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The following is a general method to find indefinite integrals that look like the result of a chain rule.
- When to use it: We use the method of substitution for indefinite integrals which look like the result of a chain rule. In particular, try to use this method when you see a composition of two functions.
- How to use it: In this method, we go from integrating with respect to $x$ to integrating with respect to a new variable, $u$ , which makes the integral much easier.
- Find inside the integral the composition of two functions and set $u=$ ``the inner function''.
- We also write $du=\frac{du}{dx}dx$ .
- Substitute everything in the integral that depends on $x$ in terms of $u$ .
- Integrate with respect to $u$ .
- Once we have the result of integration in terms of $u$ ($+ C$ ), substitute back in terms of $x$ .
The method is best explained through examples:
Example 0.1 We want to find $\int e^{2x} dx $ . The integrand is $e^{2x}$ , which is a composition of two functions. The inner function is $2x$ so we set: $$u=2x,\quad du=2dx$$ Thus, $$x=u/2,\quad dx=du/2$$ Substitute into the integral: $$\int e^{2x}dx= \int e^u \frac{du}{2}=\frac{1}{2}\int e^u du = \frac{1}{2} e^u +C=\frac{1}{2}e^{2x} + C$$
The following are typical examples where we use the subsitution method:
Example 0.2 $$\int xe^{3x^2+7} dx$$ The inner function is $u=3x^2+7$ and $du=6x dx$ . Thus $dx=du/(6x)$ . Substitute: $$\int xe^{3x^2+7}dx = \int \frac{x e^u }{6x}du =\int \frac{e^u}{6} du = \frac{e^u}{6} +C = \frac{e^{3x^2+7}}{6} +C.$$
Example 0.3 $$\int \sin (3x+7) dx$$ The inner function is $u=3x+7$ and $du=3 dx$ . Therefore: $$\int \sin (3x+7) dx= \int \frac{\sin(u)}{3} du = -\frac{\cos(u)}{3} +C = -\frac{\cos(3x+7)}{3}+C.$$
Example 0.4 $$\int (2x+3)\sqrt{x^2+3x+20}\ dx $$ Inner $u=x^2+3x+20$ and $du=(2x+3)dx$ . Thus: $$\int (2x+3)\sqrt{x^2+3x+20}\ dx = \int \sqrt{u} du= \int u^{1/2} du = \frac{2u^{3/2}}{3} + C = \frac{2(x^2+3x+20)^{3/2}}{3} +C.$$
Now another integral which is a little more difficult:
Example 0.5 $$\int \frac{\cos (\ln x)}{x} dx$$ The inner function here is $u=\ln x$ and $du= \frac{1}{x} dx$ . $$\int \frac{\cos(\ln x)}{x} dx = \int \cos (u) \cdot \frac{1}{x} dx= \int \cos (u) du = \sin (u) +C= \sin(\ln x) +C.$$
Example 0.6 $$\int \frac{3x^2+14x+1}{x^3+7x^2+x+115} dx $$ This function is also a typical example of integration with substitution. Whenever there is a fraction, and the numerator looks like the derivative of the denominator, we set $u$ to be the denominator: $$u=x^3+7x^2+x+115, \quad du = (3x^2 +14x +1) dx$$ Thus: $$\int \frac{3x^2+14x+1}{x^3+7x^2+x+115} dx = \int \frac{1}{u} du = \ln u + C= \ln (x^3+7x^2+x+115) + C.$$
Example 0.7
$$\int \frac{7}{1+3x} dx $$ As in the example above, we set $u=1+3x$ , $du=3 dx $ : $$\int \frac{7}{1+3x} dx = \int \frac{7}{u} \frac{du}{3} = \frac{7}{3} \int \frac{1}{u} du = \frac{7}{3} \ln u + C = \frac{7}{3} \ln (1+3x) + C.$$
Example 0.8 $$\int t^3(t^4-50)^{700} dt$$ Here the inner function is $u=t^4-50$ and $du=4t^3 dt$ . Thus $$\int t^3(t^4-50)^{700} dt = \int \frac{u^{700}}{4} du=\frac{1}{4} \frac{u^{701}}{701} +C= \frac{(t^4-50)^{701}}{4\cdot 701} +C.$$
Some other examples (solve them!): $$ \int e^x \sin (e^x) dx,\quad \int \frac{e^x}{e^x+1} dx, \quad \int \frac{1}{x\ln x } dx$$
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