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Question: How do you integrate \( \int \dfrac{{{x^2}}}{{\sqrt {{x^2} - 4} }}dx \) using trigonometric substitu...

How do you integrate ∫x2x2−4dx\int \dfrac{{{x^2}}}{{\sqrt {{x^2} - 4} }}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.

Complete step-by-step answer :
Let, I=∫x2x2−4dxI = \int \dfrac{{{x^2}}}{{\sqrt {{x^2} - 4} }}dx .
Now, we can substitute:
x=2.sec⁡yx = 2.\sec y
Differentiate the above equation,
⇒dxdy=2sec⁡y.tan⁡y ⇒dx=2sec⁡y.tan⁡y.dy   \Rightarrow \dfrac{{dx}}{{dy}} = 2\sec y.\tan y \\\ \Rightarrow dx = 2\sec y.\tan y.dy \;
Put the value of xx in II :
∴I=∫4sec⁡2y4sec⁡2y−4.2sec⁡y.tan⁡ydy =∫4sec⁡2y2tan⁡y.2sec⁡y.tan⁡ydy =4∫sec⁡3ydy =4∫sec⁡y.sec⁡2ydy =4∫tan⁡2y+1.sec⁡2ydy   \therefore I = \int \dfrac{{4{{\sec }^2}y}}{{\sqrt {4{{\sec }^2}y - 4} }}.2\sec y.\tan ydy \\\ = \int \dfrac{{4{{\sec }^2}y}}{{2\tan y}}.2\sec y.\tan ydy \\\ = 4\int {\sec ^3}ydy \\\ = 4\int \sec y.{\sec ^2}ydy \\\ = 4\int \sqrt {{{\tan }^2}y + 1} .{\sec ^2}ydy \;
Now, we substitute: tan⁡y=t\tan y = t ;
Differentiate the above substitution:
⇒sec⁡2ydy=dt\Rightarrow {\sec ^2}ydy = dt
∴I=4∫t2+1dt =4t2t2+1+12ln⁡∣t+t2+1∣ =2tan⁡y.sec⁡y+ln⁡∣tan⁡y+sec⁡y∣   \therefore I = 4\int \sqrt {{t^2} + 1} dt \\\ = 4\\{ \dfrac{t}{2}\sqrt {{t^2} + 1} + \dfrac{1}{2}\ln |t + \sqrt {{t^2} + 1} |\\} \\\ = 2\\{ \tan y.\sec y + \ln |\tan y + \sec y|\\} \;
(∵t=tan⁡y)(\because t = \tan y)
Now, returning to:
sec⁡y=x2\sec y = \dfrac{x}{2} , so that: tan⁡y=x24−1\tan y = \sqrt {\dfrac{{{x^2}}}{4} - 1}
SO, we have now:
∴I=2x2.x2−42+ln⁡∣x2+x2−42∣ ⇒I=x2.x2−4+2ln⁡∣x2+x2−42∣+C   \therefore I = 2\\{ \dfrac{x}{2}.\dfrac{{\sqrt {{x^2} - 4} }}{2} + \ln |\dfrac{x}{2} + \dfrac{{\sqrt {{x^2} - 4} }}{2}|\\} \\\ \Rightarrow I = \dfrac{x}{2}.\sqrt {{x^2} - 4} + 2\ln |\dfrac{x}{2} + \dfrac{{\sqrt {{x^2} - 4} }}{2}| + C \;
,or,

I=x2x2−4+2ln⁡∣x+x2−4∣+cI = \dfrac{x}{2}\sqrt {{x^2} - 4} + 2\ln |x + \sqrt {{x^2} - 4} | + c , c=C−2ln⁡2c = C - 2\ln 2
However, the integral can easily be dealt with without using the substitution as follows:
I = \int \dfrac{{{x^2}}}{{\sqrt {{x^2} - 4} }}dx \\\ = \int \\{ \dfrac{{({x^2} - 4) + 4}}{{\sqrt {{x^2} - 4} }}dx\, \\\ = \int \\{ \dfrac{{({x^2} - 4)}}{{\sqrt {{x^2} - 4} }} + \dfrac{4}{{\sqrt {{x^2} - 4} }}\\} dx \\\ = \int \sqrt {{x^2} - 4} dx + 4\int \dfrac{1}{{\sqrt {{x^2} - 4} }}dx \\\ = \\{ \dfrac{x}{2}\sqrt {{x^2} - 4} - \dfrac{4}{2}\ln |x + \sqrt {{x^2} - 4} |\\} + 4\ln |x + \sqrt {{x^2} - 4} | \;
∴I=x2x2−4+2ln⁡∣x+x2−4∣+C1\therefore I = \dfrac{x}{2}\sqrt {{x^2} - 4} + 2\ln |x + \sqrt {{x^2} - 4} | + {C_1}
So, the correct answer is “ x2x2−4+2ln⁡∣x+x2−4∣+C1\dfrac{x}{2}\sqrt {{x^2} - 4} + 2\ln |x + \sqrt {{x^2} - 4} | + {C_1} ”.

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.