Blog do Professor Hulk: Método de integração

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← Integration techniques/Partial Fraction Decomposition	Calculus	Integration techniques/Tangent Half Angle →
	Integration techniques/Trigonometric Substitution

If the integrand contains a single factor of one of the forms $\sqrt{a^2-x^2} \mbox{ or } \sqrt{a^2+x^2} \mbox{ or } \sqrt{x^2-a^2}$ we can try a trigonometric substitution.

If the integrand contains $\sqrt{a^2-x^2}$ let x = asinθ and use the identity 1 − sin²θ = cos²θ.

If the integrand contains $\sqrt{a^2+x^2}$ let x = atanθ and use the identity 1 + tan²θ = sec²θ.

If the integrand contains $\sqrt{x^2-a^2}$ let x = asecθ and use the identity sec²θ − 1 = tan²θ.

Trigonometric Substitutions

[hide]

1 Sine substitution

2 Tangent substitution

3 Secant substitution
- 3.1 Example 1
- 3.2 Example 2

[edit] Sine substitution

This substitution is easily derived from a triangle, using the Pythagorean Theorem.

If the integrand contains a piece of the form $\sqrt{a^2-x^2}$ we use the substitution

$x=a\sin \theta \quad dx=a \cos \theta d\theta$

This will transform the integrand to a trigonometic function. If the new integrand can't be integrated on sight then the tan-half-angle substitution described below will generally transform it into a more tractable algebraic integrand.

Eg, if the integrand is √(1-x²),

$\begin{matrix} \int_0^1 \sqrt{1-x^2} dx & = & \int_0^{\pi/2} \sqrt{1-\sin^2 \theta} \cos \theta \, d\theta \\ & = & \int_0^{\pi/2} \cos^2 \theta \, d\theta \\ & = & \frac{1}{2} \int_0^{\pi/2} 1+ \cos 2\theta \, d\theta \\ & = & \frac{\pi}{4} \end{matrix}$

If the integrand is √(1+x)/√(1-x), we can rewrite it as

$\sqrt{\frac{1+x}{1-x}} = \sqrt{\frac{1+x}{1+x}\frac{1+x}{1-x}} =\frac{1+x}{\sqrt{1-x^2}}$

Then we can make the substitution

$\begin{matrix} \int_0^a \frac{1+x}{\sqrt{1-x^2}} dx & = & \int_0^\alpha \frac{1+\sin \theta}{\cos \theta} \cos \theta \, d\theta & 0 <a < 1 \\ & = & \int_0^\alpha 1+ \sin \theta \, d\theta & \alpha = \sin^{-1} a \\ & = & \alpha + \left[ - \cos \theta \right]_0^\alpha & \\ & = & \alpha + 1 - \cos \alpha & \\ & = & 1+ \sin^{-1} a - \sqrt{1-a^2} & \\ \end{matrix}$

[edit] Tangent substitution

This substitution is easily derived from a triangle, using the Pythagorean Theorem.

When the integrand contains a piece of the form $\sqrt{a^2+x^2}$ we use the substitution

$x = a \tan \theta \quad \sqrt{x^2+a^2} = a \sec \theta \quad dx = a \sec^2 \theta d\theta$

E.g, if the integrand is (x²+a²)^-3/2 then on making this substitution we find

$\begin{matrix} \int_0^z \left( x^2+a^2 \right)^{-\frac{3}{2}}dx & = & a^{-2} \int_0^\alpha \cos \theta \, d\theta & z>0 \\ & = & a^{-2} \left[ \sin \theta \right]_0^\alpha & \alpha = \tan^{-1} (z/a) \\ & = & a^{-2} \sin \alpha & \\ & = & a^{-2} \frac{z/a}{\sqrt{1+z^2/a^2}} & = \frac{1}{a^2} \frac{z}{\sqrt{a^2+z^2}} \\ \end{matrix}$

