यदि I(x) = int e^{sin^2 x} (cos x sin 2x - si | vedclass

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

Question

(C) दिया गया है $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$. $\sin 2x = 2 \sin x \cos x$ का उपयोग करने पर: $I(x) = \int e^{\sin^2 x} (2 \s

Vedclass · Accepted Answer

$\frac{1}{2} e^{\frac{3}{4}}$

Vedclass · Answer

(C) दिया गया है $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$. $\sin 2x = 2 \sin x \cos x$ का उपयोग करने पर: $I(x) = \int e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) dx$. यहाँ,$\frac{d}{dx} (e^{\sin^2 x} \cos x) = e^{\sin^2 x} (2 \sin x \cos x) \cos x + e^{\sin^2 x} (-\sin x) = e^{\sin^2 x} (2 \sin x \cos^2 x - \sin x) = e^{\sin^2 x} (\cos x \sin 2x - \sin x)$ होता है। अतः,$I(x) = e^{\sin^2 x} \cos x + C$. चूँकि $I(0) = 1$,इसलिए $1 = e^{\sin^2 0} \cos 0 + C \Rightarrow 1 = 1 \cdot 1 + C \Rightarrow C = 0$. अतः,$I(x) = e^{\sin^2 x} \cos x$. इसलिए $I\left(\frac{\pi}{3}\right) = e^{\sin^2(\pi/3)} \cos(\pi/3) = e^{3/4} \cdot \frac{1}{2} = \frac{1}{2} e^{\frac{3}{4}}$.

यदि $I(x) = \int e^{\sin^2 x} (\cos x \sin 2x - \sin x) dx$ और $I(0) = 1$ है,तो $I\left(\frac{\pi}{3}\right)$ का मान ज्ञात कीजिए।

Similar Questions

यदि $\int \frac{1-(\cot x)^{2019}}{\tan x+(\cot x)^{2020}} dx = \frac{1}{n} \ln |(f(x))^n + (g(x))^n| + c$ है,तो $n[(f(x))^4 + (g(x))^4]_{x=\frac{\pi}{3}}$ का मान ज्ञात कीजिए।

$\int \frac{3\sin x + 2\cos x}{3\cos x + 2\sin x} \, dx = $

यदि $\tan \alpha = \frac{4}{3}$ है,तो $\int \frac{1}{3 \cos x - 4 \sin x} dx = $

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

$\int \frac{dx}{4+3 \cot x} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module