Question

Calculate the following integrals: (a) \(\int_{-1}^{1} \frac{x d x}{x^{2}+x+1}\) (b) \(\int_{1}^{e}(x \ell n x)^{2} d x\) (c) \(\int_{0}^{3} \sin ^{-1} \sqrt{\frac{x}{1+x}} d x\) (d) \(\int_{0}^{2 \pi} \frac{d x}{(2+\cos x)(3+\cos x)}\) (e) \(\int_{0}^{\pi / 2} \sin x \sin 2 x \sin 3 x d x\) (I) \(\int_{0}^{\pi}(\mathrm{x} \sin \mathrm{x})^{2} \mathrm{dx}\)

Short answer

^3(u)+2\cos(u))]\bigg|_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}} -\frac{2}{3}(3\cos^2(u)-2) du$ #tag_title#Step 3: Simplify and Integrate#tag_content#Simplifying the integral on the right-hand side and integrating: \(-\frac{2}{3}[u(\cos^3(u)+2\cos(u))]\bigg|_0^{\frac{\pi}{4}} + 2\int_0^{\frac{\pi}{4}} \cos^2(u) du - \frac{4}{3}\int_0^{\frac{\pi}{4}} du\) Using the identity \(\cos^2(u) = \frac{1 + \cos(2u)}{2}\): \(-\frac{2}{3}[u(\cos^3(u)+2\cos(u))]\bigg|_0^{\frac{\pi}{4}} + \int_0^{\frac{\pi}{4}} (1 + \cos(2u)) du - \frac{4}{3}\int_0^{\frac{\pi}{4}} du\) #tag_title#Step 4: Evaluate the Integral#tag_content#Integrate and evaluate the integrals: \(\frac{\pi}{3}\left[ 1 - \frac{4}{3}}\right] + \frac{1}{2}\sin(2u)\bigg|_0^{\frac{\pi}{4}} = \boxed{\frac{\pi}{6}}\) (d) #tag_title#Step 1: Substitution and Double Angle Formula#tag_content#Substitute \(x = \tan(u)\) where \(-\frac{\pi}{4} \leq u \leq \frac{\pi}{4}\). Then, \(dx = \sec^2(u)du\) and: \(\int_{-1}^{1} \sqrt{1-x^2} dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(u) du\) Now, use the double angle formula \(\cos^2(u) = \frac{1 + \cos(2u)}{2}\), so: \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \cos^2(u) du = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1 + \cos(2u)}{2} du\) #tag_title#Step 2: Integration and Evaluation#tag_content#Integrate and evaluate the expression: \(\frac{1}{2}\left[u + \frac{1}{2}\sin(2u)\right]\bigg|_{-\frac{\pi}{4}}^{\frac{\pi}{4}} = \boxed{\frac{\pi}{4}}\) (e) #tag_title#Step 1: Product-to-Sum Identity#tag_content#Use the product-to-sum identity \(\sin(u)\sin(v) = \frac{1}{2}[\cos(u-v) - \cos(u+v)]\): \(\int_0^{\frac{\pi}{2}} \sin(2x)\sin(3x) dx = \frac{1}{2}\int_0^{\frac{\pi}{2}} (\cos(x) - \cos(5x)) dx\) #tag_title#Step 2: Integration and Evaluation#tag_content#Now, integrate and evaluate the integrals: \(\frac{1}{2}\left[\sin(x) - \frac{1}{5}\sin(5x)\right]\bigg|_0^{\frac{\pi}{2}} = \boxed{\frac{1}{2}}\) (I) #tag_title#Step 1: Integration by Parts (First Iteration)#tag_content#Use integration by parts twice. For the first iteration, let \(u = x^2\), \(dv = e^{2x} dx\). Then, \(du = 2x dx\), \(v = \frac{1}{2}e^{2x}\). Applying integration by parts: \(\int_0^1 x^2 e^{2x} dx = x^2 \frac{1}{2}e^{2x}\bigg|_0^1 - \int_0^1 2x \frac{1}{2}e^{2x} dx\) #tag_title#Step 2: Integration by Parts (Second Iteration)#tag_content#For the second iteration, let \(u = x\), \(dv = e^{2x} dx\). Then, \(du = dx\), \(v = \frac{1}{2}e^{2x}\). Applying integration by parts: \(\int_0^1 x e^{2x} dx = x \frac{1}{2}e^{2x}\bigg|_0^1 - \int_0^1 \frac{1}{2}e^{2x} dx\) #tag_title#Step 3: Combine and Evaluate#tag_content#Combine the two results and evaluate: \(\int_0^1 x^2 e^{2x} dx = \left[\frac{1}{2}e^{2x}(x^2 - x + \frac{1}{4})\right]\bigg|_0^1 = \boxed{\frac{1}{2}(e^2 - \frac{3}{4})}\)

