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Question: How do you integrate \(\int \dfrac{{\sqrt {{x^2} - 25} }}{x}dx\) using trigonometric substitution?...

How do you integrate ∫x2−25xdx\int \dfrac{{\sqrt {{x^2} - 25} }}{x}dx using trigonometric substitution?

Explanation

Solution

Hint : To solve this question, first we will assume any of the set of variables or constant be another variable to get the expression easier. And conclude until the non-operational state is not achieved. And finally substitute the assumed value. Here we substitute x= secu by looking at the nature of expression given.

Complete step by step solution:
Let us suppose that, I=∫x2−25xdxI = \int \dfrac{{\sqrt {{x^2} - 25} }}{x}dx , and
let x=5sec⁡ux = 5\sec u .
Differentiate x=5sec⁡ux = 5\sec u :
⇒dx=5sec⁡utan⁡udu\Rightarrow dx = 5\sec u\tan u du
Now,
∴I=∫25sec⁡2u−255sec⁡u.5sec⁡utan⁡udu =∫5tan⁡u.tan⁡udu =5∫tan⁡2udu =5∫(sec⁡2u−1)du =5(tan⁡u−u)   \therefore I = \int \dfrac{{\sqrt {25{{\sec }^2}u - 25} }}{{5\sec u}}.5\sec u\tan udu \\\ = \int 5\tan u.\tan udu \\\ = 5\int {\tan ^2}udu \\\ = 5\int ({\sec ^2}u - 1)du \\\ = 5(\tan u - u) \;
Here,
5tan⁡u=x2−255\tan u = \sqrt {{x^2} - 25} and
x=5sec⁡u ⇒u=arc sec⁡(x5)   x = 5\sec u \\\ \Rightarrow u = arc\,\sec (\dfrac{x}{5}) \;
Hence,
⇒I=x2−25−5arc sec⁡(x5)+C\Rightarrow I = \sqrt {{x^2} - 25} - 5arc\,\sec (\dfrac{x}{5}) + C .
So, the correct answer is “I=x2−25−5arc sec⁡(x5)+C I = \sqrt {{x^2} - 25} - 5arc\,\sec (\dfrac{x}{5}) + C ”.

Note : Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.