Question

Applying the formula for integration by parts calculate the following integrals: (a) \(\int_{0}^{\ln 2} \mathrm{xe}^{-\mathrm{x}} \mathrm{dx}\) (b) \(\int_{0}^{2 \pi} x^{2} \cos x d x\) (c) \(\int_{0}^{1} \cos ^{-1} x d x\)

Short answer

#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) \(\int_{0}^{1} \cos ^{-1} x d x = \frac{\pi}{2} - 1\)

Step-by-step solution

Choose 'u' and 'dv' for integration by parts

Let \(u = x\) and \(dv = e^{-x} dx\).

Compute 'du' and 'v'

Differentiating u with respect to x, we get \(du = dx\). Now, integrating 'dv' with respect to x, we get \(v = \int e^{-x} dx = -e^{-x}\).

Apply the integration by parts formula and evaluate the integral

Substituting 'u', 'v', 'du', and 'dv' into the integration by parts formula, we get $$\int_{0}^{\ln 2} xe^{-x} dx = -xe^{-x}\Big|_{0}^{\ln2} + \int_{0}^{\ln 2} e^{-x} dx$$ Now, we evaluate the definite integral of \(e^{-x}\) from 0 to \(\ln2\), $$\int_{0}^{\ln 2} e^{-x} dx = -e^{-x} \Big|_{0}^{\ln2} = -e^{\ln \frac{1}{2}} + 1 = \frac{1}{2} + 1 = \frac{3}{2}$$ Thus, $$ \int_{0}^{\ln 2} xe^{-x} dx = -xe^{-x}\Big|_{0}^{\ln2} + \frac{3}{2} = -1 + \frac{3}{2} = \frac{1}{2}$$ ##(b) \(\int_{0}^{2 \pi} x^{2} \cos x d x\)##

Choose 'u' and 'dv' for integration by parts

Let \(u = x^2\) and \(dv = \cos{x} dx\).

Compute 'du' and 'v'

Differentiating u with respect to x, we get \(du = 2x\;dx\). Now, integrating 'dv' with respect to x, we get \(v = \int \cos{x} dx = \sin{x}\).

Apply the integration by parts formula and evaluate the integral

Substituting 'u', 'v', 'du', and 'dv' into the integration by parts formula, we get $$\int_{0}^{2 \pi} x^{2} \cos x dx = x^{2}\sin x\Big|_{0}^{2\pi} - 2\int_{0}^{2 \pi} x\sin{x} dx$$ Now, we need to compute the definite integral of \(x\sin{x}\) from 0 to \(2\pi\). Applying integration by parts again, Let \(u = x\) and \(dv = \sin{x}dx\). Then, \(du = dx\) and \(v = \int \sin{x} dx = -\cos x\). So, $$\int_{0}^{2 \pi} x\sin{x} dx = -x\cos x\Big|_{0}^{2\pi} + \int_{0}^{2 \pi} \cos x dx = 0$$ Thus, our final answer is: $$\int_{0}^{2 \pi} x^{2} \cos x dx = x^{2}\sin x\Big|_{0}^{2\pi} - 0 = 0$$ ##(c) \(\int_{0}^{1} \cos ^{-1} x d x\)## Integrating the arccosine function can be challenging even using integration by parts. A better approach would be to use the following property of the arccosine function: $$\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$$ Now, we can rewrite the given integral as: $$\int_{0}^{1} \cos ^{-1} x d x = \int_{0}^{1} (\frac{\pi}{2} - \sin^{-1} x)dx$$ This can be computed by splitting the integral and integrating each part separately.