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Use substitution to find the antiderivative of $\int 3{x}^{2}{\left({x}^{3}-3\right)}^{2}dx}.$
$\int 3{x}^{2}{\left({x}^{3}-3\right)}^{2}dx}=\frac{1}{3}{\left({x}^{3}-3\right)}^{3}+C$
Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are substituting.
Use substitution to find the antiderivative of $\int z\sqrt{{z}^{2}-5}dz}.$
Rewrite the integral as $\int z{\left({z}^{2}-5\right)}^{1\text{/}2}dz}.$ Let $u={z}^{2}-5$ and $du=2z\phantom{\rule{0.2em}{0ex}}dz.$ Now we have a problem because $du=2z\phantom{\rule{0.2em}{0ex}}dz$ and the original expression has only $z\phantom{\rule{0.2em}{0ex}}dz.$ We have to alter our expression for du or the integral in u will be twice as large as it should be. If we multiply both sides of the du equation by $\frac{1}{2}.$ we can solve this problem. Thus,
Write the integral in terms of u , but pull the $\frac{1}{2}$ outside the integration symbol:
Integrate the expression in u :
Use substitution to find the antiderivative of $\int {x}^{2}{\left({x}^{3}+5\right)}^{9}dx}.$
$\frac{{\left({x}^{3}+5\right)}^{10}}{30}+C$
Use substitution to evaluate the integral $\int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}t}{{\text{cos}}^{3}t}dt}.$
We know the derivative of $\text{cos}\phantom{\rule{0.1em}{0ex}}t$ is $\text{\u2212}\text{sin}\phantom{\rule{0.1em}{0ex}}t,$ so we set $u=\text{cos}\phantom{\rule{0.1em}{0ex}}t.$ Then $du=\text{\u2212}\text{sin}\phantom{\rule{0.1em}{0ex}}tdt.$ Substituting into the integral, we have
Evaluating the integral, we get
Putting the answer back in terms of t , we get
Use substitution to evaluate the integral $\int \frac{\text{cos}\phantom{\rule{0.1em}{0ex}}t}{{\text{sin}}^{2}t}dt.$
$-\frac{1}{\text{sin}\phantom{\rule{0.1em}{0ex}}t}+C$
Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we are done, u should be the only variable in the integrand. In some cases, this means solving for the original variable in terms of u . This technique should become clear in the next example.
Use substitution to find the antiderivative of $\int \frac{x}{\sqrt{x-1}}dx.$
If we let $u=x-1,$ then $du=dx.$ But this does not account for the x in the numerator of the integrand. We need to express x in terms of u . If $u=x-1,$ then $x=u+1.$ Now we can rewrite the integral in terms of u :
Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result. Thus,
Use substitution to evaluate the indefinite integral $\int {\text{cos}}^{3}t\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}t\phantom{\rule{0.2em}{0ex}}dt}.$
$-\frac{{\text{cos}}^{4}t}{4}+C$
Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.
Let $u=g\left(x\right)$ and let ${g}^{\text{\u2032}}$ be continuous over an interval $\left[a,b\right],$ and let f be continuous over the range of $u=g\left(x\right).$ Then,
Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for indefinite integrals, if $F\left(x\right)$ is an antiderivative of $f\left(x\right),$ we have
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