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Question: How do you integrate \[\int {\dfrac{{2 - {x^2}}}{{\sqrt {{x^2} - 4} }}} dx\] using trigonometric sub...

How do you integrate ∫2−x2x2−4dx\int {\dfrac{{2 - {x^2}}}{{\sqrt {{x^2} - 4} }}} dx using trigonometric substitution?

Explanation

Solution

Here we are asked to integrate the given function using the trigonometric substitution. When we have complicated terms in the function during the integration, we used to make it simply by substituting a temporary variable to those terms but here instead of a temporary variable we have to use a suitable trigonometric function. Then by integrating that function will give us the result.
Formula: Some of the formulas that we need to know before solving the problem:
d(cosh⁡(x))=sinh⁡(x)d\left( {\cosh (x)} \right) = \sinh (x)
∫cf(x)dx=c∫f(x)dx\int {cf(x)dx} = c\int {f(x)dx} where ccis constant.
∫cosh⁡(x)dv=sinh⁡(x)\int {\cosh (x)dv} = \sinh (x)
sinh⁡2(x)=cosh⁡(2x)2−12{\sinh ^2}(x) = \dfrac{{\cosh (2x)}}{2} - \dfrac{1}{2}
sinh⁡(2cosh⁡−1(x))=2xx−1x+1\sinh (2{\cosh ^{ - 1}}(x)) = 2x\sqrt {x - 1} \sqrt {x + 1}

Complete step-by-step solution:
It is given that ∫2−x2x2−4dx\int {\dfrac{{2 - {x^2}}}{{\sqrt {{x^2} - 4} }}} dx. Let us rewrite the numerator for our convenience as 2−x2=−(x2−4)−22 - {x^2} = - ({x^2} - 4) - 2
Let us substitute it in the given function.
\int {\dfrac{{2 - {x^2}}}{{\sqrt {{x^2} - 4} }}} dx$$$$ = \int {\dfrac{{ - ({x^2} - 4) - 2}}{{\sqrt {{x^2} - 4} }}} dx
=∫−x2−4−2x2−4dx= \int { - \sqrt {{x^2} - 4} } - \dfrac{2}{{\sqrt {{x^2} - 4} }}dx
Let us now split the integral.
∫2−x2x2−4dx\int {\dfrac{{2 - {x^2}}}{{\sqrt {{x^2} - 4} }}} dx=(−∫2x2−4dx−∫4sinh⁡2(u)du) \left(- {\int {\dfrac{2}{{\sqrt {{x^2} - 4} }}dx - } \int {4{{\sinh }^2}(u)du} } \right)
We are asked to solve this problem by using trigonometric identities. So, let us substitute x=2cosh⁡(u)x = 2\cosh (u).
Therefore, dx=2sinh⁡(u)dudx = 2\sinh (u)du(using the formula cosh⁡(u)′=sinh⁡(u)du\cosh (u)' = \sinh (u)du).
It also implies that u=cosh⁡−1(x2)u = {\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right).
Now let us substitute x=2cosh⁡(u)x = 2\cosh (u) in x2−4\sqrt {{x^2} - 4}
x2−4=4cosh⁡2(u)−4\sqrt {{x^2} - 4} = \sqrt {4{{\cosh }^2}(u) - 4}
=2cosh⁡2(u)−1= 2\sqrt {{{\cosh }^2}(u) - 1}
Using the formula, cosh⁡2(x)−1=sinh⁡2(x){\cosh ^2}(x) - 1 = {\sinh ^2}(x) we get
=2sinh⁡2(u)= 2\sqrt {{{\sinh }^2}(u)}
x2−4=2sinh⁡(u)\sqrt {{x^2} - 4} = 2\sinh (u)
Let us substitute it in the equation(1)(1).
\left(- {\int {\dfrac{2}{{\sqrt {{x^2} - 4} }}dx - } \int {\sqrt {{x^2} - 4} dx} } \right)$$$$ = \left(- {\int {\dfrac{2}{{\sqrt {{x^2} - 4} }}dx - } \int {4{{\sinh }^2}(u)du} } \right)
Using the formula, ∫cf(x)dx=c∫f(x)dx\int {cf(x)dx} = c\int {f(x)dx} we get
=(−∫2x2−4dx−4∫sinh⁡2(u)du)= \left(- {\int {\dfrac{2}{{\sqrt {{x^2} - 4} }}dx - } 4\int {{{\sinh }^2}(u)du} } \right)
Using the formula, sinh⁡2(x)=cosh⁡(2x)2−12{\sinh ^2}(x) = \dfrac{{\cosh (2x)}}{2} - \dfrac{1}{2} we get
=−∫2x2−4dx−4∫(cos⁡(2u)2−12)du= - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - 4\int {\left( {\dfrac{{\cos (2u)}}{2} - \dfrac{1}{2}} \right)} du
Again, using the formula ∫cf(x)dx=c∫f(x)dx\int {cf(x)dx} = c\int {f(x)dx} we get
=−∫2x2−4dx−4(∫(cosh⁡(2u)−1)du2)= - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - 4\left( {\dfrac{{\int {(\cosh (2u) - 1)du} }}{2}} \right)
=−∫2x2−4dx−2∫(cosh⁡(2u)−1)du= - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - 2\int {(\cosh (2u) - 1)du}
Let’s split the integral.
=−∫2x2−4dx−2(−∫1du+∫cosh⁡(2u)du)= - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - 2\left( { - \int {1du} + \int {\cosh (2u)du} } \right)
=−∫2x2−4dx−2∫cosh⁡(2u)du+2∫1du= - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - 2\int {\cosh (2u)du} + 2\int {1du}
=−∫2x2−4dx−2∫cosh⁡(2u)du+2u= - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - 2\int {\cosh (2u)du} + 2u
Let us substitutev=2uv = 2u, then we get dv = 2du$$$$ \Rightarrow du = \dfrac{{dv}}{2}
=2u−∫2x2−4dx−2∫cosh⁡(v)2dv= 2u - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}dx - 2} \int {\dfrac{{\cosh (v)}}{2}dv}
Applying the formula, ∫cf(x)dx=c∫f(x)dx\int {cf(x)dx} = c\int {f(x)dx} we get
=2u−∫2x2−4dx−∫cosh⁡(v)dv= 2u - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - \int {\cosh (v)dv}
Using the formula, ∫cosh⁡(x)dv=sinh⁡(x)\int {\cosh (x)dv} = \sinh (x) we get
=2u−∫2x2−4dx−sinh⁡(v)= 2u - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - \sinh (v)
Now let us re-substitute v=2uv = 2u we get
=2u−∫2x2−4dx−sinh⁡(2u)= 2u - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - \sinh (2u)
Let’s re-substitute u=cosh⁡−1(x2)u = {\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right) we get
=2cosh⁡−1(x2)−∫2x2−4dx−sinh⁡(2cosh⁡−1(x2))= 2{\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right) - \int {\dfrac{2}{{\sqrt {{x^2} - 4} }}} dx - \sinh \left( {2{{\cosh }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)
Applying the formula, ∫cf(x)dx=c∫f(x)dx\int {cf(x)dx} = c\int {f(x)dx} we get
=2cosh⁡−1(x2)−sinh⁡(2cosh⁡−1(x2))−2∫1x2−4dx............(2)= 2{\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right) - \sinh \left( {2{{\cosh }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right) - 2\int {\dfrac{1}{{\sqrt {{x^2} - 4} }}} dx............(2)
Here let us substitutex=2cosh⁡(u)x = 2\cosh (u). Therefore, dx=2sinh⁡(u)dudx = 2\sinh (u)du (using the formula cosh⁡(u)′=sinh⁡(u)du\cosh (u)' = \sinh (u)du).
It also implies thatu=cosh⁡−1(x2)u = {\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right). And also using the formula, cosh⁡2(x)−1=sinh⁡2(x){\cosh ^2}(x) - 1 = {\sinh ^2}(x) we get

