Integration by Substitution: 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 substitution is the integral counterpart of the chain rule.
Suppose that:
\[x=g(t)\]is a differentiable change of variable. Then:
\[dx=g'(t)\,dt.\]Therefore:
\[\int f(x)\,dx = \int f(g(t))g'(t)\,dt.\]Another common form begins with:
\[u=g(x).\]Then:
\[du=g'(x)\,dx.\]Whenever an integral contains a composition together with the derivative of the inner function, we can use:
\[\int f(g(x))g'(x)\,dx = \int f(u)\,du.\]After evaluating the new integral, the original variable must be restored.
Common Strategies
A useful substitution often reveals a simpler structure hidden inside the original integrand.
Typical choices include:
- a linear expression raised to a power;
- the argument of a logarithm;
- the denominator of a rational expression;
- the expression inside a radical;
- trigonometric substitutions for quadratic radicals;
- hyperbolic substitutions for expressions involving sums or differences of squares.
For radicals of the form:
\[\sqrt{1-x^2},\]the substitution:
\[x=\sin t\]is often useful because:
\[1-\sin^2t=\cos^2t.\]For expressions involving:
\[1+x^2,\]the substitution:
\[x=\tan t\]can be useful because:
\[1+\tan^2t=\sec^2t.\]Hyperbolic substitutions can similarly exploit:
\[\cosh^2t-\sinh^2t=1.\]Author’s note: Always transform both the integrand and the differential consistently. A substitution is not complete until every occurrence of the original variable has been removed from the transformed integral.
Exercises
Worked Exercises
The exercises retain their integration bounds and combine substitution with other methods where needed. Integrals with singular endpoints are also treated on the Improper Integrals page.
Exercise 1
Evaluate the integral:
\[\int \frac{1}{\sqrt{x^{2}-2x}} \, dx.\]Solution.
The real domain is x < 0 or x > 2. Use the source’s Euler substitution: \(\sqrt{x^2-2x}=x+t.\) Squaring and solving for x gives: \(x=-\frac{t^2}{2(1+t)},\qquad dx=-\frac{t^2+2t}{2(1+t)^2}\,dt.\) \(\sqrt{x^2-2x}=\frac{t^2+2t}{2(1+t)}.\) Substitute both factors: \(\int\frac{dx}{\sqrt{x^2-2x}} =-\int\frac{2(1+t)}{t^2+2t}\frac{t^2+2t}{2(1+t)^2}\,dt =-\int\frac{dt}{1+t}.\) \(-\int\frac{dt}{1+t}=-\log|1+t|+C.\)
Final result
\[-\log|1+\sqrt{x^2-2x}-x|+C\]Exercise 2
Evaluate the integral:
\[\int_{0}^{\frac{\pi}{3}} \frac{2 \cos x}{5-4 \cos^{2} x} \, dx.\]Solution.
First find a primitive. Set t = sin x, with dt = cos x dx, and use cos²x = 1 − sin²x: \(\int\frac{2\cos x}{5-4\cos^2x}\,dx =2\int\frac{dt}{1+4t^2} =\arctan(2t)+C.\) Restore the variable and evaluate the original bounds: \(\left[\arctan(2\sin x)\right]_0^{\pi/3} =\arctan\sqrt3-\arctan0.\)
Final result
\[\frac\pi3\]Exercise 3
Evaluate the integral:
\[\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos x}{\sin^{2} x-2 \sin x} \, dx.\]Solution.
Set t = sin x. First integrate the transformed rational function: \(\int\frac{\cos x}{\sin^2x-2\sin x}\,dx =\int\frac{dt}{t(t-2)}.\) Use partial fractions: \(\frac1{t(t-2)}=\frac At+\frac B{t-2}, \qquad A+B=0,\quad-2A=1.\) \(A=-\frac12,\quad B=\frac12, \qquad F(t)=-\frac12\log|t|+\frac12\log|t-2|.\) The bounds in t are √2/2 and one: \(I=\left[-\frac12\log|\sin x|+\frac12\log|\sin x-2|\right]_{\pi/4}^{\pi/2}.\)
Final result
\[\frac12\log\left(\frac{\sqrt2}{4-\sqrt2}\right)\]Exercise 4
Evaluate the integral:
\[\int_{0}^{1} x^{2} e^{x^{\frac{3}{2}}} \,dx.\]Solution.
