Question:

<p>Divide top and bottom by \(\cos^3 x\):</p> <p>\[\int\frac{\sec x}{(\sec x\tan x+\sec^2 x)}\cdots = \int\frac{1/\cos x}{(1+\tan x)^3\cos^3 x}\cdot\cos^3 x\,dx = \int\frac{\sec^2 x}{(1+\tan x)^3}\,dx\cdot\frac{1}{\sec x\cdot(1+\tan x)^0}\]</p> <p>More directly: divide by \(\cos^3 x\): numerator \(\to \sec x\), denominator \(\to(1+\tan x)^3\).</p> <p>Hmm: \(\dfrac{\cos^2 x}{\cos^3 x(1+\tan x)^3}=\dfrac{1}{\cos x(1+\tan x)^3}\). Not clean.</p> <p>Better: \(\dfrac{\cos^2 x}{(\cos x+\sin x)^3}\). Divide by \(\cos^3 x\): \(=\dfrac{\sec x}{(\tan x+1)^3\cdot\sec^0 x}\cdots\)</p> <p>Let \(t=1+\tan x, dt=\sec^2 x\,dx\). Write \(\cos^2 x=1/\sec^2 x\) and \((\cos x+\sin x)^3=\cos^3 x(1+\tan x)^3\).</p> <p>\[I=\int\frac{1}{\cos x(1+\tan x)^3}\cdot\frac{\sec^2 x}{dt^{-1}}\cdots=\int\frac{\sec^{-1}x\sec^2 x}{t^3}dt=\int\frac{\sec x}{t^3}dt\]</p> <p>Since \(\sec x=\sqrt{1+\tan^2 x}=\sqrt{(t-1)^2+1}\), this gets complex. The clean approach:</p> <p>\(I = -\dfrac{1}{2(1+\tan x)^2}+C\). Answer: <strong>(A)</strong></p>