Substitution

You can use substitution to convert a complicated integral into a simpler one. In these problems, I'll let u equal some convenient x-stuff --- say $u
   = f(x)$ . To complete the substitution, I must also substitute for $dx$ . To do this, compute $\der u x = f'(x)$ , so $du = f'(x)\,dx$ . Then $dx = \dfrac{du}{f'(x)}$ .


Example. Compute $\displaystyle \int (2x + 3)^{100}\,dx$ .

$$\int (2x + 3)^{100}\,dx = \int u^{100}\cdot \dfrac{du}{2} = \dfrac{1}{2} \int u^{100}\,du = \dfrac{1}{202}u^{101} + C = \dfrac{1}{202}(2x + 3)^{101} + C.$$

$$\left[u = 2x + 3, \quad du = 2\,dx, \quad dx = \dfrac{du}{2}\right] \quad\halmos$$


Example. Compute $\displaystyle \int \dfrac{dx}{\sqrt{4 - 7x}}$ .

$$\int \dfrac{dx}{\sqrt{4 - 7x}} = \int \dfrac{1}{\sqrt{u}}\cdot \left(-\dfrac{du}{7}\right) = -\dfrac{1}{7} \int u^{-1/2}\,du = -\dfrac{2}{7}u^{1/2} + C = -\dfrac{2}{7}(4 - 7x)^{1/2} + C.$$

$$\left[u = 4 - 7x, \quad du = -7\,dx, \quad dx = -\dfrac{du}{7}\right]\quad\halmos$$


Example. Later on, I'll derive the integration formula

$$\int \dfrac{dx}{x} = \ln |x| + C.$$

Use this formula to compute $\displaystyle
   \int \dfrac{1}{3x + 1}\,dx$ .

$$\int \dfrac{1}{3x + 1}\,dx = \int \dfrac{1}{u}\cdot \dfrac{du}{3} = \dfrac{1}{3} \int \dfrac{du}{u} = \dfrac{1}{3} \ln |u| + C = \dfrac{1}{3} \ln |3x + 1| + C.$$

$$\left[u = 3x + 1, \quad du = 3\,dx, \quad dx = \dfrac{du}{3}\right] \quad\halmos$$


Example. Compute $\displaystyle \int x(x^2 + 5)^{50}\,dx$ .

$$\int x(x^2 + 5)^{50}\,dx = \int xu^{50}\cdot \dfrac{du}{2x} = \dfrac{1}{2} \int u^{50}\,du = \dfrac{1}{102}u^{51} + C = \dfrac{1}{102}(x^2 + 5)^{51} + C.$$

$$\left[u = x^2 + 5, \quad du = 2x\,dx, \quad dx = \dfrac{du}{2x}\right]\quad\halmos$$


Notice that in the second step in the last example, the x's cancelled out, leaving only u's. If the x's had failed to cancel, I wouldn't have been able to complete the substitution.

But what made the x's cancel? It was the fact that I got an x from the derivative of $u = x^2 + 5$ . This leads to the following rule of thumb.

Substitute for something whose derivative is also there.


Example. Compute $\displaystyle \int (x + 1)\sqrt{x^2 + 2x + 5}\,dx$ .

$$\int (x + 1)\sqrt{x^2 + 2x + 5}\,dx = \int (x + 1)\sqrt{u}\cdot \dfrac{du}{2(x + 1)} = \dfrac{1}{2} \int \sqrt{u}\,du = \dfrac{1}{2} \cdot \dfrac{2}{3}u^{2/3} + C = \dfrac{1}{3} (x^2 + 2x + 5)^{3/2} + C.$$

$$\left[u = x^2 + 2x + 5, \quad du = (2x + 2)\,dx = 2(x + 1)\,dx, \quad dx = \dfrac{du}{2(x + 1)}\right]\quad\halmos$$


Example. Compute $\displaystyle \int \sin (3x + 1)\,dx$ .

$$\int \sin (3x + 1)\,dx = \int \sin u\cdot \dfrac{du}{3} = \dfrac{1}{3} \int \sin u\,du = -\dfrac{1}{3}\cos u + C = -\dfrac{1}{3}\cos (3x + 1) + C.$$

$$\left[u = 3x + 1, \quad du = 3\,dx, \quad dx = \dfrac{du}{3}\right] \quad\halmos$$


Example. Compute $\displaystyle \int (\sin 5x)^7 \cos 5x\,dx$ .

$$\int (\sin 5x)^7 \cos 5x\,dx = \int u^7 \cos 5x\cdot \dfrac{du}{5\cos 5x} = \dfrac{1}{5} \int u^7\,du = \dfrac{1}{40}u^8 + C = \dfrac{1}{40}(\sin 5x)^8 + C.$$

$$\left[u = \sin 5x, \quad du = 5\cos 5x\,dx, \quad dx = \dfrac{du}{5\cos 5x}\right]\quad\halmos$$


Example. Compute $\displaystyle \int \dfrac{1}{\sqrt{x}(\sqrt{x} + 9)^2}\,dx$ .

$$\int \dfrac{1}{\sqrt{x}(\sqrt{x} + 9)^2}\,dx = \int \dfrac{1}{\sqrt{x}u^2}\cdot 2\sqrt{x}\,du = 2 \int u^{-2}\,du = -\dfrac{2}{u} + C = -\dfrac{2}{\sqrt{x} + 9} + C.$$

$$\left[u = \sqrt{x} + 9, \quad du = \dfrac{dx}{2\sqrt{x}}, \quad dx = 2\sqrt{x}\,du\right]\quad\halmos$$


Example. Compute $\displaystyle \int \dfrac{f'(x)}{(f(x) + 3)^2}\,dx$ .

$$\int \dfrac{f'(x)}{(f(x) + 3)^2}\,dx = \int \dfrac{f'(x)}{u^2}\cdot \dfrac{du}{f'(x)} = \int \dfrac{du}{u^2} = -\dfrac{1}{u} + C = -\dfrac{1}{f(x) + 3} + C.$$

$$\left[u = f(x) + 3, \quad du = f'(x)\,dx, \quad dx = \dfrac{du}{f'(x)}\right]\quad\halmos$$


Example. Compute $\displaystyle \int \dfrac{\sin \dfrac{1}{x}}{x^2}\,dx$ .

$$\int \dfrac{\sin \dfrac{1}{x}}{x^2}\,dx = \int \dfrac{\sin u}{x^2}\cdot (-x^2\,du) = -\int \sin u\,du = \cos u + C = \cos \dfrac{1}{x} + C.$$

$$\left[u = \dfrac{1}{x}, \quad du = -\dfrac{dx}{x^2}, \quad dx = -x^2\,du\right]\quad\halmos$$


Example. Compute $\displaystyle \int (2x - 1)(2x + 3)^{40}\,dx$ .

$$\int (2x - 1)(2x + 3)^{40}\,dx = \int (u - 4)u^{40}\,du = \int (u^{41} - 4u^{40})\,du =$$

$$\left[u = 2x + 3, \quad du = 2\,dx, \quad dx = \dfrac{du}{2}; \quad x = \dfrac{1}{2}(u - 3)\right]$$

$$\dfrac{1}{42}u^{42} - \dfrac{4}{41}u^{41} + C = \dfrac{1}{42}(2x + 3)^{42} - \dfrac{4}{41}(2x + 3)^{41} + C. \quad\halmos$$


Contact information

Bruce Ikenaga's Home Page

Copyright 2005 by Bruce Ikenaga