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