Definite Integration
Trigonometric Identities
Grade Class 12
Question:
7. Integral of $\sqrt{1+2\cot x(\cot x+\csc x)}$ w.r.t. $x$ is
(A) $2\ln \cos\frac{x}{2}+c$
(B) $2\ln \sin\frac{x}{2}+c$
(C) $\frac{1}{2}\ln \cos\frac{x}{2}+c$
(D) $\ln \sin x - \ln(\csc x - \cot x) + c$
Step-by-Step Solution
Key Concept: Simplify the expression inside the square root using trigonometric identities: 1 + 2cot^2 x + 2cot x cosec x = 1 + 2(cos^2 x / sin^2 x) + 2(cos x / sin^2 x) = (sin^2 x + 2cos^2 x + 2cos x) / sin^2 x. This simplifies to (1 + cos^2 x + 2cos x) / sin^2 x = (1 + cos x)^2 / sin^2 x. The integral becomes integral of (1 + cos x) / sin x dx = integral of (2 cos^2(x/2)) / (2 sin(x/2) cos(x/2)) dx = integral of cot(x/2) dx = 2 ln|sin(x/2)| + c.
The expression inside the square root is $1+2\cot x(\cot x+\csc x) = 1+2\cot^2 x+2\cot x\csc x = 1+2\frac{\cos^2 x}{\sin^2 x}+2\frac{\cos x}{\sin^2 x} = \frac{\sin^2 x+2\cos^2 x+2\cos x}{\sin^2 x} = \frac{1-\cos^2 x+2\cos^2 x+2\cos x}{\sin^2 x} = \frac{1+\cos^2 x+2\cos x}{\sin^2 x} = \frac{(1+\cos x)^2}{\sin^2 x}$. Thus, the integral is $\int \frac{1+\cos x}{\sin x} dx = \int \frac{2\cos^2(x/2)}{2\sin(x/2)\cos(x/2)} dx = \int \cot(x/2) dx = 2\ln|\sin(x/2)|+c$.
Correct Answer: B