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

The integral of 1/(cos^2(b)x) is sec^2(b)ln|x|+C

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integral of 1/(cos^2(b)x)

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

sec^{2} (b) ln ∣ x ∣ + C

Solution steps

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

Take the constant out:

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

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

Use the common integral:

\int \frac{1}{x} d x = ln (∣ x ∣)

= \frac{1}{cos ^{2} ( b )} ln ∣ x ∣

Use the basic trigonometric identity:

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

= sec^{2} (b) ln ∣ x ∣

Add a constant to the solution

= sec^{2} (b) ln ∣ x ∣ + C

x \to 2 lim (- 5)

\int - cos (2 t) d t

\int \frac{1}{( 1 - x ) ( 2 - x )} d x

\frac{1}{cos ^{2} ( x )}

\frac{d x}{d t} = x

What is the integral of 1/(cos^2(b)x) ?
The integral of 1/(cos^2(b)x) is sec^2(b)ln|x|+C