The value of the integral I = int_{frac{pi}{2 | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(A) Let $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an 2x}}$ ...$(1)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{24} + \frac{5\pi}{24} = \frac{6\pi}{24} = \frac{\pi}{4}$.$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an(2(\frac{\pi}{4}-x))}}$$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{	an(\frac{\pi}{2}-2x)}}$Since $	an(\frac{\pi}{2}-	heta) = \cot 	heta$,we have:$I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\cot 2x}} = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{\sqrt[3]{	an 2x} dx}{\sqrt[3]{	an 2x} + 1}$ ...$(2)$Adding $(1)$ and $(2)$:$2I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{1+\sqrt[3]{	an 2x}}{1+\sqrt[3]{	an 2x}} dx = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} 1 dx$$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}$

The value of the integral $I = \int_{\frac{\pi}{24}}^{\frac{5\pi}{24}} \frac{dx}{1+\sqrt[3]{\tan 2x}}$ is:

Similar Questions

$\int_0^{\pi /4} {\log (1 + \tan \theta )\,d\theta = } $

The value of the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(x^2 + \log \frac{\pi-x}{\pi+x}\right) \cos x \, dx$ is equal to

$\int_{0}^{1} x(1-x)^{5} dx =$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{\pi} \frac{x \, dx}{1+\sin x}$.

Evaluate the definite integral: $\int_0^{2a} f(x) dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module