What is the integral of 5sec(x+3)tan(x+3) ?

The integral of 5sec(x+3)tan(x+3) is 5sec(x+3)+C

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integral of 5sec(x+3)tan(x+3)

\int 5 sec (x + 3) tan (x + 3) d x

5 sec (x + 3) + C

Solution steps

\int 5 sec (x + 3) tan (x + 3) d x

Take the constant out:

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

= 5 \cdot \int sec (x + 3) tan (x + 3) d x

Apply u-substitution

: \int sec (u) tan (u) d u

= 5 \cdot \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= 5 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 5 \cdot 1 \cdot v

Substitute back

= 5 \cdot 1 \cdot sec (x + 3)

Simplify

= 5 sec (x + 3)

Add a constant to the solution

= 5 sec (x + 3) + C

x y^{^{'}} + 3 y = 4 x, y (3) = 6

\frac{\partial}{\partial x} (x^{3} - 6 x y + y^{3})

\frac{\partial}{\partial z} (z^{3} - x^{2} y)

y^{^{'}} = \frac{2 t}{y + t ^{2} y}

\int x ln (2 x - 3) d x

What is the integral of 5sec(x+3)tan(x+3) ?
The integral of 5sec(x+3)tan(x+3) is 5sec(x+3)+C