What is the integral of sec^2(ax+b) ?

The integral of sec^2(ax+b) is 1/a tan(ax+b)+C

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integral of sec^2(ax+b)

\int sec^{2} (a x + b) d x

\frac{1}{a} tan (a x + b) + C

Solution steps

\int sec^{2} (a x + b) d x

Apply u-substitution

: \int \frac{sec ^{2} ( u )}{a} d u

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

Use the common integral:

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

= \frac{1}{a} tan (u)

Substitute back

u = a x + b

= \frac{1}{a} tan (a x + b)

Add a constant to the solution

= \frac{1}{a} tan (a x + b) + C

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What is the integral of sec^2(ax+b) ?
The integral of sec^2(ax+b) is 1/a tan(ax+b)+C