Integration by substitution

Integration by substitution for indefinite and definite integrals, finding du, changing the limits, substitutions of the form u = x − a that need x written in terms of u, and substituting x as a function of u.

The method

Example 1●●

Use $u = x^2 + 1$ to evaluate $\displaystyle\int_0^1 x(x^2 + 1)^3\,dx$.

$du = 2x\,dx$, so $x\,dx = \tfrac12\,du$. When $x = 0$, $u = 1$; when $x = 1$, $u = 2$.

$\displaystyle\tfrac12\int_1^2 u^3\,du = \tfrac12\Big[\tfrac{u^4}{4}\Big]_1^2 = \tfrac{16 - 1}{8} = \tfrac{15}{8}$.

Example 2: writing x in terms of u●●●

Use $u = x - 1$ to evaluate $\displaystyle\int_1^2 x\sqrt{x - 1}\,dx$.

$du = dx$ and $x = u + 1$. The limits become $u = 0$ and $u = 1$.

$\displaystyle\int_0^1 (u + 1)u^{\frac12}\,du = \int_0^1 \big(u^{\frac32} + u^{\frac12}\big)\,du = \Big[\tfrac25 u^{\frac52} + \tfrac23 u^{\frac32}\Big]_0^1$.

$= \tfrac25 + \tfrac23 = \tfrac{16}{15}$.

Example 3●●●

Use $u = x + 1$ to evaluate $\displaystyle\int_0^3 \frac{x}{\sqrt{x + 1}}\,dx$.

$x = u - 1$, $dx = du$, and the limits become $1$ and $4$.

$\displaystyle\int_1^4 \big(u^{\frac12} - u^{-\frac12}\big)\,du = \Big[\tfrac23 u^{\frac32} - 2u^{\frac12}\Big]_1^4 = \big(\tfrac{16}{3} - 4\big) - \big(\tfrac23 - 2\big) = \tfrac83$.

Flashcards

First step of a substitution

Find $du$ in terms of $dx$.

Definite integrals and substitution

Change the limits to values of $u$.

$u = x - 1$ with an $x$ left over

Write $x = u + 1$.

Practice

Use $u = x^2 + 1$ to evaluate $\displaystyle\int_0^1 x(x^2 + 1)^3\,dx$.
Use $u = x - 1$ to evaluate $\displaystyle\int_1^2 x\sqrt{x - 1}\,dx$.
Use $u = x + 1$ to evaluate $\displaystyle\int_0^3 \frac{x}{\sqrt{x + 1}}\,dx$.
Use $u = \cos x$ to evaluate $\displaystyle\int_0^{\frac{\pi}{2}} \sin x\cos^2 x\,dx$.
Use $u = \ln x$ to evaluate $\displaystyle\int_1^e \frac{\ln x}{x}\,dx$.
Under $u = x^2 + 1$, what does the upper limit $x = 1$ become?
With $u = 3x - 2$, $dx = $
For an indefinite integral done by substitution, the final answer should be written in terms of:
A substitution changes the limits of $\displaystyle\int_0^3$ to $\displaystyle\int_1^4$. Then $u$ could be: