What is the integral of tan(mx)sec^2(mx) ?

The integral of tan(mx)sec^2(mx) is 1/(2m)tan^2(mx)+C

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integral of tan(mx)sec^2(mx)

\int tan (m x) sec^{2} (m x) d x

\frac{1}{2 m} tan^{2} (m x) + C

Solution steps

\int tan (m x) sec^{2} (m x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int v d v

= \frac{1}{m} \cdot \int v d v

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{m} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{m} \cdot \frac{tan ^{2} ( m x )}{2}

Simplify

\frac{1}{m} \cdot \frac{tan ^{2} ( m x )}{2} : \frac{1}{2 m} tan^{2} (m x)

= \frac{1}{2 m} tan^{2} (m x)

Add a constant to the solution

= \frac{1}{2 m} tan^{2} (m x) + C

x \to 3 lim (\frac{\frac{3}{x} - 1}{x - 3})

\int_{1}^{\infty} \frac{2}{x ^{3}} d x

2 y^{^{''}} + 4 y = 0

\frac{d}{d x} (\frac{1}{x ( 1 + ln ( x ) )})

What is the integral of tan(mx)sec^2(mx) ?
The integral of tan(mx)sec^2(mx) is 1/(2m)tan^2(mx)+C