What is the integral of sec^2(7-4x) ?

The integral of sec^2(7-4x) is -1/4 tan(7-4x)+C

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integral of sec^2(7-4x)

\int sec^{2} (7 - 4 x) d x

- \frac{1}{4} tan (7 - 4 x) + C

Solution steps

\int sec^{2} (7 - 4 x) d x

Apply u-substitution

: \int - \frac{1}{4} sec^{2} (u) d u

= \int - \frac{1}{4} sec^{2} (u) d u

Take the constant out:

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

= - \frac{1}{4} \cdot \int sec^{2} (u) d u

Use the common integral:

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

= - \frac{1}{4} tan (u)

Substitute back

u = 7 - 4 x

= - \frac{1}{4} tan (7 - 4 x)

Add a constant to the solution

= - \frac{1}{4} tan (7 - 4 x) + C

\int (9 x^{2} + 6 x) d x

\int_{0}^{3} \frac{1}{( x - 1 ) ^{2}} d x

y = 6 x - x^{2}, y = 2 x

x \to 0 lim (\frac{1}{x} ln (1 - 6 x))

What is the integral of sec^2(7-4x) ?
The integral of sec^2(7-4x) is -1/4 tan(7-4x)+C