In this section, you will practice integration by u-substitution on some slightly harder problems. These problems will require a careful choice of \(u\) in order to remove all occurences of \(x.\) Before attempting the problems in this section, you should know how to integrate using u-substition for problems where the choice of \(u\) is obvious.
Integration by u-substitution (more difficult)
\(\displaystyle \int e^{x + e^x}\,dx\)
| \(\begin{align} & \int e^{x + e^x}\,dx \\[0.4em] =\ & \int e^x \cdot e^{e^x}\,dx \\[0.4em] =\ & \int e^u\,du \\[0.4em] =\ & e^u + C \\[0.4em] =\ & e^{e^x} + C \end{align}\) | \(\begin{align} & u = e^x \\ & du = e^x\,dx \end{align}\) |
\(\displaystyle \int \cot x[\ln(\sin x)]\,dx\)
| \(\begin{align} & \int \cot x[\ln(\sin x)]\,dx \\[0.4em] =\ & \int u\,du \\[0.4em] =\ & \dfrac{u^2}{2} + C \\[0.4em] =\ & \dfrac{[\ln(\sin x)]^2}{2} + C \end{align}\) | \(\begin{align} & u = \ln(\sin x) \\[0.4em] & du = \dfrac{1}{\sin x}\cos x\,dx \\[0.4em] & du = \cot x\,dx \end{align}\) |
\(\displaystyle \int \dfrac{x}{1 + x^4}\,dx\)
| \(\begin{align} & \int \dfrac{x}{1 + x^4}\,dx \\[0.75em] =\ & \int \dfrac{x}{1 + \left(x^2\right)^2}\,dx \\[0.75em] =\ & \int \dfrac{x}{1 + u^2} \cdot \dfrac{du}{2x} \\[0.75em] =\ & \int \dfrac{1}{1 + u^2} \cdot \dfrac{du}{2} \\[0.75em] =\ & \dfrac{1}{2} \int \dfrac{1}{1 + u^2}\,du \\[0.75em] =\ & \dfrac{1}{2}\tan^{-1} u + C \\[0.75em] =\ & \dfrac{1}{2}\tan^{-1}\left(x^2\right) + C \end{align}\) | \(\begin{align} & u = x^2 \\ & du = 2x\,dx \end{align}\) |
\(\displaystyle \int \tan x\ln(\cos x)\,dx\)
| \(\begin{align} & \int \tan x\ln(\cos x)\,dx \\[0.75em] =\ & \int (\tan x) u\dfrac{du}{-\tan x} \\[0.75em] =\ & -\int u\,du \\[0.75em] =\ & \dfrac{-1}{2}u^2 + C \\[0.75em] =\ & \dfrac{-1}{2}[\ln(\cos x)]^2 + C \end{align}\) | \(\begin{align} & u = \ln(\cos x) \\[0.4em] & du = \dfrac{1}{\cos x} \cdot (-\sin x)\,dx \\[0.4em] & du = -\tan x\,dx \\[0.4em] & \dfrac{du}{-\tan x} = dx \end{align}\) |
\(\displaystyle \int \dfrac{1}{1 + \sqrt{x}}\,dx\)
| \(\begin{align} & \int \dfrac{1}{1 + \sqrt{x}}\,dx \\[0.75em] =\ & \int \dfrac{1}{u}2(u - 1)\,du \\[0.75em] =\ & 2\int \dfrac{u - 1}{u}\,du \\[0.75em] =\ & 2\int \dfrac{u}{u} - \dfrac{1}{u}\,du \\[0.75em] =\ & 2\int 1 - \dfrac{1}{u}\,du \\[0.75em] =\ & 2(u - \ln \lvert u \rvert + C) \\[0.75em] =\ & 2u - 2\ln \lvert u \rvert + C \\[0.75em] =\ & 2(1 + \sqrt{x}) - 2\ln \lvert 1 + \sqrt{x} \rvert + C \\[0.75em] =\ & 2\sqrt{x} - 2\ln(1 + \sqrt{x}) + C \end{align}\) | \(\begin{align} u = 1 + \sqrt{x} \\[0.4em] u - 1 = \sqrt{x} \\[0.4em] du = \dfrac{1}{2\sqrt{x}}\,dx \\[0.4em] 2(u - 1)\,du = dx \end{align}\) |
\(\displaystyle \int \dfrac{e^x + 1}{e^x}\,dx\)
\(x - e^{-x} + C\)
\(\displaystyle \int 3^{\sin \theta}\cos \theta\,d\theta\)
\(\dfrac{3^{\sin \theta}}{\ln 3} + C\)
YouTube videos
- Integrating Exponential Functions By Substitution - Antiderivatives - CalculusThe Organic Chemistry Tutor