What is the integral of 1/2 sec^2(5x) ?

The integral of 1/2 sec^2(5x) is 1/10 tan(5x)+C

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integral of 1/2 sec^2(5x)

\int \frac{1}{2} sec^{2} (5 x) d x

\frac{1}{10} tan (5 x) + C

Solution steps

\int \frac{1}{2} sec^{2} (5 x) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int sec^{2} (5 x) d x

Apply u-substitution

: \int sec^{2} (u) \frac{1}{5} d u

= \frac{1}{2} \cdot \int sec^{2} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{5} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{5} tan (u)

Substitute back

u = 5 x

= \frac{1}{2} \cdot \frac{1}{5} tan (5 x)

Simplify

\frac{1}{2} \cdot \frac{1}{5} tan (5 x) : \frac{1}{10} tan (5 x)

= \frac{1}{10} tan (5 x)

Add a constant to the solution

= \frac{1}{10} tan (5 x) + C

x + \frac{2}{x}

36 = 9 x^{2} - 4 y^{2}

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What is the integral of 1/2 sec^2(5x) ?
The integral of 1/2 sec^2(5x) is 1/10 tan(5x)+C