\( 1 \leftarrow \quad \) Use integral tables to evaluate the integral. \( \quad \int 5 x \cos ^{-1} x d x \)

To evaluate the integral \( \int 5x \cos^{-1}(x) \, dx \), we can use integration by parts. **Step 1: Choose \( u \) and \( dv \)** Let: - \( u = \cos^{-1}(x) \) Then, \( du = -\frac{1}{\sqrt{1 - x^2}} \, dx \) - \( dv = 5x \, dx \) Then, \( v = \frac{5x^2}{2} \) **Step 2: Apply Integration by Parts Formula** The integration by parts formula is: \[ \int u \, dv = uv - \int v \, du \] Applying this: \[ \int 5x \cos^{-1}(x) \, dx = \frac{5x^2}{2} \cos^{-1}(x) - \int \frac{5x^2}{2} \left( -\frac{1}{\sqrt{1 - x^2}} \right) dx \] \[ = \frac{5x^2}{2} \cos^{-1}(x) + \frac{5}{2} \int \frac{x^2}{\sqrt{1 - x^2}} \, dx \] **Step 3: Simplify the Remaining Integral** To evaluate \( \int \frac{x^2}{\sqrt{1 - x^2}} \, dx \), use the substitution \( x^2 = 1 - (1 - x^2) \): \[ \int \frac{x^2}{\sqrt{1 - x^2}} \, dx = \int \frac{1 - (1 - x^2)}{\sqrt{1 - x^2}} \, dx \] \[ = \int \frac{1}{\sqrt{1 - x^2}} \, dx - \int \sqrt{1 - x^2} \, dx \] \[ = \sin^{-1}(x) - \left( \frac{x \sqrt{1 - x^2}}{2} + \frac{\sin^{-1}(x)}{2} \right) + C \] \[ = \frac{\sin^{-1}(x)}{2} - \frac{x \sqrt{1 - x^2}}{2} + C \] **Step 4: Combine Everything** Substitute back into the expression: \[ \int 5x \cos^{-1}(x) \, dx = \frac{5x^2}{2} \cos^{-1}(x) + \frac{5}{2} \left( \frac{\sin^{-1}(x)}{2} - \frac{x \sqrt{1 - x^2}}{2} \right) + C \] \[ = \frac{5x^2}{2} \cos^{-1}(x) + \frac{5}{4} \sin^{-1}(x) - \frac{5x \sqrt{1 - x^2}}{4} + C \] **Final Answer:** \[ \int 5x \cos^{-1}(x) \, dx = \frac{5}{2} x^{2} \cos^{-1}x + \frac{5}{4} \sin^{-1}x - \frac{5}{4} x \sqrt{1 - x^{2}} + C \] **Where \( C \) is the constant of integration.** Answer: After expanding, an antiderivative is (5⁄2)x² arccos x plus (5⁄4) arcsin x minus (5⁄4)x√(1 – x²) plus C. Thus, ∫5x cos⁻¹x dx = (5/2)x² arccos x + (5/4) sin⁻¹x – (5/4)x√(1−x²) + C