What is the integral of sqrt(36x^2-144) ?

The integral of sqrt(36x^2-144) is 24(1/8 xsqrt(x^2-4)-1/2 ln((|x+sqrt(x^2-4)|)/2))+C

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integral of sqrt(36x^2-144)

\int 36 x^{2} - 144 d x

24 (\frac{1}{8} x x^{2} - 4 - \frac{1}{2} ln (\frac{x + x ^{2} - 4}{2})) + C

Solution steps

\int 36 x^{2} - 144 d x

Apply Trigonometric Substitution

: \int 24 tan^{2} (u) sec (u) d u

= \int 24 tan^{2} (u) sec (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 24 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 24 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

(- 1 + sec^{2} (u)) sec (u) : - sec (u) + sec^{3} (u)

= 24 \cdot \int - sec (u) + sec^{3} (u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 24 (- \int sec (u) d u + \int sec^{3} (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

\int sec^{3} (u) d u = \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

= 24 (- ln ∣ tan (u) + sec (u) ∣ + \frac{1}{2} sec (u) tan (u) + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arcsec (\frac{6}{12} x)

= 24 (- ln tan (arcsec (\frac{6}{12} x)) + sec (arcsec (\frac{6}{12} x)) + \frac{1}{2} sec (arcsec (\frac{6}{12} x)) tan (arcsec (\frac{6}{12} x)) + \frac{1}{2} ln tan (arcsec (\frac{6}{12} x)) + sec (arcsec (\frac{6}{12} x)))

Simplify

24 (- ln tan (arcsec (\frac{6}{12} x)) + sec (arcsec (\frac{6}{12} x)) + \frac{1}{2} sec (arcsec (\frac{6}{12} x)) tan (arcsec (\frac{6}{12} x)) + \frac{1}{2} ln tan (arcsec (\frac{6}{12} x)) + sec (arcsec (\frac{6}{12} x))) : 24 (\frac{1}{8} x x^{2} - 4 - \frac{1}{2} ln (\frac{x + x ^{2} - 4}{2}))

= 24 (\frac{1}{8} x x^{2} - 4 - \frac{1}{2} ln (\frac{x + x ^{2} - 4}{2}))

Add a constant to the solution

= 24 (\frac{1}{8} x x^{2} - 4 - \frac{1}{2} ln (\frac{x + x ^{2} - 4}{2})) + C

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What is the integral of sqrt(36x^2-144) ?
The integral of sqrt(36x^2-144) is 24(1/8 xsqrt(x^2-4)-1/2 ln((|x+sqrt(x^2-4)|)/2))+C