What is the integral of sqrt(y^2-49) ?

The integral of sqrt(y^2-49) is 49(1/98 ysqrt(y^2-49)-1/2 ln((|y+sqrt(y^2-49)|)/7))+C

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integral of sqrt(y^2-49)

\int y^{2} - 49 d y

49 \frac{1}{98} y y^{2} - 49 - \frac{1}{2} ln \frac{y + y ^{2} - 49}{7} + C

Solution steps

\int y^{2} - 49 d y

Apply Trigonometric Substitution

= \int 49 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

= 49 \cdot \int tan^{2} (u) sec (u) d u

Rewrite using trig identities

= 49 \cdot \int (- 1 + sec^{2} (u)) sec (u) d u

Expand

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

= 49 \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

= 49 (- \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) ∣

= 49 (- 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{1}{7} y)

= 49 (- ln tan (arcsec (\frac{1}{7} y)) + sec (arcsec (\frac{1}{7} y)) + \frac{1}{2} sec (arcsec (\frac{1}{7} y)) tan (arcsec (\frac{1}{7} y)) + \frac{1}{2} ln tan (arcsec (\frac{1}{7} y)) + sec (arcsec (\frac{1}{7} y)))

Simplify

49 (- ln tan (arcsec (\frac{1}{7} y)) + sec (arcsec (\frac{1}{7} y)) + \frac{1}{2} sec (arcsec (\frac{1}{7} y)) tan (arcsec (\frac{1}{7} y)) + \frac{1}{2} ln tan (arcsec (\frac{1}{7} y)) + sec (arcsec (\frac{1}{7} y))) : 49 \frac{1}{98} y y^{2} - 49 - \frac{1}{2} ln \frac{y + y ^{2} - 49}{7}

= 49 \frac{1}{98} y y^{2} - 49 - \frac{1}{2} ln \frac{y + y ^{2} - 49}{7}

Add a constant to the solution

= 49 \frac{1}{98} y y^{2} - 49 - \frac{1}{2} ln \frac{y + y ^{2} - 49}{7} + C

What is the integral of sqrt(y^2-49) ?
The integral of sqrt(y^2-49) is 49(1/98 ysqrt(y^2-49)-1/2 ln((|y+sqrt(y^2-49)|)/7))+C