What is the integral of 1/(cos^4(3x)) ?

The integral of 1/(cos^4(3x)) is 1/3 (tan(3x)+(tan^3(3x))/3)+C

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integral of 1/(cos^4(3x))

\int \frac{1}{cos ^{4} ( 3 x )} d x

\frac{1}{3} (tan (3 x) + \frac{tan ^{3} ( 3 x )}{3}) + C

Solution steps

\int \frac{1}{cos ^{4} ( 3 x )} d x

Use the basic trigonometric identity:

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

= \int sec^{4} (3 x) d x

Apply u-substitution

: \int sec^{4} (u) \frac{1}{3} d u

= \int sec^{4} (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sec^{4} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

: \int 1 + v^{2} d v

= \frac{1}{3} \cdot \int 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

\int v^{2} d v = \frac{v ^{3}}{3}

= \frac{1}{3} (v + \frac{v ^{3}}{3})

Substitute back

= \frac{1}{3} (tan (3 x) + \frac{tan ^{3} ( 3 x )}{3})

Add a constant to the solution

= \frac{1}{3} (tan (3 x) + \frac{tan ^{3} ( 3 x )}{3}) + C

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What is the integral of 1/(cos^4(3x)) ?
The integral of 1/(cos^4(3x)) is 1/3 (tan(3x)+(tan^3(3x))/3)+C