Let's solve each part step-by-step as requested: --- ### **2) New Limits of Integration** Given: \[ \int_{0}^{\pi} \sin(7x + \pi) \, dx \] with substitution \( u = 7x + \pi \). **Find new limits:** - **Lower limit** (\(x = 0\)): \[ u = 7(0) + \pi = \pi \] - **Upper limit** (\(x = \pi\)): \[ u = 7(\pi) + \pi = 7\pi + \pi = 8\pi \] **Final Answer:** - Lower limit = \(\pi\) - Upper limit = \(8\pi\) --- ### **3) Evaluate the Indefinite Integral** \[ \int \cos(10x) \cos(\sin(10x)) \, dx \] Let \( u = \sin(10x) \), so \( du = 10\cos(10x)dx \implies \cos(10x)dx = \frac{1}{10} du \). Substitute: \[ \int \cos(10x) \cos(\sin(10x))dx = \int \cos(u) \cdot \frac{1}{10} du = \frac{1}{10}\int \cos(u) du = \frac{1}{10} \sin(u) + C \] Back substitute \( u \): \[ = \frac{1}{10} \sin(\sin(10x)) + C \] **Final Answer:** \[ \boxed{\frac{1}{10} \sin(\sin(10x)) + C} \] --- ### **4) Evaluate the Definite Integral** \[ \int_{0}^{\pi/2} \cos^8(x)\sin(x) dx \] Let \( u = \cos(x) \implies du = -\sin(x)dx \implies -du = \sin(x)dx \) Change limits: - \(x = 0\): \(u = \cos(0) = 1\) - \(x = \frac{\pi}{2}\): \(u = \cos\left(\frac{\pi}{2}\right) = 0\) Rewrite the integral: \[ \int_0^{\pi/2} \cos^8(x)\sin(x) dx = \int_{u=1}^{u=0} u^8 (-du) = \int_{u=0}^{u=1} u^8 du \] \[ = \left[ \frac{u^9}{9} \right]_{0}^{1} = \frac{1}{9} - 0 = \frac{1}{9} \] **Final Answer:** \[ \boxed{\frac{1}{9}} \] --- ### **5) Fundamental Theorem of Calculus** The correct statement (First Fundamental Theorem of Calculus) is: If \( F(x) = \int_a^x f(t)dt \), then \( F'(x) = f(x) \). --- ## **Summary of Answers** **2)** - Upper limit = \(8\pi\) - Lower limit = \(\pi\) **3)** \[ \int \cos(10x) \cos(\sin(10x))dx = \frac{1}{10} \sin(\sin(10x)) + C \] **4)** \[ \int_{0}^{\pi/2} \cos^8(x)\sin(x) dx = \frac{1}{9} \] **5)** The correct version: If \( F(x) = \int_a^x f(t)dt \), then \( F'(x) = f(x) \). Let me know if you want any part explained in more detail!