Trig Sub Solution 16

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SOLUTION 16:

$\ \$ Integrate

$\displaystyle{ \int { \sqrt{1-x} \cdot \sqrt{x+3} } \ dx }$ . Begin by rewriting the integrand.

$\displaystyle{ \int { \sqrt{1-x} \cdot \sqrt{x+3} } \ dx } = \displaystyle{ \int { \sqrt{-x^2-2x+3} } \ dx }$

$= \displaystyle{ \int { \sqrt{-(x^2+2x)+3} } \ dx }$ (Now complete the square.)

$= \displaystyle{ \int { \sqrt{-(x^2+2x+1)+3+1} } \ dx }$

$= \displaystyle{ \int { \sqrt{4-(x+1)^2} } \ dx }$ Now use the trig substitution

$x+1 = 2 \sin \theta$ so that

$dx = 2 \cos \theta \ d \theta$ Substitute into the original problem, replacing all forms of

$x$ , getting

$\displaystyle{ \int { \sqrt{4-(x+1)^2} } \ dx } = \displaystyle{ \int \sqrt{ 4- (2 \sin \theta)^2 } \cdot 2 \cos \theta \ d \theta }$

$= 2 \displaystyle{ \int \sqrt{ 4- 4 \sin^2 \theta } \cos \theta \ d \theta }$

$= 2 \displaystyle{ \int \sqrt{ 4(1- \sin^2 \theta) } \cos \theta \ d \theta }$

$= 2 \displaystyle{ \int \sqrt{ 4 \cos^2 \theta } \cos \theta \ d \theta }$

$= 2 \displaystyle{ \int 2 \cos \theta \cdot \cos \theta \ d \theta }$

$= 4 \displaystyle{ \int \cos^2 \theta \ d \theta }$ (Recall that

$\cos^2 \theta = (1/2)(1+ \cos 2 \theta)$ .)

$= 4 \displaystyle{ \int \frac{1}{2}(1+ \cos 2 \theta) \ d \theta }$

$= 2 \displaystyle{ \int (1+ \cos 2 \theta) \ d \theta }$

$= 2 \displaystyle{ ( \theta + \frac{1}{2} \sin 2 \theta ) } + C$ (Recall that

$\sin 2 \theta = 2 \sin \theta \cos \theta$ .)

$= 2 \displaystyle{ \Big( \theta + \frac{1}{2} (2 \sin \theta \cos \theta) \Big) } + C$

$= 2 \displaystyle{ ( \theta + \sin \theta \cos \theta ) } + C$

$= \displaystyle{ 2\theta + 2\sin \theta \cos \theta } + C$

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

$x$ .

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Since

$x+1 = 2 \sin \theta$ it follows that

$\sin \theta = \frac{ x+1 }{ 2 } = \displaystyle{ opposite \over hypotenuse }$ and

$\theta = \arcsin \frac{x+1}{2}$ and from the Pythagorean Theorem that

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

$(adjacent)^2 + (x+1)^2 = (2)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{4-(x+1)^2} = \sqrt{-x^2-2x+3} \ \ \longrightarrow$

$\cos \theta = \displaystyle{ adjacent \over hypotenuse} = \frac{ \sqrt{-x^2-2x+3} }{ 2 } . \Bigg)$

$= \displaystyle { 2 \arcsin \frac{x+1}{2} + 2 \cdot \frac{ x+1 }{ 2 } \frac{ \sqrt{-x^2-2x+3} }{ 2 } }$

$= \displaystyle { 2 \arcsin \frac{x+1}{2} + \frac{ x+1 }{ 2 } \sqrt{-x^2-2x+3} }$

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