Integration by Parts: Original University 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.
Theoretical Recall
Integration by parts is obtained directly from the product rule for derivatives.
For two differentiable functions f and g:
\[(fg)'=f'g+fg'.\]Integrating both sides gives:
\[\int (fg)'\,dx = \int f'g\,dx + \int fg'\,dx.\]Therefore:
\[fg = \int f'g\,dx + \int fg'\,dx.\]Rearranging:
\[\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx.\]This is the integration by parts formula.
A common alternative notation is:
\[\int u\,dv = uv-\int v\,du.\]Choosing the Functions
The goal is to choose the factor to differentiate so that it becomes simpler, while the other factor can be integrated easily.
A useful heuristic is the LIATE rule:
- L — Logarithmic functions
- I — Inverse trigonometric functions
- A — Algebraic functions
- T — Trigonometric functions
- E — Exponential functions
Functions appearing earlier in this list are often good candidates for the factor to differentiate.
This is a guideline rather than a theorem: the best choice is always the one that simplifies the resulting integral.
Author’s note: Integration by parts should not be viewed merely as a formula to memorize. Its real purpose is to transform an integral into another one whose structure is easier to handle.
Exercises
Worked Exercises
These retain the definite and improper bounds in the original statements. C denotes an arbitrary integration constant on each interval of definition.
Exercise 1
Evaluate the integral:
\[\int \sin x e^{x} \, dx\]Solution.
Integrate by parts twice, as in the source: \(I=\int e^x\sin x\,dx=e^x\sin x-\int e^x\cos x\,dx.\) \(\int e^x\cos x\,dx=e^x\cos x+\int e^x\sin x\,dx.\) Substitution gives the original integral on both sides: \(I=e^x(\sin x-\cos x)-I, \qquad2I=e^x(\sin x-\cos x).\)
Final result
\[\frac{e^x}{2}(\sin x-\cos x)+C\]Exercise 2
Evaluate the integral:
\[\int_{1}^{e} \sin( \log x) \, dx\]Solution.
Write the integrand as a product of one and sin(log x): \(I=\left[x\sin(\log x)\right]_1^e-\int_1^e\cos(\log x)\,dx.\) Apply integration by parts to the remaining integral: \(\int_1^e\cos(\log x)\,dx =\left[x\cos(\log x)\right]_1^e+\int_1^e\sin(\log x)\,dx.\) \(I=e\sin1-e\cos1+1-I.\) \(2I=e(\sin1-\cos1)+1.\)
Final result
\[\frac{e(\sin1-\cos1)+1}{2}\]Exercise 3
Evaluate the integral:
\[\int_{0}^{1} e^{2x} \log(1+e^{x}) \, dx.\]Solution.
Put t = eˣ, with dx = dt/t; the bounds become one and e: \(I=\int_1^e t\log(1+t)\,dt =\left[\frac{t^2}{2}\log(1+t)\right]_1^e -\frac12\int_1^e\frac{t^2}{1+t}\,dt.\) Polynomial division yields: \(\frac{t^2}{1+t}=t-1+\frac1{t+1}.\) \(\int\frac{t^2}{1+t}\,dt=\frac{t^2}{2}-t+\log(1+t)+C.\) Evaluate all terms at the endpoints: \(I=\frac{e^2}{2}\log(1+e)-\frac{\log2}{2} -\frac12\left[\frac{t^2}{2}-t+\log(1+t)\right]_1^e.\)
Final result
\[\frac{e^2-1}{2}\log(1+e)-\frac{e^2}{4}+\frac e2-\frac14\]Exercise 4
Evaluate the integral:
\[\int _{0}^{\frac{\pi}{2}}x \sin^{2} x \, dx.\]Solution.
The half-angle formula gives: \(\int\sin^2x\,dx=\frac x2-\frac{\sin2x}{4}+C.\) Integrate by parts with this primitive: \(I=\left[\frac{x^2}{2}-\frac{x\sin2x}{4}\right]_0^{\pi/2} -\int_0^{\pi/2}\left(\frac x2-\frac{\sin2x}{4}\right)dx.\) \(I=\frac{\pi^2}{8}-\left[\frac{x^2}{4}+\frac{\cos2x}{8}\right]_0^{\pi/2} =\frac{\pi^2}{8}-\frac{\pi^2}{16}+\frac14.\)
Final result
\[\frac{\pi^2}{16}+\frac14\]Exercise 5
Evaluate the integral:
\[\int_{0}^{1} x^{2} \log \sqrt{1-x} \, dx.\]Solution.
