Integration by substitution

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As usual with substitution the aim is to simplify the algebra. In this case simplifying the integrand to something we can more easily solved. The basic procedure is to take a substitution and use it simplify as follows:

A method of integration by Substitution:

• define the substitution (this may be given in the questions) in the form $u=f\left(x\right)$ where $f\left(x\right)$ will be an easily differentiable expression;
• if the integration is definite then use the limits in the substitution to find the limits in terms of $u$;
• find $f^{\text{'}}\left(x\right)$ and rewrite in the form $\frac{1}{f^{\text{'}}\left(x\right)}\frac{du}{dx}=1$;
• Substitute the new limits into the integration, multiply the integrand by lefthand side of the equation in the previous step and simplify ^* ;
• solve the simplified integration in terms of $u$;
• either complete the definite integration by substituting the calculated limits, or, complete an indefinite integration by substituting $f\left(x\right)$ for $u$.

^* This is the step which requires the most care.

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Example 1:

Find $\int x{(x-1)}^{9}dx$ using the substitution $u=x-1$.

Solution:

$$x=u+1$$ $$\frac{du}{dx}=1$$ $$\int x{(x-1)}^{9}dx$$ $$=\int (u+1)u^9\frac{du}{dx}dx$$ $$=\int u^{10}+u^{9}du$$ $$=\frac{u^{11}}{11}+\frac{u^{10}}{10}+C$$ $$=\frac{10u^{11}+11u^{10}}{110}+C$$ $$=\frac{u^{10}}{110}(10u+11)+C$$ $$=\frac{1}{110}{(x-1)}^{10}(10(x-1)+11)+C$$ $$=\frac{1}{110}{(x-1)}^{10}(10x+1)+C$$

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Questions:

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