Definite Integrals: Original Solved Exercises
Exercise editors and source: Prof. Antonino De Martino and Dr. Luana Manfredini, Eserciziario 2.1. The original exercises and source theory are presented here in English for Logic & Motion.
Original Bounds, Symmetry and Changes of Variable
The original integration bounds are part of each problem. Under substitution, change the bounds along with the differential. Half-angle identities and integration by parts are collected in the complete integration recall.
Exercise 1
Evaluate the definite integral:
\[\int_{-1}^{1} \sqrt{1-x^{2}} \, dx.\]Solution.
Use x = sin t, with −π/2 ≤ t ≤ π/2 and dx = cos t dt: \(I=\int_{-\pi/2}^{\pi/2}\sqrt{1-\sin^2t}\cos t\,dt =\int_{-\pi/2}^{\pi/2}|\cos t|\cos t\,dt.\) Author’s observation: cosine is positive on the open interval (−π/2,π/2), so the absolute value can be removed here. Use the half-angle formula: \(I=\int_{-\pi/2}^{\pi/2}\frac{1+\cos2t}{2}\,dt =\left[\frac t2+\frac{\sin2t}{4}\right]_{-\pi/2}^{\pi/2}.\)
Final result
\[\frac\pi2\]Exercise 2
Evaluate the definite integral:
\[\int_{-2}^{2} \sqrt{4-x^{2}} \, dx.\]Solution.
The integrand is even: \(I=2\int_0^2\sqrt{4-x^2}\,dx.\) Set x = 2 sin t. On 0 ≤ t ≤ π/2, cosine is nonnegative: \(I=2\int_0^{\pi/2}\sqrt{4-4\sin^2t}\,2\cos t\,dt =8\int_0^{\pi/2}\cos^2t\,dt.\) \(I=8\left[\frac t2+\frac{\sin2t}{4}\right]_0^{\pi/2}=2\pi.\) Author’s observation: integrals involving √(a²−x²) can be treated with x = a sin t. For a nonzero radius, choose a suitable branch and account for the absolute value of a.
Final result
\[2\pi\]Exercise 3
Evaluate the definite integral:
\[\int_{2}^{4} | \log (x-2) | \, dx.\]Solution.
The logarithm changes sign at three. Preserve the split and the improper endpoint: \(I=-\int_2^3\log(x-2)\,dx+\int_3^4\log(x-2)\,dx.\) Integration by parts on the second integral gives: \(\int_3^4\log(x-2)\,dx =\left[x\log(x-2)\right]_3^4-\int_3^4\frac{x}{x-2}\,dx.\) \(\frac{x}{x-2}=1+\frac2{x-2}, \qquad\int_3^4\log(x-2)\,dx=2\log2-1.\) Use the same primitive at the singular endpoint: \(\int_2^3\log(x-2)\,dx =\lim_{a\to2^+}\left[x\log(x-2)-x-2\log(x-2)\right]_a^3.\) \(\int_2^3\log(x-2)\,dx =\lim_{a\to2^+}\left(a-3-(a-2)\log(a-2)\right)=-1.\) Therefore I = 1 + 2 log 2 − 1.
Final result
\[\log4\]Exercise 4
Evaluate the definite integral:
\[\int_{1}^{e} x^{\log x} \frac{\log^{3} x}{x} \, dx.\]Solution.
Rewrite the variable power and substitute t = log x: \(x^{\log x}=e^{(\log x)^2},\qquad\frac{dx}{x}=dt, \qquad I=\int_0^1e^{t^2}t^3\,dt.\) Integrate by parts, writing t³ as t·t²: \(I=\left[\frac{t^2e^{t^2}}2\right]_0^1 -\frac12\int_0^1e^{t^2}2t\,dt.\) \(I=\left[\frac{t^2e^{t^2}}2-\frac{e^{t^2}}2\right]_0^1=\frac12.\)
Final result
\[\frac12\]Exercise 5
Solve the following exercise:
\[\int_{0}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \, dx.\]Solution.
Set x=sin t, with t from 0 to π/2. Since cos t≥0 on that interval, √(1−x²)=cos t and dx=cos t dt. The radical cancels: \(I=\int_0^{\pi/2}\sin^2t\,dt=\frac12\int_0^{\pi/2}(1-\cos2t)\,dt=\frac\pi4.\) The endpoint x=1 is improper; the transformed integral proves convergence.
Final result
\[\frac\pi4\]Exercise 6
Solve the following exercise:
\[\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\log \sin x) \cos x \, dx.\]Solution.
Set t=sin x, dt=cos x dx. The bounds become 1/√2 and 1: \(I=\int_{1/\sqrt2}^{1}\log t\,dt=[t\log t-t]_{1/\sqrt2}^{1}.\) Using log(1/√2)=−(log 2)/2 gives the stated value.
Final result
\[-1+\frac{1+\frac12\log2}{\sqrt2}\]Exercise 7
Solve the following exercise:
\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\log(1+\sin x)^{\sin x}}{\tan x} \, dx.\]Solution.
Interpret the numerator exactly as log((1+sin x)ˢⁱⁿ ˣ)=sin x log(1+sin x). Since 1/tan x=cos x/sin x, the sine factors cancel. Set t=1+sin x and dt=cos x dx: \(I=\int_{3/2}^{1+\sqrt3/2}\log t\,dt=[t\log t-t]_{3/2}^{1+\sqrt3/2}.\) Evaluate both endpoints.
Final result
\[\left(1+\frac{\sqrt3}2\right)\log\left(1+\frac{\sqrt3}2\right)-\frac32\log\frac32+\frac{1-\sqrt3}2\]Exercise 8
Solve the following exercise:
\[\int_{\frac{1}{2}}^{\frac{1}{\sqrt{2}}} \frac{\sqrt{1-x^{2}}}{x^{2}} \, dx\]Solution.
Set x=sin t, with t from π/6 to π/4. On this interval cos t>0, so \(I=\int_{\pi/6}^{\pi/4}\frac{\cos^2t}{\sin^2t}\,dt=\int_{\pi/6}^{\pi/4}(\csc^2t-1)\,dt=[-\cot t-t]_{\pi/6}^{\pi/4}.\) Use cot(π/6)=√3 and cot(π/4)=1.
Final result
\[\sqrt3-1-\frac\pi{12}\]