SATHEE: Indefinite Integration 1 Question 7

Indefinite Integration 1 Question 7

8. If $f (x) = \frac{2 - x \cos x}{2 + x \cos x}$ and $g (x) = \log_{e} x, (x > 0)$ then the value of the integral $\int_{- π / 4}^{π / 4} g (f (x)) d x$ is

(a) $\log_{e} 3$

(b) $\log_{e} e$

(d) $\log_{e} 1$

(2019 Main, 8 April I)

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Solution:

The given functions are

$g (x) = \log_{e} x, x > 0 and f (x) = \frac{2 - x \cos x}{2 + x \cos x}$

Let

$I = \int_{- π / 4}^{π / 4} g (f (x)) d x$

Then, $I = \int_{- π / 4}^{π / 4} \log_{e} \frac{2 - x \cos x}{2 + x \cos x} d x$

Now, by using the property

$\begin{aligned} \int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x, we get \\ I = \int_{- π / 4}^{π / 4} \log_{e} \frac{2 + x \cos x}{2 - x \cos x} d x \end{aligned}$

On adding Eqs. (i) and (ii), we get

$\begin{aligned} 2 I & = \int_{- π / 4}^{π / 4} \log_{e} \frac{2 - x \cos x}{2 + x \cos x} + \log_{e} \frac{2 + \cos x}{2 x \cos x} d x \\ = \int_{- π / 4}^{π / 4} \log_{e} \frac{2 - x \cos x}{2 + x \cos x} \times \frac{2 + x \cos x}{2 - x \cos x} d x \\ \Rightarrow & [∵ \log_{e} A + \log_{e} B = \log_{e} A B] \\ 2 I = \int_{- π / 4}^{π / 4} \log_{e} (1) d x = 0 \Rightarrow I = 0 = \log_{e} (1) \end{aligned}$

Indefinite Integration 1 Question 7

8. If $f (x) = \frac{2 - x \cos x}{2 + x \cos x}$ and $g (x) = \log_{e} x, (x > 0)$ then the value of the integral $\int_{- π / 4}^{π / 4} g (f (x)) d x$ is

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Indefinite Integration 1 Question 7

8. If f(x)=2−xcos⁡x2+xcos⁡x and g(x)=loge⁡x,(x>0) then the value of the integral ∫−π/4π/4g(f(x))dx is

Solution:

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8. If $f (x) = \frac{2 - x \cos x}{2 + x \cos x}$ and $g (x) = \log_{e} x, (x > 0)$ then the value of the integral $\int_{- π / 4}^{π / 4} g (f (x)) d x$ is