Almost direct integrals

An almost direct integral is an integral of the form: $\int f (u (x)) \cdot u^{'} (x) d x$ where $f (x)$ is a function and $u (x)$ is another function, and $u^{'} (x)$ its derivative. We realize that by using the chain rule backwards we can obtain this kind of integral. (remember the unit on derivatives)

That is, if we have a function $F (x)$ , which derivative is $f (x)$ , and we change $x$ for another function $u (x)$ , the derivative of $F (u (x))$ is $f (u (x)) \cdot u^{'} (x)$ . Then, the integral of $f (u (x)) \cdot u^{'} (x)$ will be $F (u (x))$ .

One integral of this form can be solved like a direct integral, as we will see in the following examples:

In the first case, we see for example that we have a direct integral, except for a constant, then we realize the integral by multiplying and dividing it by this constant,so that we are able to use the "chain rule":

Example

$\int e^{3 x} d x = \frac{1}{3} \int 3 c d o t e^{3 x} d x = \frac{1}{3} e^{3 x} + C$ , since $3$ is the derivative of $3 x$ .

Example

$\int \cos 15 x d x = \frac{1}{15} \int 15 \cos 15 x d x = \frac{1}{15} s i n 15 x + C$ , since $15$ is the derivative of $15 x$ .

In other cases, the procedure does not turn out to be so simple, but the problem often ia as simple as finding the way to turn the integral into a direct integral. We will do this when solving an integral whenever possible:

Example

$\int \frac{1}{4 + x^{2}} d x = \int \frac{1}{4} \cdot \frac{1}{1 + \frac{x^{2}}{4}} = \int \frac{1}{4} \cdot \frac{1}{1 + (\frac{x}{2})^{2}} d x = \frac{1}{2} \int \frac{\frac{1}{2}}{1 + (\frac{x}{2})^{2}} d x =$

$= \frac{1}{2} \arctan \frac{x}{2} + C$ , where $\frac{1}{2}$ is the derivative of $\frac{x}{2}$ .

Example

$\int \frac{e^{x}}{1 + e^{2 x}} d x = \int \frac{e^{x}}{1 + (e^{x})^{2}} d x = \arctan e^{x} + C$ , where $e^{x}$ is the derivative of $e^{x}$ .

Example

$\int \frac{\sin \sqrt{x^{3}}}{\sqrt{x^{3}}} \cdot x^{2} d x = \frac{2}{3} \int \sin \sqrt{x^{3}} \cdot \frac{3 x^{2}}{2 \sqrt{x^{3}}} d x = - \frac{2}{3} \cos \sqrt{x^{3}} + C$ , since $\frac{3 x^{2}}{2 \sqrt{x^{3}}}$ is the derivative of $\sqrt{x^{3}}$ .

Example

$\int \frac{e^{x}}{\sqrt{1 - e^{2 x}}} d x = \arcsin e^{x} + C$ , where $e^{x}$ is the derivative of $e^{x}$ .

Example

$\int \sin x^{2} \cdot 2 x d x = - \cos x^{2} + C$ , where $2 x$ is the derivative of $x^{2}$ .

Example

$\int \sin^{2} x \cdot \cos x d x = \frac{\sin^{3} x}{3} + C$ since $\cos x$ is the sin derivative $x$ .

Formulae

\int f^{n} (x) \cdot f^{'} (x) d x = \frac{f^{n + 1} (x)}{n + 1} + C

, if

n \neq 1

.

Particular cases:

\int \frac{f^{'} (x)}{\sqrt{f (x)}} d x = 2 \sqrt{f (x)} + C

\int a^{f (x)} f^{'} (x) d x = \frac{1}{\ln a} a^{f (x)} + C

\int \frac{f^{'} (x)}{f (x)} d x = \ln | x | + C

Trigonometric functions

\int \sin (f (x)) \cdot f^{'} (x) d x = - c o s f (x) + C

\int \cos f (x) \cdot f^{'} (x) d x = \sin f (x) + C

\int \frac{f^{'} (x)}{\cos^{2} f (x)} d x = \tan f (x) + C

\int \frac{f^{'} (x)}{\sqrt{1 - f (x)^{2}}} d x = \arcsin f (x) + C

\int \frac{f^{'} (x)}{1 + f (x)^{2}} d x = \arctan f (x) + C

Hyperbolic functions

\int \frac{f^{'} (x)}{\sqrt{(f^{'} (x))^{2} + 1}} d x = \sinh^{- 1} f (x) + C

\int \frac{f^{'} (x)}{\sqrt{(f^{'} (x))^{2} - 1}} d x = \cosh^{- 1} f (x) + C

\int \frac{f^{'} (x)}{1 - f (x)^{2}} d x = \tanh^{- 1} f (x) + C