Integral Calculus-1
Integral Calculus-1
Allen Star Batch
Grade 12
Question:
Let $f(x) = \frac{1}{4 - 3\cos^2 x + 5\sin^2 x}$ and its anti-derivative $F(x) = \frac{1}{3}\tan^{-1}(g(x)) + c$, then :
$g(x)$ is equal to $3\tan x$
$g\left(\frac{\pi}{4}\right)$ is equal to 3
$g'\left(\frac{\pi}{3}\right)$ is equal to 6
$g'\left(\frac{\pi}{3}\right)$ is equal to 12
Step-by-Step Solution
Key Concept: Dividing numerator and denominator by $\cos^2 x$ converts the integral to a standard arctangent form.
Rewrite $F(x) = \int \frac{1}{4 - 3\cos^2 x + 5\sin^2 x} dx = \int \frac{\sec^2 x}{9\sec^2 x - 8} dx = \int \frac{\sec^2 x}{1 + 9\tan^2 x} dx$. Substituting $u = 3\tan x$ yields $F(x) = \frac{1}{3}\tan^{-1}(3\tan x) + c$. Using $g\left(\frac{\pi}{4}\right) = 3$ determines the constant.
Correct Answer: 1,2,4