1. **Problem (a):** Calculate $$\int (1 - \frac{1}{w}) \cos(w - \ln(w)) \, dw$$ Step 1: Rewrite the integral $$\int \left(1 - \frac{1}{w}\right) \cos(w - \ln(w)) \, dw = \int \cos(w - \ln(w)) \, dw - \int \frac{1}{w} \cos(w - \ln(w)) \, dw$$ Step 2: Use substitution for inner function: Let $$u = w - \ln(w)$$ then $$\frac{du}{dw} = 1 - \frac{1}{w}$$ Step 3: Substitute to simplify: $$du = \left(1 - \frac{1}{w}\right) dw$$ so $$\int (1 - \frac{1}{w}) \cos(w - \ln(w)) \, dw = \int \cos(u) \, du = \sin(u) + C$$ Step 4: Return to $$w$$: $$\sin(w - \ln(w)) + C$$ --- 2. **Problem (b):** Calculate $$\int 3(8y - 1) e^{4y^2 - y} \, dy$$ Step 1: Notice the exponent's derivative $$\frac{d}{dy}(4y^2 - y) = 8y - 1$$ Step 2: Rewrite the integral $$\int 3(8y - 1) e^{4y^2 - y} \, dy = 3 \int (8y - 1) e^{4y^2 - y} \, dy$$ Step 3: Use substitution: Let $$u = 4y^2 - y$$ then $$du = (8y - 1) dy$$ Step 4: Substitute integral $$3 \int e^u \, du = 3 e^u + C = 3 e^{4y^2 - y} + C$$ --- 3. **Problem (c):** Calculate $$\int x^2 (3 - 10x^3)^4 \, dx$$ Step 1: Use substitution Let $$u = 3 - 10x^3$$ Step 2: Derivative of $$u$$ $$du = -30x^2 \, dx$$ implies $$x^2 \, dx = -\frac{1}{30} du$$ Step 3: Substitute into integral $$\int x^2 (3 - 10x^3)^4 \, dx = \int (u)^4 \left(-\frac{1}{30}\right) du = -\frac{1}{30} \int u^4 \, du$$ Step 4: Integrate $$-\frac{1}{30} \cdot \frac{u^{5}}{5} + C = -\frac{1}{150} u^{5} + C$$ Step 5: Substitute back $$-\frac{1}{150} (3 - 10x^3)^5 + C$$ --- 4. **Problem (d):** Calculate $$\int \frac{x}{\sqrt{1 - 4x^2}} \, dx$$ Step 1: Use substitution Let $$u = 1 - 4x^2$$ then $$du = -8x \, dx$$ implies $$x \, dx = -\frac{1}{8} du$$ Step 2: Substitute into integral $$\int \frac{x}{\sqrt{1 - 4x^2}} \, dx = \int \frac{x}{\sqrt{u}} \, dx = \int \frac{-\frac{1}{8} du}{\sqrt{u}} = -\frac{1}{8} \int u^{-\frac{1}{2}} \, du$$ Step 3: Integrate $$-\frac{1}{8} \cdot 2 u^{\frac{1}{2}} + C = -\frac{1}{4} \sqrt{u} + C$$ Step 4: Substitute back $$-\frac{1}{4} \sqrt{1 - 4x^2} + C$$ --- **Final answers:** (a) $$\sin(w - \ln(w)) + C$$ (b) $$3 e^{4y^2 - y} + C$$ (c) $$-\frac{1}{150} (3 - 10x^3)^5 + C$$ (d) $$-\frac{1}{4} \sqrt{1 - 4x^2} + C$$