Integrate
-$\int \sec ^ { 2 } ( 1 - 2 x ) d x$
A) -2 tan(1 - 2x) + C
B) 2 tan(1 - 2x) + C
C) tan(1 - 2x) + C
D) - $\frac { 1 } { 2 }$ tan(1 - 2x) + C
E) none of these

Question

Quizplus · Accepted Answer

The integral of $\sec^2(u) du$ is $tan(u) + C$. Here, $u = 1 - 2x$, so $du = -2 dx$. To adjust for the -2, we multiply the integral by $-\frac{1}{2}$, resulting in $-\frac{1}{2} tan(1 - 2x) + C$.

Integrate - ∫sec⁡2(1−2x)dx\int \sec ^ { 2 } ( 1 - 2 x ) d x∫sec2(1−2x)dx

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