Question

The value of limx→0∫0x2sec2⁡tdtxsin⁡x is

• A. (A) 3
• B. (B) 2
• C. (C) 1
• D. (D) 0

Answer 1

Answer: (C)

(C) 1

Answer 2

Explanation:
Given:The expression limx→0∫0x2sec2⁡tdtxsin⁡xWe have to evaluate the given expression.Here, we'll use fundamental integrals of trigonometric functions.limx→0∫0x2sec2⁡tdtxsin⁡x[∫sec2⁡tdt=tan⁡t]limx→0[tan⁡t]x2xsin⁡x=limx→0tan⁡x2xsin⁡x=limx→0tan⁡x2x2x2xx[ Dividing by x2 in Numerator & denominator ]limx→0(tan⁡x2x2)sin⁡xx=11=1[Using limit of trigonometric functions ]Hence, the correct option is (C).