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 $x$.

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 $$(adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(adjacent)^2 + (x+1)^2 = (2)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{4-(x+1)^2} = \sqrt{-x^2-2x+3} \ \ \longrightarrow$$ $$\cos \theta = \displaystyle{ adjacent \over hypotenuse} = \frac{ \sqrt{-x^2-2x+3} }{ 2 } . \Bigg)$$ $$= \displaystyle { 2 \arcsin \frac{x+1}{2} + 2 \cdot \frac{ x+1 }{ 2 } \frac{ \sqrt{-x^2-2x+3} }{ 2 } }$$ $$= \displaystyle { 2 \arcsin \frac{x+1}{2} + \frac{ x+1 }{ 2 } \sqrt{-x^2-2x+3} }$$