Question:

The value of \(\int_{0}^{\frac{\pi}{2}}\) log(\(\frac{4+3\,sin\,x}{4+3\,cos\,x}\))dx is

Updated On: Feb 29, 2024
  • 2

  • \(\frac{3}{4}\)

  • 0

  • -2

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The Correct Option is C

Solution and Explanation

Let I=\(\int_{0}^{\frac{\pi}{2}}\) log(\(\frac{4+3\,sin\,x}{4+3\,cos\,x}\))dx………………....(1)

⇒I=\(\int_{0}^{\frac{\pi}{2}}\)log[\(\frac{4+3\,sin\,(\frac{\pi}{2}-x)}{4+3\,cos\,(\frac{\pi}{2}-x)}\)]dx (\(\int_{0}^{a}\)ƒ(x)dx=\(\int_{0}^{a}\)ƒ(a-x)dx)

⇒I=\(\int_{0}^{\frac{\pi}{2}}\) log(\(\frac{4+3\,cos\,x}{4+3\,sin\,x}\))dx…………………...(2)

Adding(1)and(2),we obtain

2I=\(\int_{0}^{\frac{\pi}{2}}\){log(\(\frac{4+3\,sin\,x}{4+3\,cos\,x}\))+log(\(\frac{4+3\,cos\,x}{4+3\,sin\,x}\))}dx

⇒2I=\(\int_{0}^{\frac{\pi}{2}}\) log1dx

⇒2I=\(\int_{0}^{\frac{\pi}{2}}\)0dx

⇒I=0

Hence, the correct Answer is C.

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Concepts Used:

Definite Integral

Definite integral is an operation on functions which approximates the sum of the values (of the function) weighted by the length (or measure) of the intervals for which the function takes that value.

Definite integrals - Important Formulae Handbook

A real valued function being evaluated (integrated) over the closed interval [a, b] is written as :

\(\int_{a}^{b}f(x)dx\)

Definite integrals have a lot of applications. Its main application is that it is used to find out the area under the curve of a function, as shown below: 

Definite integral