Integration by Substitution

Solve by substituting a variable as a term. Whenever you think substitution, you want to find both $u$ and $u^{\prime }$ in the function. In other terms, you want $\frac{dy}{du}$ to be a constant or in terms of x

The Substitution Rule

Definite Integral

• If $f(x),g(x)$ and $f(g(x))g^{\prime }(x)$ are Continuous on their domains
• Then, ${\int }_a^{b}f(g(x))g^{\prime }(x)dx={\int }_{g(b)}^{g(a)}f(u)du$
• Where $u=g(x)$ and $g^{\prime }(x)=du$ Proof: Substitution Rule Definite Integral Form Proof

Example

Compute ${\int }_0^{\frac{\pi }{2}}{\sin }^5(x){\cos }^3(x)dx$ Note that this is equal to ${\int }_0^{\frac{\pi }{2}}{\sin }^5(x){\cos }^2(x)\cos (x)dx$ $={\int }_0^{\frac{\pi }{2}}{\sin }^5(x)(1-{\sin }^2(x))\cos (x)dx$ Then, choose $u=\sin (x)$ Then, $du=\cos (x)dx$ By alg, it follows that $f(x)=x^5(1-x^2)$ As: $x=0⟹u=\sin (0)=0$ $x=\frac{\pi }{2}⟹u=\sin \left(\frac{\pi }{2}\right)=1$ Then, ${\int }_0^{\frac{\pi }{2}}f(x)dx={\int }_0^{1}f(u)du$ by u-sub for definite integrals $={\int }_0^1{u}^5(1-u^2)du$ $={\int }_0^1(u^5-u^7)du$ $=\frac{1}{6}{u}^6-\frac{1}{8}{u}^8{|}_0^1$ $=\frac{1}{6}{\sin }^6(x)-\frac{1}{8}{\sin }^8(x){|}_0^1$ $=\frac{1}{6}-\frac{1}{8}-0-0$ $=\frac{2}{48}$ $=\frac{1}{24}$ Done

Indefinite Integral

• If $f(x),g(x)$ and $f(g(x))g^{\prime }(x)$ are Continuous on their domains
• Then. $\int f(g(x))g^{\prime }(x)dx=\int f(u)du$
• Where $u=g(x)$ and $g^{\prime }(x)=du$

Example

Compute $\int \frac{e^{\arctan (x)}}{1+x^2}dx$

• Note $f(x)=e^x$
• and $g(x)=\arctan (x)$
• allows for $f(g(x))g^{\prime }(x)=\frac{e^{\arctan x}}{1+x^2}$ Then, use Integration by substitution
• $\int f(u)du=\int e^{u}du=e^u+C$
• Note that by convention, $u=g(x),g^{\prime }(x)=du$
• Then, $e^u+C=e^{g(x)}+C=e^{\arctan x}+C$ Done

Practical uses

• Chain Rule Proof