Integration by Substitution

7.2 Handout for today

According to the Chain Rule...

$$\frac{d}{dx}f({\color{red}g(x)}) =f'(\color{red}{g(x)})\cdot\color{blue}{g'(x)}.$$ That means that... $$\int f'(\color{red}g(x)\color{black})\cdot\color{blue}g'(x)\color{black}\,dx=f(\color{red}{g(x)}\color{black}) + C.$$ where $\color{red}{g(x)}$ is the "inside function" and $\color{blue}{g'(x)}$ is the derivative of the inside function.

Example

Use the chain rule to take the derivative of $$f(x)=\frac 16 (x^2+1)^6$$ and then write the corresponding antiderivative formula.

$$ \color{black}{\frac{d}{dx}\frac 16} (\color{red}{x^2+1}\color{black})^6 = (\color{red}{x^2+1}\color{black})^5\cdot \color{blue}{2x}.$$ And therefore: $$\int (x^2+1)^52x\,dx = \frac 16(x^2+1)^6+C.$$

Substitution

To make a substitution in an integral:

Let $\color{red}{w(x)}$ be some function of $x$, then $dw=\color{blue}{w'(x)}\,dx=\color{blue}{\frac{dw}{dx}}\,dx$. Substitute these expressions into your integral, and try to solve a new (and hopefully simpler) integral in terms of $w$ and $dw$.

You may always carry out substitution with any integral, however it's particularly useful, in the context of the chain rule, where... $$\int f(w(x))\,w'(x)\,dx = \int f(w)\,dw$$

Example

Integrate: $$\int 3x^2\cos(x^3)\,dx.$$

Solution

Look for an inside function and its derivative... $$\int {\color{blue}3x^2}\cos({\color{red}x^3})\,dx.$$ Here the inside function is $\color{red}w=x^3$. So that means ${\color{blue}dw}=w'(x)\,dx = {\color{blue}3x^2\,dx}$. Now, we make these substitution in the integral: $$\int \cos({\color{red}x^3}){\color{blue}3x^2\,dx} = \int\cos({\color{red}w})\,{\color{blue}dw} = \sin(w)+C.$$ The last step, now that we have an anti-derivative, is to substitute back in $\color{red}w\to x^3$: $$\sin(w)+C=\sin(x^3)+C.$$

A constant factor

Imagine that we had been asked to integrate: $$\int \cos(x^3)x^2\,dx.$$ The inside function is $\color{red}{w=x^3}$, and $\color{blue}{dw=w'(x)\, dx=3x^2\, dx}$.

🙀¡ We have $x^2$ in the integrand, but not the 3 that we need to make $dw$! 🙀

Never fear... We can sneak a 3 in there by multiplying by $1=\frac33$ like this: $$\int \cos({\color{red}x^3})\frac 13{\color{blue}3x^2\,dx} = \frac 13 \int\cos(w)\,dw=\frac 13\sin(w)+C=\frac13\sin(x^3)+C.$$

Handout #7

$$\int x^2\sqrt{x-5}\,dx$$ In this one, it looks like an inside function might be $w(x)=x-5$. But that doesn't obviously get rid of the pesky $x^2$.

Well, try it anyway: Use $w(x)=x-5$, and $w'=$? Now solve $w=x-5$ for $x$, and substitute that in for $x^2$. Now there is a way to get an integrand that you can integrate. Can you find it?

Edfinity

Many of the Edfinity homework problems *require* that you tack on "$+C$" onto the end of your antiderivative. As in... $$\int 4x^3\,dx = x^4\color{red}+C$$

Try these

$\int f(g(x))\cdot g'(x)\,dx=\int f(w)\,dw$ where $w=g(x)$.

$\int te^{t^2}\,dt$

$\int t^2e^{t^3+1}\,dt$

$\int x^5\,\sin(2x^6+7)\,dx$

$\int \frac{(\ln z)^3}{z} \,dz$

$\int \frac{e^x+1}{e^x+x}\,dx$

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