What is the integral of sec^2(7/3 x+4) ?

The integral of sec^2(7/3 x+4) is 3/7 tan(7/3 x+4)+C

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integral of sec^2(7/3 x+4)

\int sec^{2} (\frac{7}{3} x + 4) d x

\frac{3}{7} tan (\frac{7}{3} x + 4) + C

Solution steps

\int sec^{2} (\frac{7}{3} x + 4) d x

Apply u-substitution

: \int sec^{2} (u) \frac{3}{7} d u

= \int sec^{2} (u) \frac{3}{7} d u

Take the constant out:

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

= \frac{3}{7} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= \frac{3}{7} tan (u)

Substitute back

u = \frac{7}{3} x + 4

= \frac{3}{7} tan (\frac{7}{3} x + 4)

Add a constant to the solution

= \frac{3}{7} tan (\frac{7}{3} x + 4) + C

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What is the integral of sec^2(7/3 x+4) ?
The integral of sec^2(7/3 x+4) is 3/7 tan(7/3 x+4)+C