Integral Calculus-1
Integral Calculus-1
Allen Star Batch
Grade 12
Question:
If $\int \sqrt{\cos ecx + 1}\,dx = kfog(x) + c$, where $k$ is a real constant, then :
$k = -2, f(x) = \cot^{-1} x, g(x) = \sqrt{\cos ecx - 1}$
$k = -2, f(x) = \tan^{-1} x, g(x) = \sqrt{\cos ecx - 1}$
$k = 2, f(x) = \tan^{-1} x, g(x) = \frac{\cot x}{\sqrt{\cos ecx - 1}}$
$k = 2, f(x) = \cot^{-1} x, g(x) = \frac{\cot x}{\sqrt{\cos ecx + 1}}$
Step-by-Step Solution
Key Concept: Successive substitutions transform the trigonometric integral into a standard arctangent form.
Starting with $I = \int \sqrt{\cos ecx + 1} dx = \int \frac{\cot x}{\sqrt{\cos ecx - 1}} dx$. Substitute $\cos ecx = t$ to get $I = -\int \frac{dt}{t\sqrt{t-1}}$. Then substitute $t - 1 = u^2$ to obtain $I = -\int \frac{2u du}{u(u^2+1)} = -2\tan^{-1}u + C = -2\tan^{-1}\sqrt{\cos ecx - 1} + C$.
Correct Answer: 2,4