Integration Technique #1: Substitution

1. Try integrating
(1)\displaystyle \int^{2}_{0}x\cos(x^{2}+1)\,\mathrm{d}x=???
It’s kind of hard, so we try to do something very clever: use the chain rule.

2. Recall the chain rule states, if f(x) and g(x) are differentiable functions, then
(2)\displaystyle \frac{\mathrm{d}(f(g(x))}{\mathrm{d}x}=\frac{\mathrm{d}f(g(x))}{\mathrm{d}g(x)}\frac{\mathrm{d}g(x)}{\mathrm{d}x}
We can apply the fundamental theorem of calculus to both sides
(3)\displaystyle \int\frac{\mathrm{d}(f(g(x))}{\mathrm{d}x}\,\mathrm{d}x=\int\frac{\mathrm{d}f(g(x))}{\mathrm{d}g(x)}\frac{\mathrm{d}g(x)}{\mathrm{d}x}\,\mathrm{d}x
The left hand side becomes
(4)\displaystyle \int\frac{\mathrm{d}(f(g(x))}{\mathrm{d}x}\,\mathrm{d}x=f(g(x))+C
But what about the right hand side of Equation (3)?

Let
(5)\displaystyle u=g(x) and \displaystyle \mathrm{d}u=\frac{\mathrm{d}g(x)}{\mathrm{d}x}\,\mathrm{d}x
Then the right hand side of (3) becomes
(6)\displaystyle \int\frac{\mathrm{d}f(g(x))}{\mathrm{d}g(x)}\frac{\mathrm{d}g(x)}{\mathrm{d}x}\,\mathrm{d}x=\int\frac{\mathrm{d}f(u)}{\mathrm{d}u}\,\mathrm{d}u
Observe, we can apply the fundamental theorem of calculus to this monstrosity. We get
(7)\displaystyle \int\frac{\mathrm{d}f(u)}{\mathrm{d}u}\,\mathrm{d}u=f(u)+C
We set this equal to (4), since both Equations (4) and (7) are just manipulations of Equation (3). We find a nice method to perform mildly complicated integrals!

Example 1. Recall our motivating problem:
(1)\displaystyle \int^{2}_{0}x\cos(x^{2}+1)\,\mathrm{d}x=???
Lets let
(8)\displaystyle u=x^{2}+1\quad and \displaystyle \mathrm{d}u=\frac{\mathrm{d}(x^{2}+1)}{\mathrm{d}x}=2x
We see we can plug these into Equation (1)
(10)\displaystyle \displaystyle \int^{2}_{0}x\cos(x^{2}+1)\,\mathrm{d}x=\int^{x=2}_{x=0}\cos(u)\frac{\mathrm{d}u}{2}
But we know how to integrate \cos(u), we see the integral is
(11)\displaystyle \displaystyle \int^{x=2}_{x=0}\cos(u)\frac{\mathrm{d}u}{2}=\left.\frac{\sin(u)}{2}\right|^{x=2}_{x=0}
Now we “unroll” our substitution, plugging in \displaystyle u=x^{2}+1\quad which makes (11) become
(12)\displaystyle \displaystyle \int^{x=2}_{x=0}\cos(u)\frac{\mathrm{d}u}{2}=\left.\frac{\sin(x^{2}+1)}{2}\right|^{x=2}_{x=0}
then we can evaluate the right hand side on the boundaries as usual.

Example 2. Lets consider a more complicated problem:
(12)\displaystyle \int\frac{\ln(x)}{x}\mathrm{d}x=???
We take
(13)\displaystyle u = \ln(x)\quad and \displaystyle \mathrm{d}u = \frac{\mathrm{d}x}{x}
Thus we can rewrite (12) as
(14)\displaystyle \int\frac{\ln(x)}{x}\mathrm{d}x=\int u\,\mathrm{d}u
This is a trivial integral
(14′)\displaystyle \int\frac{\ln(x)}{x}\mathrm{d}x=\frac{u^{2}}{2}+C
and we “unroll” the substitution obtaining
(15)\displaystyle \int\frac{\ln(x)}{x}\mathrm{d}x=\frac{\bigl(\ln(x)\bigr)^{2}}{2}+C
where C is a constant of integration.

Example 3 [Math.SE]. Evaluate
(16)\displaystyle \int \frac{\sin(x)}{3\cos^{3}(x)+\sin^{2}(x)\cos(x)} \mathrm{d}x.

