Trig Sub Solution 6

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SOLUTION 6:

$\ \$ Integrate

$\displaystyle{ \int { \sqrt{1-x^2} \over x } \ dx }$ . Use the trig substitution

$x = \sin \theta$ so that

$dx = \cos \theta \ d \theta$ Substitute into the original problem, replacing all forms of

$x$ , getting

$\displaystyle { \int \frac{\sqrt{1-x^{2}}}{x} \ dx = \int \frac{\sqrt {1-\sin^{2} \theta}}{\sin \theta} \ \cos \theta \ d \theta }$

$= \displaystyle { \int \frac{\sqrt{\cos^{2} \theta}}{\sin \theta} \ \cos \ d \theta }$

$= \displaystyle { \int \frac{\cos \theta}{\sin \theta} \cos \theta \ d \theta }$

$= \displaystyle { \int \frac{\cos^{2} \theta}{\sin \theta} \ d \theta }$

$= \displaystyle { \int \frac{1-\sin^{2}\theta}{\sin \theta} \ d \theta }$

$= \displaystyle { \int (\frac{1}{\sin \theta} -\sin \theta) \ d \theta }$

$= \displaystyle { \int \Big(\csc \theta - \sin \theta \Big) \ d \theta }$ (Recall that

$\displaystyle { \int \csc \theta \ d \theta = \ln \Big|\csc \theta - \cot \theta \Big| + C }$ )

$= \displaystyle { \ln \Big|\csc \theta - \cot \theta \Big| + \cos \theta + C }$

$\Big($ We need to write our final answer in terms of

$x$ .

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Since

$x = \sin \theta$ it follows that

$\sin \theta = \displaystyle{ x \over 1 } = \displaystyle{ opposite \over hypotenuse }$ and from the Pythagorean Theorem that

$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$

$(adjacent)^2 + (x)^2 = (1)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{1-x^2} \ \ \longrightarrow$

$\cos \theta = \displaystyle{ adjacent \over hypotenuse }= \displaystyle{ \sqrt{1-x^2} \over 1 }$

$\cot \theta = \displaystyle{ adjacent \over opposite }= \displaystyle{ \sqrt{1-x^2} \over x }$ and

$\csc \theta = \displaystyle{ hypotenuse \over opposite }= \displaystyle{ 1 \over x } . \Big)$

$= \displaystyle { \ln \Big|\frac{1}{x} - \frac{\sqrt{1-x^{2}}}{x} \Big| + \sqrt{1-x^{2}} + C }$

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