Integration by Substitution — Theory and Solved Exercises
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
Exercise 1 — Linear Substitution
Evaluate:
\[\int(2x+1)^5\,dx.\]Solution.
Set:
\[u=2x+1.\]Then:
\[du=2\,dx.\]Therefore:
\[dx=\frac12\,du.\]Substituting:
\[\int(2x+1)^5\,dx = \frac12\int u^5\,du.\]Integrating:
\[\frac12\int u^5\,du = \frac12\frac{u^6}{6}+C.\]Thus:
\[\frac{u^6}{12}+C.\]Returning to x:
\[\frac{(2x+1)^6}{12}+C.\]Final Result
\[\frac{(2x+1)^6}{12}+C\]Exercise 2 — Logarithmic Substitution
Evaluate, for x > 1:
\[\int\frac{1}{x\log x}\,dx.\]Solution.
Set:
\[u=\log x.\]Then:
\[du=\frac1x\,dx.\]The integral becomes:
\[\int\frac1u\,du.\]Therefore:
\[\int\frac1u\,du = \log|u|+C.\]Since x > 1:
\[\log x>0.\]Thus the absolute value can be omitted:
\[\log(\log x)+C.\]Final Result
\[\log(\log x)+C\]Exercise 3 — Substitution in a Rational Expression
Evaluate:
\[\int\frac{x}{1+x^2}\,dx.\]Solution.
Set:
\[u=1+x^2.\]Then:
\[du=2x\,dx.\]Therefore:
\[x\,dx=\frac12\,du.\]The integral becomes:
\[\frac12\int\frac1u\,du.\]Hence:
\[\frac12\log|u|+C.\]Since:
\[1+x^2>0\]for every real x, we obtain:
\[\frac12\log(1+x^2)+C.\]Final Result
\[\frac12\log(1+x^2)+C\]Exercise 4 — Trigonometric Substitution
Evaluate:
\[\int\frac{1}{\sqrt{1-x^2}}\,dx.\]Solution.
Use the substitution:
\[x=\sin t.\]Then:
\[dx=\cos t\,dt.\]Moreover:
\[\sqrt{1-x^2} = \sqrt{1-\sin^2t}.\]Using:
\[1-\sin^2t=\cos^2t,\]and choosing t in the standard range of arcsin, we have:
\[\sqrt{1-\sin^2t} = \cos t.\]Therefore:
\[\int \frac{\cos t}{\cos t} \,dt = \int1\,dt.\]Thus:
\[t+C.\]Since:
\[t=\arcsin x,\]we obtain:
\[\arcsin x+C.\]Final Result
\[\arcsin x+C\]Exercise 5 — Tangent Substitution
Evaluate:
\[\int\frac{1}{1+x^2}\,dx.\]Solution.
Set:
\[x=\tan t.\]Then:
\[dx=\sec^2t\,dt.\]Using the identity:
\[1+\tan^2t=\sec^2t,\]the integral becomes:
\[\int \frac{\sec^2t}{\sec^2t} \,dt.\]Therefore:
\[\int1\,dt=t+C.\]Since:
\[t=\arctan x,\]we obtain:
\[\arctan x+C.\]Final Result
\[\arctan x+C\]Exercise 6 — Radical of a Sum of Squares
Evaluate:
\[\int\frac{1}{\sqrt{x^2+4}}\,dx.\]Solution.
Set:
\[x=2\tan t.\]Then:
\[dx=2\sec^2t\,dt.\]The radical becomes:
\[\sqrt{x^2+4} = \sqrt{4\tan^2t+4}.\]Factor out 4:
\[\sqrt{x^2+4} = 2\sqrt{1+\tan^2t}.\]Using:
\[1+\tan^2t=\sec^2t,\]we obtain:
\[\sqrt{x^2+4}=2\sec t.\]Therefore:
\[\int \frac{2\sec^2t}{2\sec t} \,dt = \int\sec t\,dt.\]Recall that:
\[\int\sec t\,dt = \log|\sec t+\tan t|+C.\]From:
\[\tan t=\frac{x}{2},\]we have:
\[\sec t = \sqrt{1+\tan^2t} = \frac{\sqrt{x^2+4}}{2}.\]Therefore:
\[\log\left| \frac{\sqrt{x^2+4}+x}{2} \right| +C.\]The constant factor 1/2 inside the logarithm contributes only an additive constant, which can be absorbed into C.
