What is the integral of sec^2(1-2x) ?

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

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integral of sec^2(1-2x)

\int sec^{2} (1 - 2 x) d x

\frac{1}{2} tan (- 1 + 2 x) + C

Solution steps

\int sec^{2} (1 - 2 x) d x

Rewrite using trig identities

= \int \frac{1}{cos ^{2} ( 1 - 2 x )} d x

Factor

1 - 2 x : - (- 1 + 2 x)

= \int \frac{1}{cos ^{2} ( - ( - 1 + 2 x ) )} d x

Simplify

\frac{1}{cos ^{2} ( - ( - 1 + 2 x ) )} : sec^{2} (- 1 + 2 x)

= \int sec^{2} (- 1 + 2 x) d x

Apply u-substitution

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

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

Use the common integral:

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

= \frac{1}{2} tan (u)

Substitute back

u = - 1 + 2 x

= \frac{1}{2} tan (- 1 + 2 x)

Add a constant to the solution

= \frac{1}{2} tan (- 1 + 2 x) + C

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