SOLUTION 16: $\ \$ Integrate $\displaystyle{ \int { \sqrt{1-x} \cdot \sqrt{x+3} } \ dx }$. Begin with the power substitution. $$1-x = u^{2}$$ so that $$x = 1-u^2$$ and $$dx = -2u \ du$$ Substitute into the original problem, replacing all forms of $x$, getting $$\displaystyle { \int \sqrt{1-x} \cdot \sqrt{x+3} \ dx = \int \sqrt{u^{2}} \cdot \sqrt{(1-u^{2})+3} \cdot (-2u) \ du}$$ $$= \displaystyle { \int u \cdot \sqrt{4-u^{2}} \cdot (-2u) \ du}$$ $$= -2 \displaystyle { \int u^2 \cdot \sqrt{4-u^{2}} \ du}$$ Now use the trig substitution $$u = 2 \sin \theta$$ so that $$du = 2 \cos \theta \ d \theta$$ Substitute into the original problem, replacing all forms of $u$, getting $$-2 \displaystyle { \int u^2 \cdot \sqrt{4-u^{2}} \ du} = -2 \displaystyle { \int (2 \sin \theta)^2 \cdot \sqrt{4-4 \sin^{2} \theta} \cdot 2 \cos \theta \ d \theta }$$ $$= -4 \displaystyle { \int 4 \sin^2 \theta \cdot \sqrt{4 (1- \sin^{2} \theta ) } \cdot \cos \theta \ d \theta }$$ $$= -16 \displaystyle { \int \sin^2 \theta \cdot 2 \sqrt{\cos^{2} \theta} \cdot \cos \theta \ d \theta }$$ $$= -32 \displaystyle { \int \sin^2 \theta \cdot \cos \theta \cdot \cos \theta \ d \theta }$$ $$= -32 \displaystyle { \int \sin^2 \theta \cos^2 \theta \ d \theta }$$ $$= -32 \displaystyle { \int (\sin \theta \cos \theta)^2 \ d \theta }$$ (Recall that $\sin 2 \theta = 2 \sin \theta \cos \theta$ so that $\sin \theta \cos \theta = \displaystyle{ \frac {1}{2} } \sin 2 \theta$ .) $$= -32 \displaystyle { \int \Big(\frac {1}{2} \sin 2 \theta \Big)^2 \ d \theta }$$ $$= -32 \displaystyle { \int \frac {1}{4} \sin^2 2 \theta \ d \theta }$$ (Recall that $\cos 2 \beta = 1 - 2 \sin^2 \beta$ so that $\sin^2 \beta = \displaystyle{ \frac {1}{2} } (1-\cos 2 \beta)$ .) $$= -8 \displaystyle { \int { \frac {1}{2} } (1-\cos 2 (2 \theta )) \ d \theta }$$ $$= -4 \displaystyle { \int (1-\cos 4 \theta ) \ d \theta }$$ $$= \displaystyle { -4 \Big( \theta - \frac {1}{4 } \sin 4 \theta \Big) + C }$$ $$= \displaystyle { -4 \theta + \sin 4 \theta + C }$$ (Recall that $\sin 2 \beta = 2 \sin \beta \cos \beta$ .) $$= \displaystyle { -4 \theta + \sin 2(2 \theta) + C }$$ $$= \displaystyle { -4 \theta + 2 (\sin 2 \theta )( \cos 2 \theta ) + C }$$ (Recall that $\cos 2 \theta = \cos^2 \theta - \sin^2 \theta$ .) $$= \displaystyle { -4 \theta + 2 ( 2 \sin \theta \cos \theta )(\cos^2 \theta - \sin^2 \theta ) + C }$$ $$= \displaystyle { -4 \theta + 4 \sin \theta \cos^3 \theta - 4 \sin^3 \theta \cos \theta + C }$$ $\Big($ We need to write our final answer in terms of $x$.

Since $$u = 2 \sin \theta$$ it follows that $$\sin \theta = \frac{ u }{ 2 } = \displaystyle{ opposite \over hypotenuse }$$ and $$\theta = \arctan \frac{u}{2}$$ and from the Pythagorean Theorem that $$(adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(adjacent)^2 + (u)^2 = (2)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{4-u^2} \ \ \longrightarrow$$ $$\cos \theta = \displaystyle{ adjacent \over hypotenuse} = \frac{ \sqrt{4- u^2} }{ 2 } . \Bigg)$$ $$= \displaystyle { -4 \arctan \frac{u}{2} + 4 \frac{ u }{ 2 } \Bigg( \frac{ \sqrt{4- u^2} }{ 2 } \Bigg)^3 - 4 \Bigg( \frac{ u }{ 2 } \Bigg)^3 \frac{ \sqrt{4- u^2} }{ 2 } + C }$$ (Now use the fact that $u^2=1-x$ and $u= \sqrt{1-x}$.) $$= \displaystyle { -4 \arctan \frac{\sqrt{1-x} }{2} + 4 \frac{ \sqrt{1-x} }{ 2 } \Bigg( \frac{ \sqrt{4- (1-x )} }{ 2 } \Bigg)^3 - 4 \Bigg( \frac{ \sqrt{1-x} }{ 2 } \Bigg)^3 \frac{ \sqrt{4- (1-x )} }{ 2 } + C }$$ $$= \displaystyle { -4 \arctan \frac{\sqrt{1-x} }{2} + \frac{1}{4} \sqrt{1-x}(x+3)^{3/2} - \frac{1}{4} (1-x)^{3/2} \sqrt{x+3} + C }$$