Find $du$ and $v$. $du = dx$ and $v = \int \cos x\,dx = \sin x$. Step 3: Apply the integration by parts formula. $\int x \cos x\,dx = x \sin x - \int \sin x\,dx$ Step 4: Integrate $\int \sin x\,dx$. $x \sin x - (-\cos x) + C = x \sin x + \cos x + C$ $x \sin x + \cos x + C$

5. For $\int e^x \sin x\,dx$: Step 1: Use integration by parts, $\int u\,dv = uv - \int v\,du$. Let $I = \int e^x \sin x\,dx$. Step 2: First application of integration by parts: Let $u = \sin x$ and $dv = e^x\,dx$. Then $du = \cos x\,dx$ and $v = e^x$. $I = e^x \sin x - \int e^x \cos x\,dx$ Step 3: Second application of integration by parts for $\int e^x \cos x\,dx$: Let $u = \cos x$ and $dv = e^x\,dx$. Then $du = -\sin x\,dx$ and $v = e^x$. $\int e^x \cos x\,dx = e^x \cos x - \int e^x (-\sin x)\,dx = e^x \cos x + \int e^x \sin x\,dx$ Step 4: Substitute this back into the equation for $I$. $I = e^x \sin x - (e^x \cos x + I)$ Step 5: Solve for $I$. $I = e^x \sin x - e^x \cos x - I$ $2I = e^x \sin x - e^x \cos x$ $I = \frac{1}{2} (e^x \sin x - e^x \cos x) + C$ \frac{e^x{2} (\sin x - \cos x) + C}

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