How do you integrate $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} dx$ by trigonometric substitution?
Answer
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Hint: Here we are asked to find the value of the given integral by trigonometric substitution. Trigonometric substitution is nothing but substituting the variable by any expression which will help us to make the process of finding the integral easy. Also, the integrals are classified into two types, definite integral and indefinite integral: a definite integral contains upper and lower limits whereas an indefinite integral does not contain upper and lower limits; here, we are given an indefinite integral.
Formula to be used:
$\dfrac{d}{{dx}}\sec x = \sec x\tan x$
$\sqrt {{{\sec }^2}x - 1} = \tan x$
$\tan t = \dfrac{{\sin t}}{{\cos t}}$
$\sec t = \dfrac{1}{{\cos t}}$
$\dfrac{d}{{dt}}\left( {\sin x} \right) = \cos x$
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
$\sin t = \sqrt {1 - {{\cos }^2}t} $
Complete answer:
Let us consider $\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}$
We shall substitute $x = a\sec t$
Now, we need to differentiate $x = a\sec t$with respect to $x$
Thus, we have $dx = a\sec t\tan tdt$ (Here we have applied the formula $\dfrac{d}{{dx}}\sec x = \sec x\tan x$)
Now, we shall substitute $x = a\sec t$in $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $
Thus, we have $\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}} = \dfrac{{\sqrt {{{\left( {a\sec t} \right)}^2} - {a^2}} }}{{{{\left( {a\sec t} \right)}^4}}}$
$ = \dfrac{{\sqrt {{a^2}{{\sec }^2}t - {a^2}} }}{{{a^4}{{\sec }^4}t}}$
$ = \dfrac{{a\sqrt {{{\sec }^2}t - 1} }}{{{a^4}{{\sec }^4}t}}$
$ = \dfrac{{a\tan t}}{{{a^4}{{\sec }^4}t}}$ (Here we have applied $\sqrt {{{\sec }^2}x - 1} = \tan x$)
$ = \dfrac{{\tan t}}{{{a^3}{{\sec }^4}t}}$
$ = \dfrac{{\dfrac{{\sin t}}{{\cos t}}}}{{{a^3}{{\left( {\dfrac{1}{{\cos }}} \right)}^4}}}$ (Here we have applied $\tan t = \dfrac{{\sin t}}{{\cos t}}$ and $\sec t = \dfrac{1}{{\cos t}}$ )
$ = \dfrac{1}{{{a^3}}}\dfrac{{\sin t}}{{\cos t}}{\cos ^4}t$
$ = \dfrac{1}{{{a^3}}}\sin t{\cos ^3}t$
Now, we shall substitute the above equation in $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} dx$
$\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} dx$$ = \int {\dfrac{{\sin t{{\cos }^3}t}}{{{a^3}}}a\sec t \cdot \tan tdt} $
(Here we have applied $dx = a\sec t\tan tdt$)
$ = \dfrac{1}{{{a^2}}}\int {\sin t{{\cos }^3}t\sec t \cdot \tan tdt} $
$ = \dfrac{1}{{{a^2}}}\int {\sin t{{\cos }^3}t\dfrac{1}{{\cos t}}\dfrac{{\sin t}}{{\cos t}}dt} $
(Here we have applied $\tan t = \dfrac{{\sin t}}{{\cos t}}$ and $\sec t = \dfrac{1}{{\cos t}}$ )
$ = \dfrac{1}{{{a^2}}}\int {{{\sin }^2}t} \cos tdt$ …………….$\left( 1 \right)$
Now, we shall assume $y = \sin t$ and we need to differentiate it with respect to $y$
Thus, $dy = \cos tdt$ (We have applied $\dfrac{d}{{dt}}\left( {\sin x} \right) = \cos x$)
We need to apply $y = \sin t$and $dy = \cos tdt$ in the equation $\left( 1 \right)$
Thus, we have $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $$ = \dfrac{1}{{{a^2}}}\int {{y^2}} dy$ (We shall apply the formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $ to integrate )
$ = \dfrac{1}{{{a^2}}}\dfrac{{{y^{2 + 1}}}}{{2 + 1}}$
$ = \dfrac{1}{{{a^2}}}\dfrac{{{y^3}}}{3}$
Since we have assumed $y = \sin t$, we shall apply it in the above.
$\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $$ = \dfrac{1}{{{a^2}}}\dfrac{{{{\left( {\sin t} \right)}^3}}}{3}$ ……………..$\left( 2 \right)$
Earlier, we substituted $x = a\sec t$and $dx = a\sec t\tan tdt$.