Solution. We recall the trigonometric identity
(17)\displaystyle \sin^{2}(x)=1-\cos^{2}(x)
which makes our integral become
(18)\displaystyle \int \frac{\sin(x)}{3\cos^{3}(x)+\sin^{2}(x)\cos(x)} \mathrm{d}x=\int \frac{\sin(x)}{3\cos^{3}(x)+\bigl(1-\cos^{2}(x)\bigr)\cos(x)} \mathrm{d}x.
Now we pick
(19)\displaystyle u = \cos(x) and \displaystyle \mathrm{d}u = -\sin(x)\,\mathrm{d}x.
Our integral (18) becomes
(20)\displaystyle \int \frac{\sin(x)}{3\cos^{3}(x)+\bigl(1-\cos^{2}(x)\bigr)\cos(x)} \mathrm{d}x=\int \frac{-\mathrm{d}u}{3u^{3}+\bigl(1-u^{2}\bigr)u} .
Now we do the following trick:
(21)\displaystyle \begin{aligned}\frac{2u}{2u^{2}+1}-\frac{1}{u}  &=\frac{2u^{2}}{u(2u^{2}+1)}-\frac{2u^{2}+1}{u(2u^{2}+1)}\\  &=\frac{2u^{2}-(2u^{2}+1)}{u(2u^{2}+1)}\\  &=\frac{-1}{u(2u^{2}+1)}\end{aligned}
Thus our integral (20) simplifies to
(22)\displaystyle \int \frac{-\mathrm{d}u}{3u^{3}+\bigl(1-u^{2}\bigr)u} = \int\left(\frac{2u}{2u^{2}+1}-\frac{1}{u}\right)\mathrm{d}u.
But these are two easy integrals! We see that
(23)\displaystyle \int\frac{1}{u}\,\mathrm{d}u = \ln(u)+C
where C is a constant of integration.

The other integral
(24)\displaystyle \int\frac{2u}{2u^{2}+1}\mathrm{d}u
requires another substitution, namely,
(25)\displaystyle v=u^{2} and \displaystyle\mathrm{d}v=2u\,\mathrm{d}u
So (24) becomes
(26)\displaystyle \begin{aligned}\int\frac{u}{u^{2}+1}\mathrm{d}u&=\int\frac{\mathrm{d}v}{2v+1}\\  &=\ln(2v+1)+C'\\  &=\ln(2u^{2}+1)+C'\end{aligned}
where we simply unroll the substitution.

Now we see that
(27)\displaystyle \int\left(\frac{2u}{2u^{2}+1}-\frac{1}{u}\right)\mathrm{d}u = \ln(2u^{2}+1)-\ln(u)+C.
WARNING: the constant of integration C in (27) is different than in (23). We just are sloppy/lazy and use the same letter.

But we’re not done yet! We still have to plug in u=\cos(x). So we have
(28)\displaystyle \ln(2u^{2}+1)-\ln(u)+C=\ln\bigl(2\cos^{2}(x)+1\bigr)-\ln\bigl(\cos(x)\bigr)+C.
Thus we have
(29)\displaystyle \int \frac{\sin(x)}{3\cos^{3}(x)+\sin^{2}(x)\cos(x)} \mathrm{d}x=\ln\bigl(2\cos^{2}(x)+1\bigr)-\ln\bigl(\cos(x)\bigr)+C.
This concludes our example.

Exercise 1. Calculate \displaystyle\int \frac{1}{\sqrt{2-x^2}} \,\mathrm{d}x = ???

Exercise 2 [Math.SE]. Calculate \displaystyle\int\frac{1}{(1-x^{2})^{3/2}}\,\mathrm{d}x

Exercise 3 [Math.SE]. Calculate \displaystyle\int \frac{x^{3}}{(x^{2} + 1)^{1/3}}\,\mathrm{d}x

Exercise 4. Calculate \displaystyle\int \frac {\mathrm{d}x}{x^{2}\sqrt{4-x^2}}= ???

Exercise 5. Evaluate \displaystyle\int x\ln(1+x)\,\mathrm{d}x.

Exercise 6. Compute \displaystyle\int \frac{\mathrm{d}u}{u \sqrt{5-u^2}}

Exercise 7. Calculate \displaystyle\int\mathrm{e}^{\cos(t)}\sin(2t)\,\mathrm{d}t

Exercise 8. Consider \displaystyle\int\frac{\mathrm{e}^{\arctan(x)}}{1+x^2}\,\mathrm{d}x = ???

Exercise 9 [Math.SE]. Evaluate \displaystyle\int \sqrt{x^2 + 2x}\,\mathrm{d}x = ???

Exercise 10 [Very Trick!, Math.SE] . Compute \displaystyle \int \frac{x+4}{x^2 + 2x + 5}\,\mathrm{d}x

Exercise 11 [Math.SE]. Calculate the indefinite integral \displaystyle\int\frac {x^{2}}{x^{6} + 9}\,\mathrm{d}x

Exercise 12. Compute \displaystyle \int\sec(\theta)\,\mathrm{d}\theta. [Hint: multiply by 1=f(\theta)/f(\theta) where f(\theta) is some function involving only secant and tangent of \theta.]

Exercise 13. Find \displaystyle\int\sqrt{1-x^{2}}\,\mathrm{d}x.

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About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Calculus, Definite Integral, Integral, Integration Techniques. Bookmark the permalink.

3 Responses to Integration Technique #1: Substitution

  1. Pingback: Area of a Surface of Revolution (Part 1) | My Math Blog

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