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How to Do Integration by Substitution

Substitution in integration is the chain rule reversed. The formula is as follows:

Formula

Substitution

\begin{array}{l} \int f (g (x)) \cdot g^{'} (x) d x & = \int f (u) d u \\ = F (u) \\ = F (g (x)) \end{array}

You use substitution when the expression contains two functions where one of the functions is the derivative of the other. Choose $u$ as the function that is not differentiated! Next, you differentiate both sides with respect to $x$ so that you have $\frac{d u}{d x}$ . Then, you multiply by $d x$ on both sides of the equation to cancel the $d x$ in the denominator, such that you have solved for $d u$ .

Example 1

$\begin{array}{l} \int 2 x e^{x^{2}} d x & \overset{*}{=} \int e^{u} d u \\ = e^{u} + C \\ = e^{x^{2}} + C \end{array}$

$\int 2 x e^{x^{2}} d x \overset{*}{=} \int e^{u} d u = e^{u} + C = e^{x^{2}} + C$

*

$\begin{aligned} u & = x^{2} \\ \frac{d u}{d x} & = 2 x \\ d u & = 2 x d x \end{aligned}$

Example 2

Compute $\int \frac{2 x + 1}{x^{2} + x} d x$

\begin{array}{l} \int \frac{2 x + 1}{x^{2} + x} d x & \overset{*}{=} \int \frac{1}{u} d u \\ = \ln | u | + C \\ = \ln | x^{2} + x | + C \end{array}

\begin{aligned} u & = x^{2} + x \\ \frac{d u}{d x} & = 2 x + 1 \\ d u & = 2 x + 1 d x \end{aligned}

Example 3

Compute $\int \frac{\sin x}{\cos^{2} x} d x$

\begin{array}{l} \int \frac{\sin x}{\cos^{2} x} d x & \overset{*}{=} \int - \frac{1}{u^{2}} d u \\ = - \int u^{- 2} d u \\ = u^{- 1} + C \\ = \frac{1}{u} + C \\ = \frac{1}{\cos x} + C \end{array}

\begin{aligned} u & = \cos x \\ \frac{d u}{d x} & = - \sin x \\ d u & = - \sin x d x \end{aligned}

Example 4

Compute $\int \cos x e^{\sin x} d x$

\begin{array}{l} \int \cos x e^{\sin x} d x & \overset{*}{=} \int e^{u} d u \\ = e^{u} + C \\ = e^{\sin x} + C \end{array}

\begin{aligned} u & = \sin x \\ \frac{d u}{d x} & = \cos x \\ d u & = \cos x d x \end{aligned}

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