What is the integral of sec(pi/4-2x) ?

The integral of sec(pi/4-2x) is 1/2 ln|tan(-pi/4+2x)+sec(-pi/4+2x)|+C

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integral of sec(pi/4-2x)

\int sec (\frac{π}{4} - 2 x) d x

\frac{1}{2} ln tan (- \frac{π}{4} + 2 x) + sec (- \frac{π}{4} + 2 x) + C

Solution steps

\int sec (\frac{π}{4} - 2 x) d x

Factor

\frac{π}{4} - 2 x : - (- \frac{π}{4} + 2 x)

= \int sec (- (- \frac{π}{4} + 2 x)) d x

Simplify

sec (- (- \frac{π}{4} + 2 x)) : sec (- \frac{π}{4} + 2 x)

= \int sec (- \frac{π}{4} + 2 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = - \frac{π}{4} + 2 x

= \frac{1}{2} ln tan (- \frac{π}{4} + 2 x) + sec (- \frac{π}{4} + 2 x)

Add a constant to the solution

= \frac{1}{2} ln tan (- \frac{π}{4} + 2 x) + sec (- \frac{π}{4} + 2 x) + C

x \to 3 + lim (x - 2)

f (x) = 5 x^{2}, at x = 10

\frac{d ^{3}}{d x ^{3}} (tan (x))

y^{^{''}} + y^{^{'}} = sin (t), y (0) = 1, y^{^{'}} (0) = 2

\int - 8 x d x

What is the integral of sec(pi/4-2x) ?
The integral of sec(pi/4-2x) is 1/2 ln|tan(-pi/4+2x)+sec(-pi/4+2x)|+C