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Limit of f from ∫(x−sinxcosx)/(x²cos²x)dx | JEE

JEE Maths question with a full step-by-step solution.

Question
If ∫x−sin⁡xcos⁡xx2cos⁡2x dx=f(x)+c\displaystyle\int \dfrac{x-\sin x\cos x}{x^{2}\cos^{2}x}\,dx=f(x)+c, then lim⁡x→0f(x)\displaystyle\lim_{x\to 0}f(x) equals
A11correct
B2\sqrt{2}
C12\dfrac{1}{\sqrt{2}}
D3\sqrt{3}
Solution
Step 1: Split the integrand into two fractions:
x−sin⁡xcos⁡xx2cos⁡2x=xx2cos⁡2x−sin⁡xcos⁡xx2cos⁡2x=sec⁡2xx−tan⁡xx2\dfrac{x-\sin x\cos x}{x^{2}\cos^{2}x}=\dfrac{x}{x^{2}\cos^{2}x}-\dfrac{\sin x\cos x}{x^{2}\cos^{2}x}=\dfrac{\sec^{2}x}{x}-\dfrac{\tan x}{x^{2}}
 Step 2: Recognise an exact derivative. By the quotient rule,
ddx ⁣(tan⁡xx)=xsec⁡2x−tan⁡xx2=sec⁡2xx−tan⁡xx2\dfrac{d}{dx}\!\left(\dfrac{\tan x}{x}\right)=\dfrac{x\sec^{2}x-\tan x}{x^{2}}=\dfrac{\sec^{2}x}{x}-\dfrac{\tan x}{x^{2}}
 which equals the integrand. Step 3: Therefore
f(x)=∫ddx ⁣(tan⁡xx)dx=tan⁡xx+cf(x)=\int\dfrac{d}{dx}\!\left(\dfrac{\tan x}{x}\right)dx=\dfrac{\tan x}{x}+c
 Step 4: Take the limit (standard limit tan⁡x/x→1\tan x/x\to 1):
lim⁡x→0f(x)=lim⁡x→0tan⁡xx+c∣c=0=1\lim_{x\to 0}f(x)=\lim_{x\to 0}\dfrac{\tan x}{x}+c\Big|_{c=0}=1
 Correct answer: (1)
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