If the integral is

$I= \int_0^z \sqrt{x^2+a^2} \quad z>0$

then on making this substitution we find

$\begin{matrix} I & = & a^2 \int_0^\alpha \sec^3 \theta \, d\theta & & & \alpha = \tan^{-1} (z/a) \\ & = & a^2 \int_0^\alpha \sec \theta \, d\tan \theta & & & \\ & = & a^2 [ \sec \theta \tan \theta ]_0^\alpha & - & a^2 \int_0^\alpha \sec \theta \tan^2 \theta \, d\theta & \\ & = & a^2 \sec \alpha \tan \alpha & - & a^2 \int_0^\alpha \sec^3 \theta \, d\theta & + a^2 \int_0^\alpha \sec \theta \, d\theta \\ & = & a^2 \sec \alpha \tan \alpha & - & I & + a^2 \int_0^\alpha \sec \theta \, d\theta \\ \end{matrix}$

After integrating by parts, and using trigonometric identities, we've ended up with an expression involving the original integral. In cases like this we must now rearrange the equation so that the original integral is on one side only

$\begin{matrix} I & = & \frac{1}{2}a^2 \sec \alpha \tan \alpha & + & \frac{1}{2}a^2 \int_0^\alpha \sec \theta \, d\theta \\ & = & \frac{1}{2}a^2 \sec \alpha \tan \alpha & + & \frac{1}{2}a^2 \left[ \ln \left( \sec \theta + \tan \theta \right) \right]_0^\alpha \\ & = & \frac{1}{2}a^2 \sec \alpha \tan \alpha & + & \frac{1}{2}a^2 \ln \left( \sec \alpha + \tan \alpha \right) \\ & = & \frac{1}{2}a^2 \left( \sqrt{1+\frac{z^2}{a^2}} \right) \frac{z}{a} & + & \frac{1}{2}a^2 \ln \left( \sqrt{1+\frac{z^2}{a^2}}+\frac{z}{a} \right) \\ & = & \frac{1}{2}z\sqrt{z^2+a^2} & + & \frac{1}{2}a^2 \ln \left(\frac{z}{a} + \sqrt{1+\frac{z^2}{a^2}} \right) \\ \end{matrix}$

As we would expect from the integrand, this is approximately z²/2 for large z.

[edit] Secant substitution

This substitution is easily derived from a triangle, using the Pythagorean Theorem.

If the integrand contains a factor of the form $\sqrt{x^2-a^2}$ we use the substitution

$x = a \sec \theta \quad dx = a \sec \theta \tan \theta d\theta \quad \sqrt{x^2-a^2} = a \tan \theta.$

[edit] Example 1

Find $\int_1^z \frac{\sqrt{x^2-1}}{x}dx.$

$\begin{matrix} \int_1^z \frac{\sqrt{x^2-1}}{x}dx & = & \int_1^\alpha \frac{\tan \theta }{\sec \theta }\sec \theta \tan \theta \,d\theta & z>1 \\ & = & \int_0^\alpha \tan^2 \theta \, d\theta & \alpha = \sec^{-1} z \\ & = & \left[ \tan \theta -\theta \right]_0^\alpha & \tan \alpha = \sqrt{\sec^2 \alpha -1} \\ & =& \tan \alpha -\alpha & \tan \alpha = \sqrt{z^2-1} \\ & =& \sqrt{z^2-1} - \sec^{-1} z & \\ \end{matrix}$

[edit] Example 2

Find $\int_1^z \frac{\sqrt{x^2-1}}{x^2} dx.$

$\begin{matrix} \int_1^z \frac{\sqrt{x^2-1}}{x^2}dx & = & \int_1^\alpha \frac{\tan \theta}{\sec^2 \theta}\sec \theta \tan \theta \, d\theta & z>1 \\ & = & \int_0^\alpha \frac{\sin^2 \theta}{\cos \theta} d\theta & \alpha = \sec^{-1} z \\ \end{matrix}$

We can now integrate by parts

$\begin{matrix} \int_1^z \frac{\sqrt{x^2-1}}{x^2}dx & = & -\left[ \tan \theta \cos \theta \right]_0^\alpha + \int_0^\alpha \sec \theta \, d\theta \\ & = & -\sin \alpha +\left[ \ln (\sec \theta + \tan \theta ) \right]_0^\alpha \\ & = & \ln (\sec \alpha + \tan \alpha ) - \sin \alpha \\ & = & \ln (z+ \sqrt{z^2-1} ) - \frac{\sqrt{z^2-1}}{z}\\ \end{matrix}$