# Almost direct integrals

An almost direct integral is an integral of the form:$$\displaystyle \int f(u(x)) \cdot u'(x) \ dx$$ 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":

$\displaystyle \int e^{3x} \ dx = \frac{1}{3} \int 3 \ cdot e^{3x} \ dx = \frac{1}{3} e^{3x}+C$, since $3$ is the derivative of $3x$.

$\displaystyle \int \cos 15x \ dx = \frac{1}{15} \int 15 \cos 15x \ dx = \frac{1}{15} \ sin 15 x +C$ , since $15$ is the derivative of $15x$.

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:

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

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

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

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

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

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

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

 Formulae $\displaystyle \int f^n (x) \cdot f'(x) \ dx = \frac {f^{n+1}(x)}{n+1}+C$, if $n \neq 1$.Particular cases: $\displaystyle \int \frac{f'(x)}{\sqrt{f(x)}} \ dx = 2\sqrt{f(x)}+C$ $\displaystyle \int a^{f(x)} f'(x) \ dx= \frac{1}{\ln a} a^{f(x)}+C$ $\displaystyle \int \frac{f'(x)}{f(x)} \ dx = \ln |x| +C$ Trigonometric functions $\displaystyle \int \sin (f(x)) \cdot f'(x) \ dx= -cos f(x) +C$ $\displaystyle \int \cos f(x) \cdot f'(x) \ dx= \sin f(x) +C$ $\displaystyle \int \frac{f'(x)}{\cos^2 f(x)} \ dx = \tan f(x) +C$ $\displaystyle \int \frac{f'(x)}{\sqrt{1-f(x)^2}} \ dx= \arcsin f(x)+C$ $\displaystyle \int \frac{f'(x)}{1+f(x)^2} \ dx = \arctan f(x) +C$ Hyperbolic functions $\displaystyle \int \frac{f'(x)}{\sqrt{(f'(x))^2+1}} \ dx= \sinh^{-1} f(x)+C$ $\displaystyle \int \frac{f'(x)}{\sqrt{(f'(x))^2-1}} \ dx= \cosh^{-1} f(x)+C$ $\displaystyle \int \frac{f'(x)}{1-f(x)^2} \ dx= \tanh^{-1} f(x)+C$