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

$I=0$

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}

\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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