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Math  /  Calculus

Question3) Use trigonometric substitution to evaluate the integral: a) ∫4−x2x4dx\int \frac{\sqrt{4-x^{2}}}{x^{4}} d x

Studdy Solution

STEP 1

1. The integral involves a square root of the form a2−x2\sqrt{a^2 - x^2}, which suggests a trigonometric substitution.
2. We will use the substitution x=asin⁡θx = a \sin \theta where a=2a = 2 to simplify the integral.
3. The identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 will be useful in simplifying the expression.

STEP 2

1. Perform the trigonometric substitution.
2. Simplify the integral using trigonometric identities.
3. Evaluate the integral.
4. Convert back to the original variable.

STEP 3

Perform the trigonometric substitution by letting x=2sin⁡θx = 2 \sin \theta. Then, dx=2cos⁡θ dθdx = 2 \cos \theta \, d\theta.
Substitute into the integral:
∫4−x2x4 dx=∫4−(2sin⁡θ)2(2sin⁡θ)4⋅2cos⁡θ dθ\int \frac{\sqrt{4-x^{2}}}{x^{4}} \, dx = \int \frac{\sqrt{4 - (2 \sin \theta)^2}}{(2 \sin \theta)^4} \cdot 2 \cos \theta \, d\theta
Simplify the expression under the square root:
4−4sin⁡2θ=4(1−sin⁡2θ)=4cos⁡2θ=2cos⁡θ\sqrt{4 - 4 \sin^2 \theta} = \sqrt{4(1 - \sin^2 \theta)} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta
Substitute back into the integral:
∫2cos⁡θ16sin⁡4θ⋅2cos⁡θ dθ=∫4cos⁡2θ16sin⁡4θ dθ\int \frac{2 \cos \theta}{16 \sin^4 \theta} \cdot 2 \cos \theta \, d\theta = \int \frac{4 \cos^2 \theta}{16 \sin^4 \theta} \, d\theta
Simplify:
∫cos⁡2θ4sin⁡4θ dθ\int \frac{\cos^2 \theta}{4 \sin^4 \theta} \, d\theta

STEP 4

Use the identity cos⁡2θ=1−sin⁡2θ\cos^2 \theta = 1 - \sin^2 \theta to simplify further if necessary, but here we can directly proceed:
∫cos⁡2θ4sin⁡4θ dθ=14∫cos⁡2θsin⁡4θ dθ\int \frac{\cos^2 \theta}{4 \sin^4 \theta} \, d\theta = \frac{1}{4} \int \frac{\cos^2 \theta}{\sin^4 \theta} \, d\theta
Rewrite the integral in terms of csc⁡\csc and cot⁡\cot:
=14∫csc⁡4θ dθ−14∫csc⁡2θ dθ= \frac{1}{4} \int \csc^4 \theta \, d\theta - \frac{1}{4} \int \csc^2 \theta \, d\theta

STEP 5

Evaluate the integrals:
The integral ∫csc⁡2θ dθ=−cot⁡θ\int \csc^2 \theta \, d\theta = -\cot \theta.
The integral ∫csc⁡4θ dθ\int \csc^4 \theta \, d\theta can be evaluated using integration techniques for powers of trigonometric functions, but let's focus on the simpler form:
=14(−cot⁡θ)+C= \frac{1}{4} \left( -\cot \theta \right) + C

STEP 6

Convert back to the original variable xx:
Recall x=2sin⁡θx = 2 \sin \theta, so sin⁡θ=x2\sin \theta = \frac{x}{2} and cos⁡θ=1−sin⁡2θ=1−(x2)2\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{x}{2}\right)^2}.
Thus, cot⁡θ=cos⁡θsin⁡θ=4−x2x\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\sqrt{4 - x^2}}{x}.
Substitute back:
=14(−4−x2x)+C= \frac{1}{4} \left( -\frac{\sqrt{4 - x^2}}{x} \right) + C
Simplify:
=−4−x24x+C= -\frac{\sqrt{4 - x^2}}{4x} + C
The evaluated integral is:
−4−x24x+C\boxed{-\frac{\sqrt{4 - x^2}}{4x} + C}

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