int_{-frac{pi}{8092}}^{frac{pi}{8092}} frac{s | vedclass

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

Question

(B) ધારો કે $I = \int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec (2023 x)}{1+(2023)^{(2023 x)}} d x$. ગુણધર્મ $\int_{-a}^{a} f(x) d x = \int_{0

Vedclass · Accepted Answer

$\frac{\log (\sqrt{2}+1)}{2023}$

Vedclass · Answer

(B) ધારો કે $I = \int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec (2023 x)}{1+(2023)^{(2023 x)}} d x$. ગુણધર્મ $\int_{-a}^{a} f(x) d x = \int_{0}^{a} [f(x) + f(-x)] d x$ નો ઉપયોગ કરતા: $f(x) = \frac{\sec (2023 x)}{1+(2023)^{(2023 x)}}$ $f(-x) = \frac{\sec (-2023 x)}{1+(2023)^{(-2023 x)}} = \frac{\sec (2023 x)}{1+\frac{1}{(2023)^{(2023 x)}}} = \frac{(2023)^{(2023 x)} \sec (2023 x)}{(2023)^{(2023 x)}+1}$ $f(x)$ અને $f(-x)$ નો સરવાળો કરતા: $f(x) + f(-x) = \frac{\sec (2023 x) [1 + (2023)^{(2023 x)}]}{1+(2023)^{(2023 x)}} = \sec (2023 x)$ તેથી,$I = \int_{0}^{\frac{\pi}{8092}} \sec (2023 x) d x$ $= \left[ \frac{1}{2023} \log |\sec (2023 x) + 	an (2023 x)| ight]_{0}^{\frac{\pi}{8092}}$ $= \frac{1}{2023} [\log |\sec(\frac{\pi}{4}) + 	an(\frac{\pi}{4})| - \log |\sec(0) + 	an(0)|]$ $= \frac{1}{2023} [\log (\sqrt{2} + 1) - \log (1 + 0)]$ $= \frac{\log (\sqrt{2} + 1)}{2023}$

$\int_{-\frac{\pi}{8092}}^{\frac{\pi}{8092}} \frac{\sec (2023 x)}{1+(2023)^{(2023 x)}} d x=$

Similar Questions

$\int_{-4}^{4} \log \left(\frac{8-x}{8+x}\right) d x=$

જો $h(a) = h(b)$ હોય,તો સંકલન $\int_a^b {[f(g(h(x)))]^{-1} f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) \, dx} = $ નું મૂલ્ય શોધો.

જો $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos^2 x}{1+e^x} dx = \pi(\alpha \pi^2 + \beta)$,જ્યાં $\alpha, \beta \in \mathbb{Z}$,તો $(\alpha + \beta)^2$ ની કિંમત શોધો:

જો ${I_n} = \int_0^{\pi /4} {{\tan ^n}\theta \,d\theta }$ હોય,તો ${I_8} + {I_6}$ ની કિંમત શોધો.

$\int_0^{\frac{\pi}{4}} \log \left(\frac{\sin x+\cos x}{\cos x}\right) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module