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.

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

$$\int 2x\cos\left(x^2\right)\, \mathrm{d}x$$ 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 $$\int \cos(u)\,\mathrm{d}u \to \sin(u) + C = \boxed{\sin(x^2) + C}$$

$$\int \sin^5(x)\cos^5(x) \, \mathrm{d}x$$ $$\int \sin^5(x)\left(1 -\sin^2(x)\right)^2\boxed{\cos(x) \, \mathrm{d}x}$$ 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 $$\frac{\sin^{10}(x)}{10} - \frac{\sin^8(x)}{4} + \frac{\sin^6(x)}{6} + C$$

$$\int \cos(10x + 5)\, \mathrm{d}x$$ 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 $$\dfrac{1}{10}\int \cos(u)\, \mathrm{d}u \to \boxed{\dfrac{1}{10}\sin(10x+5) + C}$$