Factor x² as x¹ᐟ² x³ᐟ² and put t = x³ᐟ²: \(dt=\frac32\sqrt x\,dx, \qquad I=\frac23\int_0^1te^t\,dt.\) Integration by parts gives: \(\frac23\int te^t\,dt=\frac23\left(te^t-\int e^t\,dt\right) =\frac23e^t(t-1)+C.\) \(I=\left[\frac23e^{x^{3/2}}(x^{3/2}-1)\right]_0^1=\frac23.\)
Final result
\[\frac23\]Exercise 5
Evaluate the integral:
\[\int \frac{x^{3}}{\sqrt{1-x^{2}}} \, dx\]Solution.
For −1 < x < 1 choose x = sin t with −π/2 < t < π/2, so cos t > 0 and dx = cos t dt: \(\int\frac{x^3}{\sqrt{1-x^2}}\,dx =\int\frac{\sin^3t}{\cos t}\cos t\,dt =\int(1-\cos^2t)\sin t\,dt.\) \(\int\sin t\,dt+\int\cos^2t(-\sin t)\,dt =-\cos t+\frac{\cos^3t}{3}+C.\) \(-\sqrt{1-x^2}+\frac{(1-x^2)^{3/2}}3+C =-\frac{(2+x^2)\sqrt{1-x^2}}3+C.\)
Final result
\[-\frac{(2+x^2)\sqrt{1-x^2}}3+C\]Exercise 6
Evaluate the integral:
\[\int_{0}^{\frac{\pi}{2}} \cos x \log(1+\sin x) \, dx.\]Solution.
First integrate by parts: \(I=\left[\sin x\log(1+\sin x)\right]_0^{\pi/2} -\int_0^{\pi/2}\frac{\sin x\cos x}{1+\sin x}\,dx.\) For the remaining primitive put t = sin x: \(\int\frac{\sin x\cos x}{1+\sin x}\,dx =\int\frac t{1+t}\,dt=t-\log|1+t|+C.\) \(I=\left[\sin x\log(1+\sin x)-\sin x+\log(1+\sin x)\right]_0^{\pi/2}.\)
Final result
\[\log4-1\]Exercise 7
Solve the following exercise:
\[\int x^{3} (1+x^{2})^{\frac{1}{3}} \, dx.\]Solution.
Put t=1+x², so dt=2x dx and x³dx=(t−1)dt/2. The transformed integral is \(\frac12\int(t-1)t^{1/3}\,dt=\frac12\int(t^{4/3}-t^{1/3})\,dt.\) Integrate the two powers and replace t by 1+x².
Final result
\[\frac3{14}(1+x^2)^{7/3}-\frac38(1+x^2)^{4/3}+C\]Exercise 8
Solve the following exercise:
\[\int \frac{1}{x^{4} \sqrt{x^{3}+1}} \, dx.\]Solution.
Set t=√(x³+1), so 2t dt=3x²dx and x³=t²−1. The integrand transforms to \(\int\frac{dx}{x^4\sqrt{x^3+1}}=\frac23\int\frac{dt}{(t^2-1)^2}.\) Resolve the repeated factors: \(\frac1{(t^2-1)^2}=\frac14\left(\frac1{(t-1)^2}+\frac1{(t+1)^2}-\frac1{t-1}+\frac1{t+1}\right).\) Integrating yields −t/[3(t²−1)]+(1/6)log|(t+1)/(t−1)|. Substitute t back. The real integrand requires x>−1 and x≠0.
Final result
\[-\frac{\sqrt{x^3+1}}{3x^3}+\frac16\log\left|\frac{\sqrt{x^3+1}+1}{\sqrt{x^3+1}-1}\right|+C\]