Integral Calculus-1
Integral Calculus-1
Allen Star Batch
Grade 12
Question:
A function $f(x)$ continuous on $\mathbb{R}$ and periodic with $2\pi$ satisfies $f(x) + (\sin x) f(x + \pi) = \sin^2 x$ then,
$$f(x) = \frac{\sin^2 x(1 + \sin^2 x)}{(1 - \sin x)}$$
$$f(x) = \frac{\sin^2 x(1 - \sin x)}{(1 + \sin^2 x)}$$
$$\int f(x)\mu x = x + \cos x - \frac{1}{\sqrt{2}}\tan^{-1}(\sqrt{2}\tan x) + \frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{2} - \cos x}{\sqrt{2} + \cos x}\right) + c$$
None of these
Step-by-Step Solution
Key Concept: Use the periodicity condition f(x+2π)=f(x) and the given functional equation to create a system by substituting x→x+π, then solve simultaneously to eliminate f(x+π) and find f(x)=sin²x(1-sinx)/(1+sin²x). The integral requires decomposing this rational function using partial fractions with appropriate trigonometric substitutions.
Given $f(x) + \sin x \cdot f(x+\pi) = \sin^2 x$, substitute $x \to x+\pi$ to get $f(x+\pi) - \sin x \cdot f(x) = \sin^2 x$. Solving these two equations simultaneously yields $f(x) = \frac{\sin^2 x(1-\sin x)}{1+\sin^2 x}$. The integral is then evaluated by splitting into partial fractions and standard forms, yielding $x - \frac{1}{2}\sqrt{2}\tan^{-1}(\sqrt{2}\tan x) - \frac{1}{2\sqrt{2}}\ln\left(\frac{\sqrt{2}+\cos x}{\sqrt{2}-\cos x}\right) + \cos x + C$.
Correct Answer: 2,3