int frac{sec ^2 x}{sin ^7 x} d x - int frac{7 | vedclass

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Question

(A) ધારો કે $I = \int \frac{\sec ^2 x}{\sin ^7 x} d x - \int \frac{7}{\sin ^7 x} d x = \int \frac{1}{\cos ^2 x \sin ^7 x} d x - \int \frac{7}{\sin ^7

Vedclass · Accepted Answer

$\frac{1}{\sin ^6 x \cos x} + c$

Vedclass · Answer

(A) ધારો કે $I = \int \frac{\sec ^2 x}{\sin ^7 x} d x - \int \frac{7}{\sin ^7 x} d x = \int \frac{1}{\cos ^2 x \sin ^7 x} d x - \int \frac{7}{\sin ^7 x} d x$. $f(x) = \frac{	an x}{\sin ^6 x}$ નું વિકલન લેતા,ભાગાકારના નિયમ મુજબ,$\frac{d}{dx} \left( \frac{	an x}{\sin ^6 x} ight) = \frac{\sec^2 x \cdot \sin^6 x - 	an x \cdot 6 \sin^5 x \cos x}{\sin^{12} x}$. $= \frac{\frac{1}{\cos^2 x} \sin^6 x - 6 \frac{\sin x}{\cos x} \sin^5 x \cos x}{\sin^{12} x} = \frac{\frac{\sin^6 x}{\cos^2 x} - 6 \sin^6 x}{\sin^{12} x} = \frac{1}{\cos^2 x \sin^6 x} - \frac{6}{\sin^6 x}$. આમ,સંકલનનો ઉકેલ $\frac{	an x}{\sin^6 x} + c$ મળે છે,જે $\frac{1}{\sin^5 x \cos x} + c$ તરીકે લખી શકાય છે. વિકલ્પોને જોતા,સાચો વિકલ્પ $A$ છે.

$\int \frac{\sec ^2 x}{\sin ^7 x} d x - \int \frac{7}{\sin ^7 x} d x =$

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