समाकलन I = int_{frac{pi}{24}}^{frac{5pi}{24}} | vedclass

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Question

(A) माना $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an 2x}}$ ...$(1)$गुणधर्म $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(A) माना $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an 2x}}$ ...$(1)$गुणधर्म $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ का उपयोग करने पर, जहाँ $a+b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$.$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an(2(\frac{\pi}{4}-x))}}$$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an(\frac{\pi}{2}-2x)}}$चूंकि $	an(\frac{\pi}{2}-	heta) = \cot 	heta$, इसलिए:$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\cot 2x}} = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{\sqrt[3]{	an 2x} dx}{\sqrt[3]{	an 2x} + 1}$ ...$(2)$$(1)$ और $(2)$ को जोड़ने पर:$2I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{1+\sqrt[3]{	an 2x}}{1+\sqrt[3]{	an 2x}} dx = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} 1 dx$$2I = [x]_{\frac{\pi}{24}}^{\frac{5\pi}{24}} = \frac{5\pi}{24} - \frac{\pi}{24} = \frac{4\pi}{24} = \frac{\pi}{6}$$I = \frac{\pi}{12}$

समाकलन $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan 2x}}$ का मान है:

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