91Ó°ÊÓ

#tag_title# Step 1: Split the integral and integrate each part#tag_content# The integral becomes: $$\int_{0}^{1} \frac{\pi}{2} dx - \int_{0}^{1} \sin^{-1} x dx$$ The first integral can be evaluated directly: $$\int_{0}^{1} \frac{\pi}{2} dx = \frac{\pi}{2}x\Big|_{0}^{1} = \frac{\pi}{2}$$ For the second integral, we use integration by parts. Let \(u = \sin^{-1} x\) and \(dv = dx\). Then, \(du = \frac{1}{\sqrt{1-x^2}}dx\) and \(v = x\). #tag_title# Step 2: Apply the integration by parts formula and evaluate the second integral#tag_content# Substituting 'u', 'v', 'du', and 'dv' into the integration by parts formula, we get $$\int_{0}^{1} \sin^{-1} x dx = x\sin^{-1}x\Big|_{0}^{1} - \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} dx$$ The first part is evaluated to be \(x\sin^{-1}x\Big|_{0}^{1} = 0\). The second part is a standard integral, $$\int_{0}^{1} \frac{x}{\sqrt{1-x^2}} dx = \int_{0}^{1} \frac{d(\frac{1}{2}(1-x^2))}{\sqrt{1-x^2}} dx = \int_{0}^{1} \frac{d(\frac{1}{2}(1-x^2))}{\sqrt{1-x^2}} \times \frac{d(\frac{1}{2}(1-x^2))}{-d(\frac{1}{2}(1-x^2))} dx = - \int_{0}^{1} d(\frac{1}{2}(1-x^2))^{\frac{1}{2}} = \sqrt{1-x^2}\Big|_{0}^{1} = 1 - 0 = 1$$ #tag_title# Step 3: Combine the evaluated integrals#tag_content# Now, we can combine the evaluated integrals: $$\int_{0}^{1} \cos ^{-1} x d x = \frac{\pi}{2} - 1$$ ##Final Answers## (a) \(\int_{0}^{\ln 2} \mathrm{xe}^{-\mathrm{x}} \mathrm{dx} = \frac{1}{2}\) (b) \(\int_{0}^{2 \pi} x^{2} \cos x d x = 0\) (c) \(\int_{0}^{1} \cos ^{-1} x d x = \frac{\pi}{2} - 1\)