What is the integral of 1/(cos^2(7x)) ?

The integral of 1/(cos^2(7x)) is 1/7 tan(7x)+C

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integral of 1/(cos^2(7x))

\int \frac{1}{cos ^{2} ( 7 x )} d x

\frac{1}{7} tan (7 x) + C

Solution steps

\int \frac{1}{cos ^{2} ( 7 x )} d x

Rewrite using trig identities

= \int sec^{2} (7 x) d x

Apply u-substitution

: \int sec^{2} (u) \frac{1}{7} d u

= \int sec^{2} (u) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{7} tan (u)

Substitute back

u = 7 x

= \frac{1}{7} tan (7 x)

Add a constant to the solution

= \frac{1}{7} tan (7 x) + C

(6 x^{2} + y - 1) d x - (14 y - x) d y = 0, y (1) = 0

x^{2} \frac{d y}{d x} + 2 x y = 2 x

\frac{x}{6 x + 2}

\frac{x ^{2}}{x - 1}

\int \frac{1}{1 + 2 x ^{2} + x ^{4}} d x

What is the integral of 1/(cos^2(7x)) ?
The integral of 1/(cos^2(7x)) is 1/7 tan(7x)+C