Integral Calculus
Integral via substitution; evaluating f(θ)-g(θ)
MJMT_Full_Test_09
Grade 12
Question:
Let $I=\displaystyle\int\frac{\sin^2\theta\cdot\log_e(e\cot\theta)}{\cos^4\theta((\log_e\tan\theta)^4-(\tan\theta)^4)}d\theta=\frac{1}{4}\log_e|f(\theta)|-\frac{1}{2}\tan^{-1}(g(\theta))+c$. The value of $f(\theta)-g(\theta)$ at $\theta=\pi/3$ is
0
$\dfrac{\ln3+2\sqrt3}{\ln3-2\sqrt3}-\dfrac{\ln3}{2\sqrt3}$
$\dfrac{\ln3-2\sqrt3}{\ln3+2\sqrt3}-\dfrac{\ln3}{2\sqrt3}$
$\dfrac{\ln3}{2\sqrt3}$
Step-by-Step Solution
Key Concept: Substitute $t=\tan\theta$: integral becomes $\int\frac{(1+\ln t)\,dt}{t(\ln^4 t-t^4)}$. Partial fractions with $\ln t=tu$: split into log and arctan parts.
$f(\theta)-g(\theta)=\dfrac{\ln3-2\sqrt3}{\ln3+2\sqrt3}-\dfrac{\ln3}{2\sqrt3}$.
Correct Answer: 3