Integrate the function.
-$$\int _ { - 1 } ^ { 1 } \frac { 4 } { 1 + 16 t ^ { 2 } } d t$$
A) $2 \sin ^ { - 1 } 4$
B) $2 \tan ^ { - 1 } \left( \frac { 1 } { 4 } \right)$
C) $\frac { \pi } { 2 }$
D) $2 \tan ^ { - 1 } 4$

Question

Quizplus · Accepted Answer

This integral can be solved by recognizing it as a standard form integral. The integral $\int \frac{1}{1 + a^2x^2} dx = \frac{1}{a} \tan^{-1}(ax) + C$, where $a$ is a constant. In this case, $a = 4$, so the integral becomes $\frac{1}{4} \tan^{-1}(4t)$ evaluated from $-1$ to $1$. This evaluates to $\frac{1}{4} [\tan^{-1}(4) - \tan^{-1}(-4)] = \frac{1}{4} [2\tan^{-1}(4)] = \frac{1}{2} \cdot 2\tan^{-1}(4) = 2\tan^{-1}(4)$, which matches option D.

Integrate the Function ∫−1141+16t2dt\int _ { - 1 } ^ { 1 } \frac { 4 } { 1 + 16 t ^ { 2 } } d t∫−11​1+16t24​dt

Integrate the Function $\int _ { - 1 } ^ { 1 } \frac { 4 } { 1 + 16 t ^ { 2 } } d t$