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Q21:
Choose the correct answer in Exercises 20 and 21. 21. The value of $\int_{0}^{\frac{\pi}{2}} \log (\frac{4+3\sin x}{4+3\cos x}) dx$ is

Solution :

Step 1: Definition of the Definite Integral

Let the given integral be $I$:

$I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{4+3\sin x}{4+3\cos x} \right) dx$ [Equation 1]

Step 2: Application of King's Property

Utilize the definite integral property: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$. Here, $a=0$ and $b=\frac{\pi}{2}$.

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

Applying trigonometric co-function identities $\sin(\frac{\pi}{2}-x) = \cos x$ and $\cos(\frac{\pi}{2}-x) = \sin x$ [Per Trigonometric Identity Axioms]:

$I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{4+3\cos x}{4+3\sin x} \right) dx$ [Equation 2]

Step 3: Summation of Equations

Add Equation 1 and Equation 2:

$2I = \int_{0}^{\frac{\pi}{2}} \left[ \log \left( \frac{4+3\sin x}{4+3\cos x} \right) + \log \left( \frac{4+3\cos x}{4+3\sin x} \right) \right] dx$

Employ the logarithmic property $\log(a) + \log(b) = \log(ab)$ [Per Logarithmic Laws]:

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

Step 4: Algebraic Simplification

The terms within the logarithm cancel out to yield unity:

$2I = \int_{0}^{\frac{\pi}{2}} \log(1) dx$

Since $\log(1) = 0$ [By logarithmic definition]:

$2I = \int_{0}^{\frac{\pi}{2}} 0 \cdot dx$ $2I = 0$ $I = 0$

Visual Representation of the Integrand

x y O f(x) = log((4+3sin x)/(4+3cos x))

Final Solution: 0


More Questions from Class 12 Mathematics Integrals EXERCISE 7.10


CBSE Solutions for Class 12 Mathematics Integrals


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