1x2−4=14cosh⁡2(u)−4  =12cosh⁡2(u)−1  =12sinh⁡2(u)  =12sinh⁡(u)  \dfrac{1}{{\sqrt {{x^2} - 4} }} = \dfrac{1}{{\sqrt {4{{\cosh }^2}(u) - 4} }} \\\ {\text{ }} = \dfrac{1}{{2\sqrt {{{\cosh }^2}(u) - 1} }} \\\ {\text{ }} = \dfrac{1}{{2\sqrt {{{\sinh }^2}(u)} }} \\\ {\text{ }} = \dfrac{1}{{2\sinh (u)}} \\\

Substituting this in the equation (2)(2) we get
=2cosh⁡−1(x2)−sinh⁡(2cosh⁡−1(x2))−2∫12sinh⁡(u)(2sinh⁡(u))du= 2{\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right) - \sinh \left( {2{{\cosh }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right) - 2\int {\dfrac{1}{{2\sinh (u)}}(2\sinh (u))du}
=2cosh⁡−1(x2)−sinh⁡(2cosh⁡−1(x2))−2(cosh⁡−1(x2))= 2{\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right) - \sinh \left( {2{{\cosh }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right) - 2\left( {{{\cosh }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)
On integrating the above expression, we get
=2cosh⁡−1(x2)−sinh⁡(2cosh⁡−1(x2))−2u= 2{\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right) - \sinh \left( {2{{\cosh }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right) - 2u
Now let us re-substitute u=cosh⁡−1(x2)u = {\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right) we get
=2cosh⁡−1(x2)−sinh⁡(2cosh⁡−1(x2))−2(cosh⁡−1(x2))= 2{\cosh ^{ - 1}}\left( {\dfrac{x}{2}} \right) - \sinh \left( {2{{\cosh }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right) - 2\left( {{{\cosh }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)
Therefore, \int {\dfrac{{2 - {x^2}}}{{\sqrt {{x^2} - 4} }}} dx$$$$ = - \sinh \left( {2{{\cosh }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)
Now using the formula, sinh⁡(2cosh⁡−1(x))=2xx−1x+1\sinh (2{\cosh ^{ - 1}}(x)) = 2x\sqrt {x - 1} \sqrt {x + 1} we get
\int {\dfrac{{2 - {x^2}}}{{\sqrt {{x^2} - 4} }}} dx$$$$ = - x\sqrt {\dfrac{x}{2} - 1} \sqrt {\dfrac{x}{2} + 1}
Let us simplify it further.
\int {\dfrac{{2 - {x^2}}}{{\sqrt {{x^2} - 4} }}} dx$$$$ = - \dfrac{{x\sqrt {x - 2} \sqrt {x + 2} }}{2}
Thus, the integration of the given function is \int {\dfrac{{2 - {x^2}}}{{\sqrt {{x^2} - 4} }}} dx$$$$ = - \dfrac{{x\sqrt {x - 2} \sqrt {x + 2} }}{2} + C.

Note: In both differentiation and integration, whenever we use a temporary variable while doing the substitution method, we have to re-substitute it again to get the answer in the original variable. Here we have used temporary variablesu&vu\& v . After the calculation, we have re-substituted their values back to the original variable.