यदि int frac{sin ^3 x}{left(cos ^4 x+3 cos ^2 | vedclass

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Question

(A) माना $I = \int \frac{\sin ^3 x}{(\cos ^4 x + 3 \cos ^2 x + 1) 	an ^{-1}(\sec x + \cos x)} dx$ है। माना $t = 	an ^{-1}(\sec x + \cos x)$ है। तब $

Vedclass · Accepted Answer

$	an ^{-1}(\sec x+\cos x)$

Vedclass · Answer

(A) माना $I = \int \frac{\sin ^3 x}{(\cos ^4 x + 3 \cos ^2 x + 1) 	an ^{-1}(\sec x + \cos x)} dx$ है। माना $t = 	an ^{-1}(\sec x + \cos x)$ है। तब $dt = \frac{1}{1 + (\sec x + \cos x)^2} (\sec x 	an x - \sin x) dx$ होगा। $dt = \frac{1}{1 + (\frac{1}{\cos x} + \cos x)^2} (\frac{\sin x}{\cos ^2 x} - \sin x) dx$। $dt = \frac{\cos ^2 x}{\cos ^2 x + (1 + \cos ^2 x)^2} \cdot \frac{\sin x (1 - \cos ^2 x)}{\cos ^2 x} dx$। $dt = \frac{\sin x \cdot \sin ^2 x}{\cos ^2 x + 1 + 2 \cos ^2 x + \cos ^4 x} dx = \frac{\sin ^3 x}{\cos ^4 x + 3 \cos ^2 x + 1} dx$। अतः,$I = \int \frac{dt}{t} = \ln |t| + C = \ln |	an ^{-1}(\sec x + \cos x)| + C$। चूंकि $f(x) = \ln |	an ^{-1}(\sec x + \cos x)|$ है,इसलिए $e^{f(x)} = 	an ^{-1}(\sec x + \cos x)$।

यदि $\int \frac{\sin ^3 x}{\left(\cos ^4 x+3 \cos ^2 x+1\right) \tan ^{-1}(\sec x+\cos x)} d x=f(x)+C$ है,तो $e^{f(x)}=$

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