More on U-Substitution

U-substitution is not only a technique that allows us to reverse the chain rule. It can also be a powerful means to guide the transformation of an integral into a more evaluatable form.

The following provides several examples of this more general application.


Evaluate $\displaystyle{\int \frac{4}{x^2+6x+9}\,dx}$


One's natural first thought for the choice of $u$ is likely to be the denominator. However, the derivative of that quadratic is not present or manufacturable in the numerator. However, we can factor the denominator which suggests a better choice of $u$.

$$\displaystyle{\int \frac{4}{x^2+6x+9}\,dx = \int \frac{4}{(x+3)^2}\,dx}$$

Now pick $u=x+3$, making $du = dx$. This gives us $$\displaystyle{\int \frac{4}{u^2}\,du = -4u^{-1} + C = \frac{-4}{x+3} + C}$$


Evaluate $\displaystyle{\int \frac{x}{\sqrt{1-4x^4}}\,dx}$


Here again, one's first thought for the choice of $u$ might be the denominator, or what is seen inside the square root. However, again we forsee difficulties manufacturing the derivative of either of these. Noting the form the denominator takes -- i.e., one minus a perfect square inside a square root -- suggests another way to look at this integral, one where the Arcsine function plays a role:

Let $u = 2x^2$, so that $u^2 = 4x^4$ and $du = 4x\,dx$. Then,

$$\displaystyle{\int \frac{x}{\sqrt{1-4x^4}}\,dx = \frac{1}{4} \int \frac{du}{\sqrt{1-u^2}} = \frac{1}{4}\textrm{Arcsin }(2x^2) + C}$$


Evaluate $\displaystyle{\int \cos^3(2x)\,dx}$


Here, the first choice for $u$ might be $\cos (2x)$, but its derivative of $-\sin (2x) \cdot 2$ is nowhere in sight. However, we can manufacture the appearance of the $\sin x$ and $\cos x$ pair by "stealing" a $\cos^2 x$ factor in the following way:

$$\int \cos^3 2x\,dx = \int \cos^2 2x \cdot \cos 2x\,dx = \int (1-\sin^2 2x) \cdot \cos 2x\,dx = \int (\cos 2x - \sin^2 2x \cos 2x)\,dx$$

Now we can split the integral into two, with the first evaluated with $u=2x$ and the second with $u = \sin 2x$ to get

$$\frac{1}{2} \sin 2x - \frac{1}{6} \sin^3 2x + C$$


Evaluate $\displaystyle{\int \sin^2 3x\,dx}$


Here, the first choice of $u$ might be $\sin 3x$, but we don't have its derivative $3\cos 3x$, and we don't have enough $\sin 3x$ factors to pull the trick used in the previous example.

Help comes in the form of the double-angle formula from trigonometry, $\cos 2\theta = 1 - 2\sin^2 \theta$.

We can solve this to find a replacement for $\sin^2 \theta$:

$$\sin^2 \theta = \frac{1-\cos 2\theta}{2}$$

Applying this to the integral above, we have

$$\int \sin^2 3x\,dx = \int \frac{1-\cos 6x}{2}\,dx = \frac{1}{2} \int(1-\cos 6x)\,dx$$

At this point, the integral can be easily found with $u = 6x$, yielding

$$\frac{1}{2}x - \frac{1}{12}\sin 6x + C$$


Evaluate $\displaystyle{\int x^2 \csc^4 x^3\,dx}$


Here, we do see the derivative of $x^3$ can be manufactured, so we start with a substitution of $u=x^3$ and $du = 3x^2\,dx$

$$\displaystyle{\int x^2 \csc^4 x^3\,dx = \frac{1}{3}\int \csc^4 u\,du}$$

Still, we are left with an expression where substituting $w = \csc u$ won't get us anywhere.

Instead, let us steal a $\csc^2 u$ factor, knowing we can translate this into something involving $\cot^2 u$. (Remember, the derivative of $\cot u$ is $-\csc^2 u$.) This gives us

$$\frac{1}{3}\int (1+\cot^2 u)(\csc^2 u)\,du$$

Now substitute $w = \cot u$ and $dw = -(\csc^2 u)\,du$ to get

$$-\frac{1}{3} \int (1 + w^2)\,dw = -\frac{1}{3}(w + \frac{1}{3}w^3) + C = -\frac{1}{3} \left(\cot u + \frac{1}{3}\cot^3 u \right)+ C$$

Finally, replace the $u$ with the appropriate expression in the original variable $x$ to complete the evaluation of the original integral

$$-\frac{1}{3} \left(\cot x^3 + \frac{1}{3}\cot^3 x^3 \right)+ C$$

We can use $u$-substitution with definite integrals too.

Although, we must remember that the limits of integration are values of the variable of integration. Thus, when we make our substitution that changes the variable of integration, we will also need to change the limits of integration to the corresponding values of this new variable.

As a benefit, however, this extra step removes the need to rewrite things in terms of our original variable after integrating, as the following examples demonstrate.


Evaluate $\displaystyle{\int_0^{\sqrt{2}/2} \frac{\textrm{Arcsin }t}{\sqrt{1-t^2}}\,dt}$


Let $u = \textrm{Arcsin }t$, so $\displaystyle{du = \frac{1}{\sqrt{1-t^2}}\,dt}$

Also, when $t=0$, $u = \textrm{Arcsin } 0 = 0$, and when $\displaystyle{t=\frac{\sqrt{2}}{2}}$, $\displaystyle{u = \textrm{Arcsin }\frac{\sqrt{2}}{2} = \frac{\pi}{4}}$

Then, making these substitutions, we have

$$\int_0^{\sqrt{2}/2} \frac{\textrm{Arcsin }t}{\sqrt{1-t^2}}\,dt = \int_0^{\pi/4} u\,du = \frac{1}{2} u^2 \bigg\rvert_0^{\pi/4} = \frac{1}{2} \left( \frac{\pi}{4} \right)^2 - \frac{1}{2} (0)^2 = \frac{\pi^2}{32}$$


Evaluate $\displaystyle{\int_1^4 \frac{s}{\sqrt{5-s}}\,ds}$


Let $u=5-s$, so $du = -ds$ and $s = 5-u$.

Also, when $s=1$, $u = 5-1 = 4$, and when $s=4$, $u=5-4=1$.

Then, making these substitutions, we have

$$\int_1^4 \frac{s}{\sqrt{5-s}}\,ds = -\int_4^1 \frac{5-u}{\sqrt{u}}\,du = -\int_4^1 (5u^{-1/2} - u^{1/2})\,du = \left(-10u^{1/2} + \frac{2}{3}u^{3/2}\right) \bigg\rvert_4^1 = -10 + \frac{14}{3} = \frac{16}{3}$$