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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(A) ધારો કે $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an 2x}}$ ...$(1)$ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા,જ્યાં $a+b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$.$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an(2(\frac{\pi}{4}-x))}}$$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an(\frac{\pi}{2}-2x)}}$આપણે જાણીએ છીએ કે $	an(\frac{\pi}{2}-	heta) = \cot 	heta$,તેથી:$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\cot 2x}} = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{\sqrt[3]{	an 2x} dx}{\sqrt[3]{	an 2x} + 1}$ ...$(2)$$(1)$ અને $(2)$ નો સરવાળો કરતા:$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}$

સંકલન $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan 2x}}$ ની કિંમત શોધો:

Similar Questions

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x) dx =$

સંકલન $\int_{-\pi/4}^{\pi/4} \left( \frac{32 \cos^4 x}{1 + e^{\sin x}} \right) dx$ નું મૂલ્ય છે:

જો $I = \int_{0}^{1} \frac{dx}{1+x^{\pi / 2}}$ હોય,તો

$\int_0^{\pi /2} \frac{dx}{1 + \tan^3 x}$ નું મૂલ્ય શોધો.

નિશ્ચિત સંકલનનું મૂલ્ય શોધો: $\int_0^\pi \frac{x \cos^2 x}{1+\sin x} dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module