Integration by trigonometric substitution

analealsuarez (52)in #english • 4 years ago (edited)

Considering that the current situation affects the regular development of the school's experiential activity at all educational levels, moving it from the classroom to the home under the guidelines emanating from school institutions, the contribution of those of us who have teaching experience in different areas of knowledge is necessary; this through electronic means, since this is the form or tactical tool for it; all this in collaboration with society.

Particularly I am a professional of mathematics and in this sense in that area will be my contribution.

In my post The indefinite integral of the Absolute Value function I developed the integral of an absolute value function La integral indefinida de la función Valor Absoluto, following the same logic, I will continue contributing on this topic until I think it is necessary to change it.

Source

The topic to be developed today is:

Integral by trigonometric substitution

This type of integration is applied when the integrand has any of these three forms:

$\sqrt[]{a^{2}-x^{2}}$ in this case a new variable is introduced θ such that x=asenθ
$\sqrt[]{a^{2}+x^{2}}$ in this case x=atanθ
$\sqrt[]{x^{2}-a^{2}}$ this time the variale truck will be x=asecθ

Below is an application example:

Calculate the following integral:

$\int \frac{dx}{x^{2}\sqrt[]{x^{2}+2}}$

This integral can be rearranged like this:

$\int \frac{dx}{x^{2}\sqrt[]{2+ x^{2}}}$ = $\int \frac{dx}{x^{2}\sqrt[]{(\sqrt{2})^{2}+ x^{2}}}$

It is observed that the integrand has the second form, therefore:

x= √2tanθ→ dx= √2 sec²θdθ

Then:

$\sqrt{2+x^{2}}=\sqrt{\sqrt{2}^{2}+\sqrt{2}^{2}tan^{2}\Theta}=\sqrt{2+2tan^{2}\Theta}=\sqrt{2(1+tan^{2}\Theta)}=\sqrt{2}\sqrt{1+tan^{2}\Theta }=\sqrt{2}\sqrt{sec^{2}\Theta }=\sqrt{2}sec\Theta$

Where:

sustitución trigonometrica.jpg
Then:

$\int \frac{\sqrt{2}sec^{2}\Theta d\Theta}{2tan^{2}\Theta \sqrt{2}sec\Theta }=\frac{1}{2}\int \frac{sec\Theta d\Theta }{tan^{2}\Theta }$

$= \frac{1}{2}\int \frac{\frac{1}{cos\Theta}}{\frac{sen^{2}\Theta}{cos^{2}\Theta }}d\Theta =\frac{1}{2}\int \frac{cos\Theta }{sen^{2}\Theta }d\Theta$

$=\frac{1}{2}\int \frac{cos\Theta }{sen\Theta }\frac{1}{sen\Theta }d\Theta=\frac{1}{2} \int cot\Theta csc\Theta d\Theta$

$=\frac{1}{2}\(-csc\Theta )+C= \frac{1}{2}(\frac{-1}{sen\Theta })+C$ Where C is a constant of integration.

Finally, using the resource of the given triangles, the sine of the angle is calculated θ

This is:

senθ = $\frac{x}{\sqrt{x^{2}+2}}$

Substituting this value in the previous result, the final result of the integral is obtained:

$\int \frac{dx}{x^{2}\sqrt[]{x^{2}+2}}$ = $=\frac{1}{2}(\frac{-1}{\frac{x}{\sqrt{x^{2}+2}}})+C=-\frac{1}{2}\frac{\sqrt{x^{2}+2}}{x}+C=-\frac{\sqrt{x^{2}+2}}{2x}+C$