begin{aligned} & int frac{dx}{(2 sin x+sec x) | vedclass

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(D) Given the integral $I = \int \frac{dx}{(2 \sin x + \sec x)^4}$.Multiply numerator and denominator by $\sec^4 x$:$I = \int \frac{\sec^4 x}{(2 \sin

Vedclass · Accepted Answer

$\frac{-16}{105}$

Vedclass · Answer

(D) Given the integral $I = \int \frac{dx}{(2 \sin x + \sec x)^4}$.Multiply numerator and denominator by $\sec^4 x$:$I = \int \frac{\sec^4 x}{(2 \sin x \cos x + 1)^4} dx = \int \frac{\sec^4 x}{(\sin 2x + 1)^4} dx$.Alternatively,rewrite the denominator: $2 \sin x + \sec x = \frac{2 \sin x \cos x + 1}{\cos x} = \frac{\sin 2x + 1}{\cos x}$.Thus,$I = \int \frac{\cos^4 x}{(\sin 2x + 1)^4} dx = \int \frac{\cos^4 x}{((\sin x + \cos x)^2)^4} dx = \int \frac{\cos^4 x}{(\sin x + \cos x)^8} dx$.Divide numerator and denominator by $\cos^8 x$:$I = \int \frac{\sec^4 x}{(1 + 	an x)^8} dx$.Let $	an x = t$,then $\sec^2 x dx = dt$. Since $\sec^2 x = 1 + t^2$,we have:$I = \int \frac{1 + t^2}{(1 + t)^8} dt = \int \frac{(1 + t)^2 - 2t}{(1 + t)^8} dt = \int \frac{(1 + t)^2 - 2(1 + t - 1)}{(1 + t)^8} dt$$I = \int \frac{dt}{(1 + t)^6} - 2 \int \frac{dt}{(1 + t)^7} + 2 \int \frac{dt}{(1 + t)^8}$$I = -\frac{1}{5}(1 + t)^{-5} + \frac{2}{6}(1 + t)^{-6} - \frac{2}{7}(1 + t)^{-7} + k$$I = -\frac{1}{5}(1 + 	an x)^{-5} + \frac{1}{3}(1 + 	an x)^{-6} - \frac{2}{7}(1 + 	an x)^{-7} + k$.Comparing with the given form,$A = -\frac{1}{5}$,$B = \frac{1}{3}$,$C = -\frac{2}{7}$.$A + B + C = -\frac{1}{5} + \frac{1}{3} - \frac{2}{7} = \frac{-21 + 35 - 30}{105} = -\frac{16}{105}$.

$\begin{aligned} & \int \frac{dx}{(2 \sin x+\sec x)^4}=A(1+\tan x)^{-5} \\ & +B(1+\tan x)^{-6}+C(1+\tan x)^{-7}+k, \text{ then } \\ & A+B+C= \end{aligned}$

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