What is the integral of sec^2(-0.578t) ?

The integral of sec^2(-0.578t) is 1.73010…tan(0.578t)+C

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integral of sec^2(-0.578t)

\int sec^{2} (- 0.578 t) d t

1.73010 \dots tan (0.578 t) + C

Solution steps

\int sec^{2} (- 0.578 t) d t

Simplify

sec^{2} (- 0.578 t) : sec^{2} (0.578 t)

= \int sec^{2} (0.578 t) d t

Apply u-substitution

: \int sec^{2} (u) \frac{1}{0.578} d u

= \int sec^{2} (u) \frac{1}{0.578} d u

Take the constant out:

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

= \frac{1}{0.578} \cdot \int sec^{2} (u) d u

Simplify

= 1.73010 \dots \cdot \int sec^{2} (u) d u

Use the common integral:

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

= 1.73010 \dots tan (u)

Substitute back

u = 0.578 t

= 1.73010 \dots tan (0.578 t)

Add a constant to the solution

= 1.73010 \dots tan (0.578 t) + C

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What is the integral of sec^2(-0.578t) ?
The integral of sec^2(-0.578t) is 1.73010…tan(0.578t)+C