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

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

Question

(A) Let $I = \int \frac{\sin ^3 x}{(\cos ^4 x + 3 \cos ^2 x + 1) 	an ^{-1}(\sec x + \cos x)} dx$. Let $t = 	an ^{-1}(\sec x + \cos x)$. Then $dt = \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{\sin ^3 x}{(\cos ^4 x + 3 \cos ^2 x + 1) 	an ^{-1}(\sec x + \cos x)} dx$. Let $t = 	an ^{-1}(\sec x + \cos x)$. Then $dt = \frac{1}{1 + (\sec x + \cos x)^2} (\sec x 	an x - \sin x) dx$. $dt = \frac{1}{1 + (\frac{1}{\cos x} + \cos x)^2} (\frac{\sin x}{\cos ^2 x} - \sin x) dx$. $dt = \frac{\cos ^2 x}{\cos ^2 x + (1 + \cos ^2 x)^2} \cdot \frac{\sin x (1 - \cos ^2 x)}{\cos ^2 x} dx$. $dt = \frac{\sin x \cdot \sin ^2 x}{\cos ^2 x + 1 + 2 \cos ^2 x + \cos ^4 x} dx = \frac{\sin ^3 x}{\cos ^4 x + 3 \cos ^2 x + 1} dx$. Thus,$I = \int \frac{dt}{t} = \ln |t| + C = \ln |	an ^{-1}(\sec x + \cos x)| + C$. Given $f(x) = \ln |	an ^{-1}(\sec x + \cos x)|$,we have $e^{f(x)} = 	an ^{-1}(\sec x + \cos x)$.

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

Similar Questions

$\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx$ is equal to-

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

$\int \frac{e^{2025+x} - e^{2025-x}}{e^{2026+x} + e^{2026-x}} dx = $ . . . . . . + $C$

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

$\int \frac{\log (\cot x)}{\sin 2 x} \,d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module