What is the integral of sqrt(1+a^2y^2) ?

The integral of sqrt(1+a^2y^2) is 1/2 ysqrt(1+a^2y^2)+1/(2a)ln|ay+sqrt(1+a^2y^2)|+C

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integral of sqrt(1+a^2y^2)

\int 1 + a^{2} y^{2} d y

\frac{1}{2} y 1 + a^{2} y^{2} + \frac{1}{2 a} ln a y + 1 + a^{2} y^{2} + C

Solution steps

\int 1 + a^{2} y^{2} d y

Apply Trigonometric Substitution

: \int \frac{sec ^{3} ( u )}{a} d u

= \int \frac{sec ^{3} ( u )}{a} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int sec^{3} (u) d u

Apply Integral Reduction

= \frac{1}{a} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

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

= \frac{1}{a} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Substitute back

u = arctan (a y)

= \frac{1}{a} (\frac{sec ^{2} ( arctan ( a y ) ) sin ( arctan ( a y ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (a y)) + sec (arctan (a y)) ∣)

Simplify

\frac{1}{a} (\frac{sec ^{2} ( arctan ( a y ) ) sin ( arctan ( a y ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (a y)) + sec (arctan (a y)) ∣) : \frac{1}{2} y 1 + a^{2} y^{2} + \frac{1}{2 a} ln a y + 1 + a^{2} y^{2}

= \frac{1}{2} y 1 + a^{2} y^{2} + \frac{1}{2 a} ln a y + 1 + a^{2} y^{2}

Add a constant to the solution

= \frac{1}{2} y 1 + a^{2} y^{2} + \frac{1}{2 a} ln a y + 1 + a^{2} y^{2} + C

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What is the integral of sqrt(1+a^2y^2) ?
The integral of sqrt(1+a^2y^2) is 1/2 ysqrt(1+a^2y^2)+1/(2a)ln|ay+sqrt(1+a^2y^2)|+C