For our convenience, we need to change as follows.
$x = a\sec t$
$\sec t = \dfrac{x}{a}$
$ \Rightarrow \dfrac{1}{{\cos t}} = \dfrac{x}{a}$ (Here we have applied $\sec t = \dfrac{1}{{\cos t}}$ )
$ \Rightarrow \cos t = \dfrac{a}{x}$
We know that $\sin t = \sqrt {1 - {{\cos }^2}t} $
Thus, $\sin t = \sqrt {1 - {{\left( {\dfrac{a}{x}} \right)}^2}} $
$ \Rightarrow \sin t = \sqrt {1 - \dfrac{{{a^2}}}{{{x^2}}}} $
$ \Rightarrow \sin t = \sqrt {\dfrac{{{x^2} - {a^2}}}{{{x^2}}}} $ $$ $$
$ \Rightarrow \sin t = \dfrac{{\sqrt {{x^2} - {a^2}} }}{x}$
Now, we shall substitute $\sin t = \dfrac{{\sqrt {{x^2} - {a^2}} }}{x}$ in the equation $\left( 2 \right)$
Thus, we have $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $$ = \dfrac{1}{{{a^2}}}\dfrac{{{{\left( {\dfrac{{\sqrt {{x^2} - {a^2}} }}{x}} \right)}^3}}}{3}$
$ = \dfrac{1}{{3{a^2}{x^3}}}{\left( {{x^2} - {a^2}} \right)^{\dfrac{3}{2}}} + C$ (Here $C$ is the constant of integration)
$ = \dfrac{{{{\left( {{x^2} - {a^2}} \right)}^{\dfrac{3}{2}}}}}{{3{a^2}{x^3}}} + C$
Therefore, $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $$ = \dfrac{{{{\left( {{x^2} - {a^2}} \right)}^{\dfrac{3}{2}}}}}{{3{a^2}{x^3}}} + C$
Note:
The steps to be followed in trigonometric substitution are as follows.
a) We shall make sure that we cannot use a simpler method to solve the integral and we need to identify that the given is a trigonometric substitution problem.
b) Then we need to decide which substitution to use. We can use sine substitution, tangent substitution, or secant substitution. In this process, we can get the values of x and dx.
c) Then we need to substitute the obtained values into the integral. We shall simplify the integral using any method.
d) At last, we need to back-substitute the integrated value back in terms of x.
Formula to be used:
$\dfrac{d}{{dx}}\sec x = \sec x\tan x$
$\sqrt {{{\sec }^2}x - 1} = \tan x$
$\tan t = \dfrac{{\sin t}}{{\cos t}}$
$\sec t = \dfrac{1}{{\cos t}}$
$\dfrac{d}{{dt}}\left( {\sin x} \right) = \cos x$
$\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
$\sin t = \sqrt {1 - {{\cos }^2}t} $
Complete answer:
Let us consider $\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}$
We shall substitute $x = a\sec t$
Now, we need to differentiate $x = a\sec t$with respect to $x$
Thus, we have $dx = a\sec t\tan tdt$ (Here we have applied the formula $\dfrac{d}{{dx}}\sec x = \sec x\tan x$)
Now, we shall substitute $x = a\sec t$in $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $
Thus, we have $\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}} = \dfrac{{\sqrt {{{\left( {a\sec t} \right)}^2} - {a^2}} }}{{{{\left( {a\sec t} \right)}^4}}}$
$ = \dfrac{{\sqrt {{a^2}{{\sec }^2}t - {a^2}} }}{{{a^4}{{\sec }^4}t}}$
$ = \dfrac{{a\sqrt {{{\sec }^2}t - 1} }}{{{a^4}{{\sec }^4}t}}$
$ = \dfrac{{a\tan t}}{{{a^4}{{\sec }^4}t}}$ (Here we have applied $\sqrt {{{\sec }^2}x - 1} = \tan x$)
$ = \dfrac{{\tan t}}{{{a^3}{{\sec }^4}t}}$
$ = \dfrac{{\dfrac{{\sin t}}{{\cos t}}}}{{{a^3}{{\left( {\dfrac{1}{{\cos }}} \right)}^4}}}$ (Here we have applied $\tan t = \dfrac{{\sin t}}{{\cos t}}$ and $\sec t = \dfrac{1}{{\cos t}}$ )
$ = \dfrac{1}{{{a^3}}}\dfrac{{\sin t}}{{\cos t}}{\cos ^4}t$
$ = \dfrac{1}{{{a^3}}}\sin t{\cos ^3}t$
