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U-Substitution

Basic idea

U-substitution is the reverse of the derivative chain rule. This is important when integrating an expression while chain rule is important while differentiating.

MathJax TeX Test Page Let's say you have the expression $$\dfrac{d}{dx} e^{\sin(x)} =\cos(x)e^{\sin(x)}$$. We know this because of the chain rule. Now, let's go backward. $$e^{\sin(x)} = \int \cos(x)e^{\sin(x)}\, \mathrm{d}x$$ In order to solve this, we have to use the opposite of the chain rule: u-substitution.

Notice that $\cos(x)$ is the derivative of $\sin(x)$. We can replace $\sin(x)$ with $u(x)$. $$\int u'(x) e^{u(x)}\, \mathrm{d}x$$ We rearrange to write this: $$\int e^{u(x)}\, \boxed{u'(x)\mathrm{d}x}$$ Because it's the opposite of the chain rule, this equals $$e^{u(x)} + C = e^{\sin(x)} + C $$

Examples

MathJax TeX Test Page $$\int 2x\cos\left(x^2\right)\, \mathrm{d}x$$ In order to solve this, we have to look for some chain rule. So, we look inside our wrapping function, $\cos$, which envelops $x^2$. Now, we say $$u = x^2 \to \,\mathrm{d}u = 2x\,\mathrm{d}x$$ We rewrite our integral as $$\int \cos(u)\,\mathrm{d}u \to \sin(u) + C = \boxed{\sin(x^2) + C}$$
MathJax TeX Test Page $$\int \sin^5(x)\cos^5(x) \, \mathrm{d}x$$ $$\int \sin^5(x)\left(1 -\sin^2(x)\right)^2\boxed{\cos(x) \, \mathrm{d}x}$$ We now set $u = \sin(x)$, so $du = \cos(x)\,\mathrm{d}x$. $$\int u^5\left(1 - u^2\right)^2\,\mathrm{d}u$$ $$\int u^5\left(u^4 - 2u^2 + 1\right)\, \mathrm{d}u = \int u^9 - 2u^7 + u^5\, \mathrm{d}u = \frac{u^{10}}{10} - \frac{u^8}{4} + \frac{u^6}{6} + C$$ Our final answer becomes $$\frac{\sin^{10}(x)}{10} - \frac{\sin^8(x)}{4} + \frac{\sin^6(x)}{6} + C$$
MathJax TeX Test Page $$\int \cos(10x + 5)\, \mathrm{d}x$$ This problem requires u-substitution, even though there is only one term. This is because the integral isn't $\sin(10x + 5)$! If we differentiate that we get $10\cos(10x + 5)$ So, how do we u-sub this integral? We create the $u'(x)$! We multiple by $\frac{10}{10}$. $$\int \cos(10x + 5)\, \mathrm{d}x = \dfrac{1}{10}\int 10 \cos(10x + 5)\, \mathrm{d}x $$ Now, we set $u(x) = 10x + 5, \mathrm{d}u = 10\, \mathrm{d}x$. The integral becomes $$\dfrac{1}{10}\int \cos(u)\, \mathrm{d}u \to \boxed{\dfrac{1}{10}\sin(10x+5) + C}$$
David Witten

Work

Partial Fractions