What is the integral of 1/(cos^2(bx)) ?

The integral of 1/(cos^2(bx)) is 1/b tan(bx)+C

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integral of 1/(cos^2(bx))

\int \frac{1}{cos ^{2} ( b x )} d x

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

Solution steps

\int \frac{1}{cos ^{2} ( b x )} d x

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

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

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{b} \cdot \int sec^{2} (u) d u

Use the common integral:

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

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

Substitute back

u = b x

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

Add a constant to the solution

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

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What is the integral of 1/(cos^2(bx)) ?
The integral of 1/(cos^2(bx)) is 1/b tan(bx)+C