Maths-
General
Easy

Question

integral fraction numerator sin invisible function application x over denominator left parenthesis a plus b c o s invisible function application x right parenthesis squared end fraction d x

  1. 1 over b left parenthesis a plus b blank c o s blank x right parenthesis plus c    
  2. fraction numerator 1 over denominator b left parenthesis a plus b blank c o s blank x right parenthesis end fraction plus c    
  3. 1 over b I n left parenthesis a plus b blank c o s blank x right parenthesis plus c    
  4. None of these    

hintHint:

We are given a function to integrate. We need to simplify the function before integration. We will use the method of substitution to solve the given integration.

The correct answer is: fraction numerator 1 over denominator b left parenthesis a plus b blank c o s blank x right parenthesis end fraction plus c


    The given function is integral fraction numerator sin x over denominator left parenthesis a space plus space b cos x right parenthesis squared end fraction d x
    We will denote it by "I". We will use substitution method to solve the question. In this method, we replace one of the component of function by some variable. Then, we try to simplify the equation.
    Let's solve the integration.
    I space equals integral fraction numerator s i n x over denominator left parenthesis a space plus space b cos x right parenthesis squared end fraction d x
space space space L e t space
space space space space space space a space plus b cos x space equals space t

space space space T a k i n g space d e r i v a t i v e
space space space space space space space minus b sin x space d x space equals space d t
space space space space space space space space space space space space space sin x space d x space equals fraction numerator negative 1 over denominator b end fraction d t

I space equals space integral 1 over t squared open parentheses fraction numerator negative 1 over denominator b end fraction close parentheses d t
space space
space space equals fraction numerator negative 1 over denominator b end fraction integral 1 over t squared d t
space
space space equals fraction numerator space minus 1 over denominator b end fraction open parentheses fraction numerator t to the power of negative 2 plus 1 end exponent over denominator negative 2 plus 1 end fraction close parentheses space plus space C
space space space space







space space space space space
    equals fraction numerator negative 1 over denominator b end fraction open parentheses fraction numerator t to the power of negative 1 end exponent over denominator negative 1 end fraction close parentheses space plus space C

equals 1 over b 1 over t space plus space C
space
    We will resubstitute the value of t.
    I space equals space fraction numerator 1 over denominator b left parenthesis a space plus space b cos x right parenthesis end fraction plus C
    This is the required answer.

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