# Proof of $\displaystyle \int{\dfrac{1}{\sqrt{x^2-a^2}}}\,dx$ formula

An integral rule for square root of the difference of squares in reciprocal form is required to find the integration of the irrational functions in the following form.

$\displaystyle \int{\dfrac{1}{\sqrt{x^2-a^2}}}\,dx$

Let’s learn how to prove the integral of one by square root of the subtraction of square of a constant $a$ from square of a variable $x$.

### Removing the irrationality from the function

It is not possible to find the integrals of the irrational functions directly in the calculus. Hence, the transformation technique is used for overcoming the issues caused by the irrationality property of the functions in integral calculus. In this case, the complexity of function can be removed by releasing the function from the square root.

It is possible if we take $x = a\sec{y}$

Now, differentiate the above equation with respect to $y$.

$\implies$ $\dfrac{d}{dy}{(x)}$ $\,=\,$ $\dfrac{d}{dy}{(a\sec{y})}$

$\implies$ $\dfrac{dx}{dy}$ $\,=\,$ $\dfrac{d}{dy}{(a \times \sec{y})}$

For calculating the derivative of the right hand side expression, separate the constant factor from the differentiation as per the constant multiple rule of the derivatives.

$\implies$ $\dfrac{dx}{dy}$ $\,=\,$ $a \times \dfrac{d}{dy}{(\sec{y})}$

Now, find the derivative of the secant function with respect to $y$ by using the derivative rule of secant.

$\implies$ $\dfrac{dx}{dy}$ $\,=\,$ $a \times \sec{y}\tan{y}$

$\implies$ $dx$ $\,=\,$ $a \times \sec{y}\tan{y} \times dy$

$\,\,\,\therefore\,\,\,\,\,\,$ $dx$ $\,=\,$ $a\sec{y}\tan{y} \times dy$

We have taken $x \,=\, a\sec{y}$ and we have derived that $dx \,=\, a\sec{y}\tan{y} \times dy$.

It is time to convert the integral of the irrational function in terms of $x$ into the integral of the function in terms of $y$ by using the above relations.

$\implies$ $\displaystyle \int{\dfrac{1}{\sqrt{x^2-a^2}}}\,dx$ $\,=\,$ $\displaystyle \int{\dfrac{1}{\sqrt{(a\sec{y})^2-a^2}}}\,(a\sec{y}\tan{y} \times dy)$

Let us try to simplify the trigonometric function before finding the indefinite integration.

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{\sqrt{(a\sec{y})^2-a^2}}}\,(a\sec{y}\tan{y} \times dy)$

$=\,\,\,$ $\displaystyle \int{\dfrac{1 \times a\sec{y}\tan{y}}{\sqrt{(a\sec{y})^2-a^2}}}\,dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec{y}\tan{y}}{\sqrt{(a\sec{y})^2-a^2}}}\,dy$

Use the power of a product rule to split the square of $a\sec{y}$ in the denominator.

$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec{y}\tan{y}}{\sqrt{a^2\sec^2{y}-a^2}}}\,dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec{y}\tan{y}}{\sqrt{a^2\sec^2{y}-a^2 \times 1}}}\,dy$

The expression under the square root has a common factor in each term. So, take the common factor out from the terms in the expression.

$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec{y}\tan{y}}{\sqrt{a^2(\sec^2{y}-1)}}}\,dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec{y}\tan{y}}{a\sqrt{\sec^2{y}-1}}}\,dy$

According to the tan squared identity, the subtraction of one from the square of secant of angle $y$ is equal to square of tan of angle $y$.

$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec{y}\tan{y}}{a\sqrt{\tan^2{y}}}}\,dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{a\sec{y}\tan{y}}{a\tan{y}}}\,dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{a \times \sec{y} \times \tan{y}}{a\tan{y}}}\,dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{a \times \tan{y} \times \sec{y}}{a\tan{y}}}\,dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{a\tan{y} \times \sec{y}}{a\tan{y}}}\,dy$

Cancel a factor $a\tan{y}$ in the numerator by the same factor of the denominator.

$=\,\,\,$ $\displaystyle \int{\dfrac{\cancel{a\tan{y}} \times \sec{y}}{\cancel{a\tan{y}}}}\,dy$

The simplification process of the trigonometric function is completed and we can move forward in finding the integration of the function.

$=\,\,\,$ $\displaystyle \int{\sec{y}}\,dy$

### Find the integration of trigonometric function

It is time to find the indefinite integration of the secant function.

$=\,\,\,$ $\displaystyle \int{(\sec{y} \times 1)}\,dy$

For calculating the integral of secant function, multiply the secant function with $\sec{y}+\tan{y}$ and divide it by the same function for maintaining the mathematical balance.

$=\,\,\,$ $\displaystyle \int{\bigg(\sec{y} \times \dfrac{\sec{y}+\tan{y}}{\sec{y}+\tan{y}}\bigg)}\,dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{\sec{y} \times (\sec{y}+\tan{y})}{\sec{y}+\tan{y}}}\,dy$

Now, multiply the factor secant function with each term of the expression in the numerator as per the distributive property of multiplication across the addition.

