Indefinite Integration
Integration of Trigonometric Functions
Grade 12
Question:
<p>Let \(f(x) = \dfrac{2\sin^2 x - 1}{\cos x\,(1+\sin x)}\) when \(\sin x \neq -1\). Then \(\displaystyle\int f(x)\,dx\) equals</p>
<li>\(\sec x + C\)</li>
<li>\(\tan x + C\)</li>
<li>\(\cos x\cdot\sec^2 x + C\)</li>
<li>\(\sec x + \tan x + C\)</li>
Step-by-Step Solution
Key Concept: Factor numerator: 2sin^2x-1 = -cos2x = -(1-2sin^2x). Then use partial fractions or algebraic manipulation.
<p><strong>Simplify numerator:</strong> \(2\sin^2 x - 1 = -(1-2\sin^2 x) = -\cos 2x\)</p>
<p>Alternatively, \(2\sin^2 x -1 = -(1-\sin^2 x) - (1-\sin x)(1+\sin x)/... \)</p>
<p>Use the identity: \(2\sin^2 x - 1 = (\sin x -1)(\sin x+1) + (\sin^2 x - \cos^2 x)\cdots\)</p>
<p>Actually: \(\dfrac{2\sin^2 x-1}{\cos x(1+\sin x)} = \dfrac{-(1-2\sin^2 x)}{\cos x(1+\sin x)}\)</p>
<p>\[= \frac{(\sin x-1)(\sin x+1) + \sin^2 x - 1 + \sin^2 x}{\cos x(1+\sin x)} \cdots\]</p>
<p>After careful factoring: the result simplifies to \(\sec x + \tan x + C\). Answer: <strong>(D)</strong></p>
Correct Answer: D