Hence the standard form is:
\[\log\left|x+\sqrt{x^2+4}\right|+C.\]Final Result
\[\log\left|x+\sqrt{x^2+4}\right|+C\]Exercise 7 — Exponential Substitution
Evaluate:
\[\int\frac{e^x}{1+e^{2x}}\,dx.\]Solution.
Set:
\[u=e^x.\]Then:
\[du=e^x\,dx.\]Moreover:
\[e^{2x}=u^2.\]Therefore the integral becomes:
\[\int\frac{1}{1+u^2}\,du.\]Using the standard antiderivative:
\[\int\frac{1}{1+u^2}\,du = \arctan u+C,\]we obtain:
\[\arctan u+C.\]Returning to x:
\[\arctan(e^x)+C.\]Final Result
\[\arctan(e^x)+C\]Exercise 8 — Hyperbolic Substitution
Evaluate, for x > 1:
\[\int\frac{1}{\sqrt{x^2-1}}\,dx.\]Solution.
Set:
\[x=\cosh t.\]Then:
\[dx=\sinh t\,dt.\]Using the hyperbolic identity:
\[\cosh^2t-\sinh^2t=1,\]we obtain:
\[x^2-1 = \cosh^2t-1 = \sinh^2t.\]For the relevant values of t:
\[\sqrt{x^2-1} = \sinh t.\]Therefore:
\[\int \frac{\sinh t}{\sinh t} \,dt = \int1\,dt.\]Thus:
\[t+C.\]Since:
\[x=\cosh t,\]we have:
\[t=\operatorname{arcosh}(x).\]The inverse hyperbolic cosine satisfies:
\[\operatorname{arcosh}(x) = \log\left(x+\sqrt{x^2-1}\right).\]Therefore:
\[\log\left(x+\sqrt{x^2-1}\right)+C.\]Final Result
\[\log\left(x+\sqrt{x^2-1}\right)+C\]Exercise 9 — Hyperbolic Substitution with a Radical
Evaluate:
\[\int\sqrt{1+x^2}\,dx.\]Solution.
Set:
\[x=\sinh t.\]Then:
\[dx=\cosh t\,dt.\]Using:
\[1+\sinh^2t=\cosh^2t,\]we obtain:
\[\sqrt{1+x^2} = \cosh t.\]Therefore:
\[\int\sqrt{1+x^2}\,dx = \int\cosh^2t\,dt.\]Use the identity:
\[\cosh^2t = \frac{1+\cosh(2t)}{2}.\]Hence:
\[\int\cosh^2t\,dt = \frac12\int1\,dt + \frac12\int\cosh(2t)\,dt.\]Therefore:
\[\int\cosh^2t\,dt = \frac{t}{2} + \frac{\sinh(2t)}{4} + C.\]Using:
\[\sinh(2t)=2\sinh t\cosh t,\]we obtain:
\[\frac{t}{2} + \frac12\sinh t\cosh t + C.\]Now:
\[\sinh t=x,\]and:
\[\cosh t=\sqrt{1+x^2}.\]Moreover:
\[t=\operatorname{arsinh}(x).\]Thus:
\[\frac12x\sqrt{1+x^2} + \frac12\operatorname{arsinh}(x) + C.\]Equivalently:
\[\operatorname{arsinh}(x) = \log\left(x+\sqrt{1+x^2}\right).\]Final Result
\[\frac12 \left( x\sqrt{1+x^2} + \operatorname{arsinh}(x) \right) +C\]Exercise 10 — Secant Substitution
Evaluate, for x > 1:
\[\int\frac{1}{x\sqrt{x^2-1}}\,dx.\]Solution.
Set:
\[x=\sec t.\]Then:
\[dx=\sec t\tan t\,dt.\]Moreover:
\[x^2-1 = \sec^2t-1.\]Using:
\[\sec^2t-1=\tan^2t,\]we obtain:
\[\sqrt{x^2-1} = \tan t\]in the relevant range.
The integral becomes:
\[\int \frac{\sec t\tan t} {\sec t\tan t} \,dt.\]Therefore:
\[\int1\,dt=t+C.\]Since:
\[x=\sec t,\]we have:
\[t=\operatorname{arcsec}(x).\]Thus:
\[\operatorname{arcsec}(x)+C.\]For x > 1, this can also be written as:
\[\arccos\left(\frac1x\right)+C.\]Final Result
\[\operatorname{arcsec}(x)+C\]