Step-by-step solution

Simplification and Substitution

For this integral, we can simplify the integrand algebraically before applying substitution: \(u = x^{2} + x + 1\) \(du = (2x + 1)dx\) Therefore, we have: \(\int_{-1}^{1} \frac{x dx}{x^{2} + x + 1} = \frac{1}{2}\int_{u(-1)}^{u(1)} \frac{du}{u}\)

Integration

Now, we integrate: \(\frac{1}{2}\int_{u(-1)}^{u(1)} \frac{du}{u} = \frac{1}{2}(\ln |u(1)| - \ln |u(-1)|) = \frac{1}{2}\ln\left(\frac{u(1)}{u(-1)}\right) \)

Substitute Back and Evaluate

Substitute the value of u back and simplify: \(\frac{1}{2}\ln\left(\frac{3}{3}\right) = \boxed{0}\) (b)

Integration by Parts

Let \(u = (x\ln x)^2\) and \(dv = dx\). \(du = 2x(\ln x)^2 + 2x\ln x\) and \(v = x\). Applying integration by parts: \(\int_{1}^{e}(x\ln x)^2 dx = x(x\ln x)^2 \bigg|_1^e - \int_1^e x(2x(\ln x)^2 + 2x\ln x) dx\)

Simplify and Integrate

Simplify the integral expression and integrate: \(\int_1^e (x\ln x)^2 dx = e - (2\int_1^e (x\ln x)^3 dx + 2\int_1^e x(\ln x)^2 dx)\) Now notice the second integral on the right-hand side is the same integral we are trying to evaluate: \(\int_1^e (x\ln x)^2 dx = e - 2\int_1^e (x\ln x)^3 dx - 2\int_1^e (x\ln x)^2 dx\)

Solve for the Integral

Move the integral to the left-hand side and solve for it: \(3\int_1^e (x\ln x)^2 dx = e\) \(\int_1^e (x\ln x)^2 dx = \boxed{\frac{e}{3}}\) (c)

Substitution

Let \(x = \tan^2(u)\) where \(0 \leq u \leq \frac{\pi}{4}\). Then, \(dx = 2\tan(u)\sec^2(u) du\) and: \(\int_0^3 \sin^{-1} \sqrt{\frac{x}{1+x}} dx = \int_0^{\frac{\pi}{4}} \sin^{-1}(\sin(u)) * 2\sin(u)\cos^2(u) du\)

Simplify and Integrate

Simplify the integrand and integrate: \(\int_0^{\frac{\pi}{4}} 2u \sin(u)\cos^2(u) du\) To integrate this expression, we can use integration by parts. Let \(u = 2u\), \(dv = \sin(u)\cos^2(u)du\) \(du=2du\), and \(v=-\frac{1}{3}(\cos^3(u)+2\cos(u))\) Applying integration by parts: $\int_0^{\frac{\pi}{4}} 2u \sin(u)\cos^2(u) du = -\frac{2}{3}[u(\cos