int_{frac{pi}{4}}^{frac{3pi}{4}} frac{dx}{1 + | vedclass

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Question

(C) ધારો કે $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} \quad (i)$ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપ

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) ધારો કે $I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} \quad (i)$ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા,જ્યાં $a+b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi$ છે:$I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos(\pi - x)} = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 - \cos x} \quad (ii)$$(i)$ અને $(ii)$ નો સરવાળો કરતા:$2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \left( \frac{1}{1 + \cos x} + \frac{1}{1 - \cos x} \right) dx$$2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1 - \cos x + 1 + \cos x}{1 - \cos^2 x} dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{2}{\sin^2 x} dx$$2I = 2 \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \csc^2 x dx$$I = [-\cot x]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$$I = -(\cot \frac{3\pi}{4} - \cot \frac{\pi}{4}) = -(-1 - 1) = 2$

$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \cos x} = \dots$

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