Indefinite Integration
Integration using substitution
Grade 12
Question:
<p>Let \(I = \displaystyle\int (\sin 4x)\, e^{\tan^2 x}\,dx\). If \(I = A^{10}\) when evaluated appropriately, find the value of \(\dfrac{A^{10}}{10}\) (in decimal). Given that the answer is of the form \(A = -2\cos^4 x \cdot e^{\tan^2 x} + C\), and \(\dfrac{A^{10}}{10} = \dfrac{1024}{100}\).</p>
Step-by-Step Solution
Key Concept: Recognize that sin(4x) = 2sin(2x)cos(2x) = 4sin(x)cos(x)(cos²x - sin²x), and use substitution u = tan²x with du = 2tan(x)sec²x dx to transform the integrand into a form matching d/dx[e^(tan²x)].
<p><strong>Step 1:</strong> Recognize that the antiderivative is given as A = -2cos⁴x·e^(tan²x) + C. We need to verify this is correct by differentiating.</p><p><strong>Step 2:</strong> Differentiate A = -2cos⁴x·e^(tan²x):</p><p>dA/dx = -2[4cos³x·(-sin x)·e^(tan²x) + cos⁴x·e^(tan²x)·2tan x·sec²x]</p><p>= -2e^(tan²x)[−4cos³x·sin x + 2cos⁴x·tan x·sec²x]</p><p>= -2e^(tan²x)[−4cos³x·sin x + 2cos⁴x·(sin x/cos x)·(1/cos²x)]</p><p>= -2e^(tan²x)[−4cos³x·sin x + 2cos x·sin x]</p><p>= -2e^(tan²x)·sin x[−4cos³x + 2cos x]</p><p>= 2e^(tan²x)·sin x·2cos x[2cos²x - 1]</p><p>= 4sin x·cos x·e^(tan²x)[2cos²x - 1]</p><p>= 2sin(2x)·e^(tan²x)·cos(2x) = sin(4x)·e^(tan²x) ✓</p><p><strong>Step 3:</strong> Given A = -2cos⁴x·e^(tan²x) (ignoring constant), we have A¹⁰ = (-2)¹⁰·(cos⁴x)¹⁰·(e^(tan²x))¹⁰ = 1024·cos⁴⁰x·e^(10tan²x)</p><p><strong>Step 4:</strong> From the problem statement, A¹⁰/10 = 1024/100 = 10.24</p><p>∴ Answer: <strong>10.24</strong></p>
Correct Answer: 10.24