Question

How do you integrate $\int {\dfrac{1}{{{{\left( {{x^2} + 4}\right)}^{\dfrac{3}{2}}}}}}$ by trigonometric substitution?

Answer 1

Here, we can replace $x$ by either $\tan$ or $\cot$ function as the equation has an addition in the conversion formula. With $\tan$ or $\cot$ function, we can integrate the function and get the solution in the form of any trigonometric identity. Then, we need to convert the solution back to an equation in the variable $x$ .

Complete step by step solution:
Here, we have to substitute the variable $x$ with a trigonometric function.
Out of the six trigonometric identities, here we can use either $\tan$ or $\cot$ function because all other identities have subtraction operations in their conversion formulas to convert to their conjugate function.
Here, we suppose that,
$x = 2\tan \theta$
Consider this as an Equation $(1)$
Differentiating on both sides of the equation
$\Rightarrow dx = 2{\sec ^2}\theta d\theta$
Consider this as an Equation $(2)$
Here, the integrating function is shown as
$I = \int {\dfrac{1}{{{{\left( {{x^2} + 4} \right)}^{\tfrac{3}{2}}}}}} dx$
Substituting the value of Equation $(1)$ and $(2)$ here,
$\Rightarrow I = \int {\dfrac{1}{{{{\left( {{{(2\tan \theta )}^2} + 4} \right)}^{\tfrac{3}{2}}}}}2{{\sec }^2}\theta d\theta }$
$\Rightarrow I = \int {\dfrac{1}{{{{\left( {4{{\tan }^2}\theta + 4} \right)}^{\tfrac{3}{2}}}}}2{{\sec }^2}\theta d\theta }$
Taking out the common factor $\;4$ from the denominator with the applied power
$\Rightarrow I = \int {\dfrac{1}{{{{(4)}^{\tfrac{3}{2}}}{{\left( {{{\tan }^2}\theta + 1} \right)}^{\tfrac{3}{2}}}}}2{{\sec }^2}\theta d\theta }$
We know that, for the conversion of $\tan$ into $\sec$ , we have
$1 + {\tan ^2}\theta = {\sec ^2}\theta$
Substituting the value and cubing the number $\;4$ ,
$\Rightarrow I = \int {\dfrac{1}{{{{(64)}^{\tfrac{1}{2}}}{{\left( {{{\sec }^2}\theta } \right)}^{\tfrac{3}{2}}}}}2{{\sec }^2}\theta d\theta }$
Here the square and square root operation gets neutralized by each other, hence the resulting power of $\sec \theta$ is $\;3$ and the square root of $64$ is $8$ .
Applying the results, we get
$\Rightarrow I = \int {\dfrac{1}{{8\left( {{{\sec }^3}\theta } \right)}}2{{\sec }^2}\theta d\theta }$
Now, $\;8$ is a constant value, and hence it can be taken out of the integration
$\Rightarrow I = \dfrac{2}{8}\int {\dfrac{{{{\sec }^2}\theta }}{{{{\sec }^3}\theta }}d\theta }$
$\Rightarrow I = \dfrac{1}{4}\int {\dfrac{1}{{\sec \theta }}d\theta }$
We know that the inverse of $\sec \theta$ is $\cos \theta$
$\Rightarrow I = \dfrac{1}{4}\int {\cos \theta d\theta }$
Now, the integration of $\cos \theta$ is $\sin \theta$
$\Rightarrow I = \dfrac{1}{4}\sin \theta + c$
Now, we need to convert back to the equation in the variable $x$ .
Here the assumed value is $x = 2\tan \theta$
$\Rightarrow \tan \theta = \dfrac{x}{2}$
We know the equation of $\tan \theta$ from the right-angled triangle is
$\Rightarrow \tan \theta = \dfrac{{{\text{opposite side}}}}{{{\text{adjacent side}}}}$
Comparing the above two equations, we get
Opposite side ($\;OPS$) = $x$ , Adjacent side ($\;ADJ$) = $\;2$
By Pythagoras theorem, where $\;HYP$ = Hypotenuse
${(HYP)^2} = {(OPS)^2} + {(ADJ)^2}$
Substituting the values,
$\Rightarrow {(HYP)^2} = {(x)^2} + {(2)^2}$
$\Rightarrow {(HYP)^2} = {x^2} + 4$
Applying root on both sides of the equation
$\Rightarrow (HYP) = \sqrt {{x^2} + 4}$
Now, the formula for $\sin \theta$ from the right-angled triangle is
$\sin \theta = \dfrac{{{\text{opposite side}}}}{{{\text{Hypotenuse}}}}$
Substituting the obtained values,
$\Rightarrow \sin \theta = \dfrac{x}{{\sqrt {{x^2} + 4} }}$
Substituting this value in the solution of integration,
$\Rightarrow I = \dfrac{1}{4} \times \dfrac{x}{{\sqrt {{x^2} + 4} }} + c$
$\Rightarrow I = \dfrac{x}{{4\sqrt {{x^2} + 4} }} + c$

$\dfrac{x}{{4\sqrt {{x^2} + 4} }} + c$ is the required solution for integration using trigonometric identities.

Note: Here, at the start of the calculations, $x$ was assumed as $\tan \theta$ . If one wants to assume the value as $\cot \theta$ there will be no change in the solution. A common mistake that might happen is not considering the factor $\;2$ while assuming the value of the variable $x$ , which is necessary to common out the factor $\;4$ otherwise, the conversion of identities will not be possible.