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  • Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 41 maths

Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 41 maths

Answers (1)

best_answer

Answer:

I=0

Hint:

To solve this equation we use \int_{0}^{a} f(x) d x   formula.

Given:

\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin x}{4+3 \cos x}\right] d x

Solution:

Let

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin x}{4+3 \cos x}\right] d x \ldots(i) \\\\ &\int_{0}^{a} f(x) d x=\int_{0}^{a}(a-x) d x \end{aligned}

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right] d x \\\\ &I=\int_{0}^{\frac{\pi}{2}} \log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] d x \ldots(i i) \end{aligned}

Adding (i) and (ii)

\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}}\left(\log \left[\frac{4+3 \cos x}{4+3 \sin x}\right]+\log \left[\frac{4+3 \sin x}{4+3 \cos x}\right]\right) d x \\\\ &2 I=\int_{0}^{\frac{\pi}{2}}\left(\log \left[\frac{4+3 \cos x}{4+3 \sin x}\right] \cdot \log \left[\frac{4+3 \sin x}{4+3 \cos x}\right]\right) d x \end{aligned}

\begin{aligned} &2 I=\int_{0}^{\frac{\pi}{2}} 1 d x \\\\ &2 I=0 \\\\ &I=0 \end{aligned}

 

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