Integration by Substitution
Example 12a
- Find \(\int x\sqrt{ 2x+5 }\, dx\)
- We let \(u=2x+5\) be our substitution.
- \(\frac{du}{dx}=2\)
- Rearrange to give \(dx=\frac{du}{2}\)
- Now the sqaure root goes to \(\sqrt{ 2x+5 }=u^{\frac{1}{2}}\)
- \(x=\frac{u-5}{2}\)
- We can now rewrite our integral in terms of \(u\): \(I = \int \frac{u-5}{2}u^{\frac{1}{2}} \times \frac{1}{2}\, du\)
- \(= \int \frac{1}{4}(u-5)u^{\frac{1}{2}} \, du\)
- \(= \int \frac{1}{4}\left( u^{\frac{3}{2}}-5u^{\frac{1}{2}} \right) \, du\)
- We now integrate: \(\frac{1}{4}\times \frac{u^{\frac{5}{2}}}{\frac{5}{2}}-\frac{5u^{\frac{3}{2}}}{4\times \frac{3}{2}}+c\)
- Simplifies to: \(\frac{u^{\frac{5}{2}}}{10}-\frac{5u^{\frac{3}{2}}}{6}+c\)
- Our final answer is: \(\int = \frac{(2x+5)^{5/2}}{10}-\frac{5(2x+5)^{\frac{3}{2}}}{6}+c\)
Example 12b
- Our integral is \(\int x\sqrt{ 2x+5 }\, dx\)
- Substitution is: \(u^2=2x+5\)
- Implicit Differentiation means that this differentiates to \(2u \frac{du}{dx}=2\)
- Rearrange: \(2u \frac{du}{2}=dx\)
- \(dx=udu\)
- \(\sqrt{ 2x+5 }=u\)
- \(x=\frac{u^2-5}{2}\)
- \(I=\int (\frac{u^2-5}{2})u^2\times du\) (simplified \(u\times u \to u^2\))
- \(= \int \left( \frac{u^4}{2} -\frac{5u^2}{2}\right) du\)
\(= \frac{u^5}{10}-\frac{5u^3}{6}+c\)
\(\int \cos x\sin x (1-\sin x)^3\, dx\)
- Using \(u=\sin x+1\)
- \(\frac{du}{dx}=\cos x\)
- \(dx=\frac{du}{\cos x}\)
- \(\int (u-1)(u^3) \, du\)
- \(\int u^4-u^3 \, du\)
- \(=\frac{u^5}{5}-\frac{u^4}{4}+c\)
- \(\frac{(\sin x+1)^5}{5}-\frac{(\sin x+1)^4}{4}+c\)
\(\int x\sqrt{ 1+x } \, dx, u=1+x\)
- \(\frac{du}{dx}=1\)
- \(du=dx\)
- \(\int (u-1)(u)^{\frac{1}{2}} \, du\)
- \(= \int u^{\frac{3}{2}}-u^{\frac{1}{2}} \, du\)
- \(=\frac{u^{\frac{5}{2}}}{\frac{5}{2}}-\frac{u^{\frac{3}{2}}}{\frac{3}{2}}\)
- \(=\frac{2u^{\frac{5}{2}}}{5}+\frac{2u^{\frac{3}{2}}}{3}+c\)
\(\int \frac{1+\sin x}{\cos x} \, dx,u=\sin x\)
- \(\frac{du}{dx}=\cos x\)
- \(dx=\frac{du}{\cos x}\)
- \(\int \frac{1+u}{\cos^2x} \, du\)
- \(\int \frac{1+u}{\cos^2x} \, du\)
- \(\int \frac{1+u}{(1+u)(1-u)} \, du\)
- \(\int \frac{1}{1-u} \, du\)
- \(=\ln |1-u|+c\)
- \(=\ln |1-\sin x|+c\)
\(\int \sin^3x \, dx,u=\cos x\)
- \(\frac{du}{dx}=-\sin x\)
- \(-du=\sin xdx\)
- \(dx=-\frac{du}{\sin x}\)
- \(\int (\sin x)(1-\cos^2x) \, dx\)
- \(\int (\cos^2x-1) \, du\)
- \(\int (u^2-1) \, du\)
- \(= \frac{u^3}{3}-x\)
- \(=\frac{\cos^3 x}{3}-\cos x+c\)