Use the logarithm rule and substitute t = 1 − x: \(I=\frac12\int_0^1x^2\log(1-x)\,dx =\frac12\int_0^1(1-t)^2\log t\,dt.\) This is improper at t = 0. Integration by parts gives the primitive: \(\int(1-t)^2\log t\,dt =-\frac{(1-t)^3}{3}\log t+\frac13\int\frac{(1-t)^3}{t}\,dt.\) \(F(t)=-\frac{(1-t)^3}{3}\log t+\frac{\log t}{3} -\frac{t^3}{9}-t+\frac{t^2}{2}.\) The logarithmic part is [1 − (1 − t)³] log t/3 and tends to zero at zero. All polynomial terms also tend to zero: \(I=\frac12\lim_{a\to0^+}[F(t)]_a^1 =\frac12\left(-\frac19-1+\frac12\right).\)
Final result
\[-\frac{11}{36}\]Exercise 6
Evaluate the integral:
\[\int_{0}^{1} x^{2} \log \frac{1+x}{1-x} \, dx.\]Solution.
Split the logarithm, retaining the source’s link with the preceding exercise: \(I=\int_0^1x^2\log(1+x)\,dx-\int_0^1x^2\log(1-x)\,dx.\) From Exercise 5, the second integral equals −11/18. For the first, integrate by parts and divide the rational term: \(\int_0^1x^2\log(1+x)\,dx =\frac{\log2}{3}-\frac13\int_0^1\frac{x^3}{1+x}\,dx.\) \(\frac{x^3}{1+x}=x^2-x+1-\frac1{1+x}.\) \(\int_0^1x^2\log(1+x)\,dx =\frac{\log2}{3}-\frac13\left[\frac{x^3}{3}-\frac{x^2}{2}+x-\log(1+x)\right]_0^1 =\frac{2\log2}{3}-\frac5{18}.\) Therefore: \(I=\frac{2\log2}{3}-\frac5{18}+\frac{11}{18}.\)
Final result
\[\frac{\log4+1}{3}\]Exercise 7
Solve the following exercise:
\[\int_{1}^{2} \log(2x^{2}-3x+1) \, dx\]Solution.
Factor the logarithm argument: 2x²−3x+1=(x−1)(2x−1). On (1,2] both factors are positive, so split the logarithm into log(x−1)+log(2x−1). Each term is integrated by parts after a linear substitution: \(\int\log t\,dt=t\log t-t.\) For the first integral use t=x−1 with bounds 0 and 1; the limit t log t→0 gives −1. For the second use u=2x−1 with bounds 1 and 3: \(\frac12\int_1^3\log u\,du=\frac12[u\log u-u]_1^3=\frac32\log3-1.\) The singularity at x=1 is logarithmic and integrable. Add the two values.
Final result
\[\frac32\log3-2\]Exercise 8
Solve the following exercise:
\[\int_{0}^{1} \log \frac{2-x^{2}}{x^{2}+3} \, dx\]Solution.
Split the logarithm into log(2−x²)−log(x²+3). Both arguments are positive on [0,1]. Applying integration by parts to each gives \(\int\log(2-x^2)\,dx=x\log(2-x^2)+2\int\frac{x^2}{2-x^2}\,dx,\) \(\int\log(x^2+3)\,dx=x\log(x^2+3)-2\int\frac{x^2}{x^2+3}\,dx.\) Divide the rational terms: \(\frac{x^2}{2-x^2}=-1+\frac2{2-x^2},\qquad\frac{x^2}{x^2+3}=1-\frac3{x^2+3}.\) The linear terms cancel in the difference. A primitive for the original integrand on [0,1] is \(F(x)=x\log\frac{2-x^2}{x^2+3}+\sqrt2\log\frac{\sqrt2+x}{\sqrt2-x}-2\sqrt3\arctan\frac{x}{\sqrt3}.\) Here F(0)=0. Evaluate at x=1 and use arctan(1/√3)=π/6.
Final result
\[-\log4+\sqrt2\log\frac{\sqrt2+1}{\sqrt2-1}-\frac\pi{\sqrt3}\]