int e^{sin x} frac{(x cos^3 x - sin x)}{cos^2 | vedclass

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

Question

(A) माना $I = \int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} ight) dx$ $= \int e^{\sin x} (x \cos x - 	an x \sec x) dx$ यहाँ,यदि हम $f

Vedclass · Accepted Answer

$e^{\sin x}(x - \sec x) + C$

Vedclass · Answer

(A) माना $I = \int e^{\sin x} \left( \frac{x \cos^3 x - \sin x}{\cos^2 x} ight) dx$ $= \int e^{\sin x} (x \cos x - 	an x \sec x) dx$ यहाँ,यदि हम $f(x) = e^{\sin x}(x - \sec x)$ लें,तो इसका अवकलन करने पर: $\frac{d}{dx} [e^{\sin x} (x - \sec x)] = e^{\sin x} \cos x (x - \sec x) + e^{\sin x} (1 - \sec x 	an x)$ $= x e^{\sin x} \cos x - e^{\sin x} + e^{\sin x} - e^{\sin x} \sec x 	an x$ $= e^{\sin x} (x \cos x - \sec x 	an x)$ अतः,समाकलन का उत्तर $e^{\sin x}(x - \sec x) + C$ है।

$\int e^{\sin x} \frac{(x \cos^3 x - \sin x)}{\cos^2 x} dx =$

Similar Questions

$\int \frac{1}{x^5 \sqrt[5]{x^5+1}} d x=$

$\int e^{\tan ^{-1} x} \cdot \frac{1+x+x^2}{1+x^2} dx$ का मान ज्ञात कीजिए।

$\int \frac{dx}{(x+1) \sqrt{x^2+1}} = $

$\int \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module