$=\,\,\,$ $\displaystyle \int{\dfrac{\sec{y} \times \sec{y}+\sec{y} \times \tan{y}}{\sec{y}+\tan{y}}}\,dy$

$=\,\,\,$ $\displaystyle \int{\dfrac{\sec^2{y}+\sec{y}\tan{y}}{\sec{y}+\tan{y}}}\,dy$

In this case, the expression in the numerator can be obtained by differentiating the expression in the denominator.

$=\,\,\,$ $\displaystyle \int{\dfrac{(\sec^2{y}+\sec{y}\tan{y}) \times dy}{\sec{y}+\tan{y}}}$

So, assume $z \,=\, \sec{y}+\tan{y}$

Differentiate the above equation with respect to $y$.

$\implies$ $\dfrac{d}{dy}{(z)}$ $\,=\,$ $\dfrac{d}{dy}{(\sec{y}+\tan{y})}$

According to the sum rule of derivatives, the derivative of sum of functions can be evaluated by the sum of their derivatives.

$\implies$ $\dfrac{dz}{dy}$ $\,=\,$ $\dfrac{d}{dy}{(\sec{y})}$ $+$ $\dfrac{d}{dy}{(\tan{y})}$

Find the derivative of secant function by the derivative rule of secant function and also find the derivative of tan function by the derivative rule of tan function.

$\implies$ $\dfrac{dz}{dy}$ $\,=\,$ $\sec{y}\tan{y}$ $+$ $\sec^2{y}$

$\implies$ $\dfrac{dz}{dy}$ $\,=\,$ $\sec^2{y}$ $+$ $\sec{y}\tan{y}$

$\,\,\,\therefore\,\,\,\,\,\,$ $dz$ $\,=\,$ $(\sec^2{y}$ $+$ $\sec{y}\tan{y})$ $\times$ $dy$

We have taken $z \,=\, \sec{y}+\tan{y}$ and derived that $dz \,=\, (\sec^2{y}+\sec{y}\tan{y})\times dy$. Use these relations to convert the function in terms of $y$ into the function in terms of $z$.

$\implies$ $\displaystyle \int{\dfrac{(\sec^2{y}+\sec{y}\tan{y}) \times dy}{\sec{y}+\tan{y}}}$ $\,=\,$ $\displaystyle \int{\dfrac{dz}{z}}$

$=\,\,\,$ $\displaystyle \int{\dfrac{1 \times dz}{z}}$

$=\,\,\,$ $\displaystyle \int{\dfrac{1}{z}} \times dz$

The integral of the reciprocal of variable $z$ is evaluated as per the reciprocal integration rule of a variable.

$=\,\,\,$ $\log_{e}{|z|}+c_1$

### Express the solution in its actual form

The integral of given function is obtained in terms of $z$ but the function is given in terms of $x$. so, lets convert the solution in terms of $x$ from $z$.

We have taken $z\, =\, \sec{y}+\tan{y}$. So, replace the $z$ by its value in the solution.

$\implies$ $\log_{e}{|z|}+c_1$ $\,=\,$ $\log_{e}{|\sec{y}+\tan{y}|}+c_1$

Now, the solution is in terms of $y$. so, substitute the value of $y$ in terms of $x$.

We have considered that $x \,=\, a\sec{y}$

$\implies$ $a\sec{y} \,=\, x$

$\,\,\,\therefore\,\,\,\,\,\,$ $\sec{y} \,=\, \dfrac{x}{a}$

Now, we have to find the tan of angle $y$ from the secant of angle $y$ and it can be done as per the tan squared identity in terms of secant.

$\tan{y} \,=\, \sqrt{\sec^2{y}-1}$

$\implies$ $\tan{y} \,=\, \sqrt{\bigg(\dfrac{x}{a}\bigg)^2-1}$

$\implies$ $\tan{y} \,=\, \sqrt{\dfrac{x^2}{a^2}-1}$

$\implies$ $\tan{y} \,=\, \sqrt{\dfrac{x^2-a^2}{a^2}}$

$\implies$ $\tan{y} \,=\, \dfrac{\sqrt{x^2-a^2}}{\sqrt{a^2}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\tan{y} \,=\, \dfrac{\sqrt{x^2-a^2}}{a}$

Now, substitute the values of secant of angle $y$ and tan of angle $y$.

$\implies$ $\log_{e}{|\sec{y}+\tan{y}|}+c_1$ $\,=\,$ $\log_{e}{\Bigg|\dfrac{x}{a}+\dfrac{\sqrt{x^2-a^2}}{a}\Bigg|}+c_1$

$=\,\,\,$ $\log_{e}{\Bigg|\dfrac{x+\sqrt{x^2-a^2}}{a}\Bigg|}+c_1$

The natural logarithmic function can be split as the difference of two log functions by the quotient rule of the logarithms.

$=\,\,\,$ $\log_{e}{\Big|x+\sqrt{x^2-a^2}\Big|}$ $-$ $\log_{e}{|a|}$ $+$ $c_1$

In this expression, the second and third terms are constants. Hence, the difference of them can be denoted by a constant $c$ simply.

$=\,\,\,$ $\log_{e}{\Big|x+\sqrt{x^2-a^2}\Big|}$ $+$ $c$

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