Question:
If I=∫\(\frac{x^2dx}{(x\,sin\,x+cos\,x)^2}\)=f(x)+tan x+c, then f(x) is
If I=∫\(\frac{x^2dx}{(x\,sin\,x+cos\,x)^2}\)=f(x)+tan x+c, then f(x) is
Updated On: Mar 29, 2025
- \(\frac{sin\,x}{xsin\,x+cos\,x}\)
- \(\frac{1}{(xsin\,x+cos\,x)^2}\)
- \(\frac{-x}{cos\,x(xsin\,x+cos\,x)}\)
- \(\frac{1}{sin\,x(xcos\,x+sin\,x)}\)
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The Correct Option is C
Solution and Explanation
The correct answer is option (C): \(\frac{-x}{cos\,x(xsin\,x+cos\,x)}\)
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Concepts Used:
Integration by Parts
Integration by Parts is a mode of integrating 2 functions, when they multiplied with each other. For two functions ‘u’ and ‘v’, the formula is as follows:
∫u v dx = u∫v dx −∫u' (∫v dx) dx
- u is the first function u(x)
- v is the second function v(x)
- u' is the derivative of the function u(x)
The first function ‘u’ is used in the following order (ILATE):
- 'I' : Inverse Trigonometric Functions
- ‘L’ : Logarithmic Functions
- ‘A’ : Algebraic Functions
- ‘T’ : Trigonometric Functions
- ‘E’ : Exponential Functions
The rule as a diagram:
