What is the integral of tan(sqrt(x-7)) 1/(sqrt(x-7)) ?

The integral of tan(sqrt(x-7)) 1/(sqrt(x-7)) is -2ln|cos(sqrt(x-7))|+C

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integral of tan(sqrt(x-7)) 1/(sqrt(x-7))

\int tan (x - 7) \frac{1}{x - 7} d x

- 2 ln cos (x - 7) + C

Solution steps

\int tan (x - 7) \frac{1}{x - 7} d x

Apply u-substitution

: \int \frac{tan ( u )}{u} d u

= \int \frac{tan ( u )}{u} d u

Apply u-substitution

: \int 2 tan (v) d v

= \int 2 tan (v) d v

Take the constant out:

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

= 2 \cdot \int tan (v) d v

Rewrite using trig identities

= 2 \cdot \int \frac{sin ( v )}{cos ( v )} d v

Apply u-substitution

: \int - \frac{1}{w} d w

= 2 \cdot \int - \frac{1}{w} d w

Take the constant out:

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

= 2 (- \int \frac{1}{w} d w)

Use the common integral:

\int \frac{1}{w} d w = ln (∣ w ∣)

= 2 (- ln ∣ w ∣)

Substitute back

= 2 (- ln cos (x - 7))

Simplify

= - 2 ln cos (x - 7)

Add a constant to the solution

= - 2 ln cos (x - 7) + C

\int e^{a x + 7} d x

\frac{\partial}{\partial y} (- x)

f (x) = x^{4} - 17 x^{2} + 16, at x = 1

sin^{2} (x), sin^{3} (x), [0, π]

y = 2 x^{3}, x = 0, x = 2

What is the integral of tan(sqrt(x-7)) 1/(sqrt(x-7)) ?
The integral of tan(sqrt(x-7)) 1/(sqrt(x-7)) is -2ln|cos(sqrt(x-7))|+C