Now, we shall substitute the above equation in $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} dx$
$\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} dx$$ = \int {\dfrac{{\sin t{{\cos }^3}t}}{{{a^3}}}a\sec t \cdot \tan tdt} $
(Here we have applied $dx = a\sec t\tan tdt$)
$ = \dfrac{1}{{{a^2}}}\int {\sin t{{\cos }^3}t\sec t \cdot \tan tdt} $
$ = \dfrac{1}{{{a^2}}}\int {\sin t{{\cos }^3}t\dfrac{1}{{\cos t}}\dfrac{{\sin t}}{{\cos t}}dt} $
(Here we have applied $\tan t = \dfrac{{\sin t}}{{\cos t}}$ and $\sec t = \dfrac{1}{{\cos t}}$ )
$ = \dfrac{1}{{{a^2}}}\int {{{\sin }^2}t} \cos tdt$ …………….$\left( 1 \right)$
Now, we shall assume $y = \sin t$ and we need to differentiate it with respect to $y$
Thus, $dy = \cos tdt$ (We have applied $\dfrac{d}{{dt}}\left( {\sin x} \right) = \cos x$)
We need to apply $y = \sin t$and $dy = \cos tdt$ in the equation $\left( 1 \right)$
Thus, we have $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $$ = \dfrac{1}{{{a^2}}}\int {{y^2}} dy$ (We shall apply the formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $ to integrate )
$ = \dfrac{1}{{{a^2}}}\dfrac{{{y^{2 + 1}}}}{{2 + 1}}$
$ = \dfrac{1}{{{a^2}}}\dfrac{{{y^3}}}{3}$
Since we have assumed $y = \sin t$, we shall apply it in the above.
$\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $$ = \dfrac{1}{{{a^2}}}\dfrac{{{{\left( {\sin t} \right)}^3}}}{3}$ ……………..$\left( 2 \right)$
Earlier, we substituted $x = a\sec t$and $dx = a\sec t\tan tdt$.
For our convenience, we need to change as follows.
$x = a\sec t$
$\sec t = \dfrac{x}{a}$
$ \Rightarrow \dfrac{1}{{\cos t}} = \dfrac{x}{a}$ (Here we have applied $\sec t = \dfrac{1}{{\cos t}}$ )
$ \Rightarrow \cos t = \dfrac{a}{x}$
We know that $\sin t = \sqrt {1 - {{\cos }^2}t} $
Thus, $\sin t = \sqrt {1 - {{\left( {\dfrac{a}{x}} \right)}^2}} $
$ \Rightarrow \sin t = \sqrt {1 - \dfrac{{{a^2}}}{{{x^2}}}} $
$ \Rightarrow \sin t = \sqrt {\dfrac{{{x^2} - {a^2}}}{{{x^2}}}} $ $$ $$
$ \Rightarrow \sin t = \dfrac{{\sqrt {{x^2} - {a^2}} }}{x}$
Now, we shall substitute $\sin t = \dfrac{{\sqrt {{x^2} - {a^2}} }}{x}$ in the equation $\left( 2 \right)$
Thus, we have $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $$ = \dfrac{1}{{{a^2}}}\dfrac{{{{\left( {\dfrac{{\sqrt {{x^2} - {a^2}} }}{x}} \right)}^3}}}{3}$
$ = \dfrac{1}{{3{a^2}{x^3}}}{\left( {{x^2} - {a^2}} \right)^{\dfrac{3}{2}}} + C$ (Here $C$ is the constant of integration)
$ = \dfrac{{{{\left( {{x^2} - {a^2}} \right)}^{\dfrac{3}{2}}}}}{{3{a^2}{x^3}}} + C$
Therefore, $\int {\dfrac{{\sqrt {{x^2} - {a^2}} }}{{{x^4}}}} $$ = \dfrac{{{{\left( {{x^2} - {a^2}} \right)}^{\dfrac{3}{2}}}}}{{3{a^2}{x^3}}} + C$
Note:
The steps to be followed in trigonometric substitution are as follows.
a) We shall make sure that we cannot use a simpler method to solve the integral and we need to identify that the given is a trigonometric substitution problem.
b) Then we need to decide which substitution to use. We can use sine substitution, tangent substitution, or secant substitution. In this process, we can get the values of x and dx.
c) Then we need to substitute the obtained values into the integral. We shall simplify the integral using any method.
d) At last, we need to back-substitute the integrated value back in terms of x.
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