int e^{-2 x}left(tan 2 x-2 sec ^2 2 x tan 2 x | vedclass

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

Question

(C) माना $I = \int e^{-2 x}(	an 2 x - 2 \sec^2 2 x 	an 2 x) dx$. फलन $f(x) = 	an 2 x$ लें। तब $f'(x) = 2 \sec^2 2 x$. $2x = t$ रखने पर, $2 dx = dt$

Vedclass · Accepted Answer

$-\frac{e^{-2 x}}{2}\left[	an 2 x-\sec ^2 2 xight]+c$

Vedclass · Answer

(C) माना $I = \int e^{-2 x}(	an 2 x - 2 \sec^2 2 x 	an 2 x) dx$. फलन $f(x) = 	an 2 x$ लें। तब $f'(x) = 2 \sec^2 2 x$. $2x = t$ रखने पर, $2 dx = dt$, अतः $dx = \frac{1}{2} dt$. $I = \frac{1}{2} \int e^{-t} (	an t - 2 \sec^2 t 	an t) dt$. $f(t) = 	an t - \sec^2 t$ लें। तब $f'(t) = \sec^2 t - 2 \sec^2 t 	an t$. अतः, $I = \frac{1}{2} \int e^{-t} (f'(t) - f(t)) dt$. सूत्र $\int e^{kt} (f'(t) + k f(t)) dt = e^{kt} f(t) + C$ का उपयोग करने पर, जहाँ $k = -1$: $I = \frac{1}{2} (-e^{-t} f(t)) + C = -\frac{1}{2} e^{-t} (	an t - \sec^2 t) + C$. $t = 2x$ प्रतिस्थापित करने पर: $I = -\frac{1}{2} e^{-2 x} (	an 2 x - \sec^2 2 x) + C = \frac{1}{2} e^{-2 x} (\sec^2 2 x - 	an 2 x) + C$.

$\int e^{-2 x}\left(\tan 2 x-2 \sec ^2 2 x \tan 2 x\right) d x=$

Similar Questions

यदि $\int e^{\alpha x}\left(\frac{1-\beta \sin x}{1-\cos x}\right) d x=-e^x \cot \frac{x}{2}+c$ है, तो $\frac{\alpha^2+\beta^2}{2 \alpha \beta}=$

$\int e^{x}\left(\frac{2}{x}-\frac{2}{x^2}\right) dx$ का मान ज्ञात कीजिए।

$\int e^{\tan x}(\sec ^{2} x+\sec ^{3} x \sin x) d x$ का मान ज्ञात कीजिए।

यदि $\int e^x \left( \frac{x^2-8x+19}{(x-1)^5} \right) dx = \frac{e^x(lx+m)}{(x-1)^4} + C$ है, तो $4l+m=$

$\int \left[\frac{1-\log x}{1+(\log x)^{2}}\right]^{2} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module