Indefinite Integration
Integration by Substitution
Grade 12
Question:
<p>\(\displaystyle\int\frac{\sec^2 x}{(\sec x+\tan x)^{9/2}}\,dx\) equals (where \(C\) is a constant of integration)</p>
<li>\(-\dfrac{1}{7}(\sec x+\tan x)^{-7/2}+\dfrac{1}{11}(\sec x+\tan x)^{-11/2}+C\)</li>
<li>\(\dfrac{2}{7}(\sec x+\tan x)^{7/2}+C\)</li>
<li>\(-\dfrac{2}{11}(\sec x+\tan x)^{-11/2}+C\)</li>
<li>\(\dfrac{1}{9}(\sec x+\tan x)^{-9/2}+C\)</li>
Step-by-Step Solution
Key Concept: Let t = secx+tanx. Then dt = (secx \cdot tanx+sec^2x)dx = secx \cdot t dx, so secx \cdot dx = dt/t. Express sec^2x in terms of t using sec^2x-tan^2x=1 ↔ (secx-tanx)=1/t.
<p><strong>Substitution:</strong> Let $t=\sec x+\tan x\Rightarrow \sec x-\tan x=\frac1t$.</p>
<p>So $\sec x=\frac12\!\left(t+\frac1t\right)$, $\tan x=\frac12\!\left(t-\frac1t\right)$.</p>
<p>$dt=\sec x(\sec x+\tan x)\,dx = \sec x\cdot t\,dx\Rightarrow \sec x\,dx=\frac{dt}{t}$.</p>
<p>$$\int\frac{\sec^2 x}{t^{9/2}}\,dx = \int\frac{\sec x}{t^{9/2}}\cdot(\sec x\,dx) = \int\frac{\frac12(t+t^{-1})}{t^{9/2}}\cdot\frac{dt}{t}$$</p>
<p>$$= \frac12\int\left(t^{-9/2-1+1}+t^{-1-9/2-1}\right)dt = \frac12\int\left(t^{-9/2}+t^{-13/2}\right)dt$$</p>
<p>$$= \frac12\!\left[\frac{t^{-7/2}}{-7/2}+\frac{t^{-11/2}}{-11/2}\right]+C = -\frac{1}{7}t^{-7/2}-\frac{1}{11}t^{-11/2}+C$$</p>
<p>Answer: <strong>(A)</strong></p>
Correct Answer: A