# Chapter 4 - Integration - 4.3 Integration By Substitution - Exercises Set 4.3 - Page 286: 4

$${\text{part a) }}\frac{2}{{3\pi }}{\left( {\sin \pi \theta } \right)^{3/2}} + C{\text{ part b)}}\frac{5}{9}{\left( {{x^2} + 7x + 3} \right)^{9/5}} + C$$

#### Work Step by Step

\eqalign{ & {\text{part a}} \cr & \cr & \int {\sqrt {\sin \pi \theta } } \cos \pi \theta d\theta \cr & {\text{substitute }}u = \sin \pi \theta ,{\text{ }}du = \pi \cos \pi \theta d\theta \cr & = \int {\sqrt u } \left( {\frac{1}{\pi }} \right)du \cr & = \frac{1}{\pi }\int {\sqrt u } du \cr & = \frac{1}{\pi }\int {{u^{1/2}}} du \cr & {\text{find antiderivative}} \cr & = \frac{1}{\pi }\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right) + C \cr & = \frac{2}{{3\pi }}{u^{3/2}} + C \cr & {\text{write in terms of }}\theta \cr & = \frac{2}{{3\pi }}{\left( {\sin \pi \theta } \right)^{3/2}} + C \cr & \cr & {\text{part b}} \cr & \int {\left( {2x + 7} \right)} {\left( {{x^2} + 7x + 3} \right)^{4/5}}dx \cr & {\text{substitute }}u = {x^2} + 7x + 3,{\text{ }}du = \left( {2x + 7} \right)dx \cr & \int {\left( {2x + 7} \right)} {\left( {{x^2} + 7x + 3} \right)^{4/5}}dx = \int {{u^{4/5}}} du \cr & {\text{find antiderivative}} \cr & = \frac{{{u^{9/5}}}}{{9/5}} + C \cr & = \frac{5}{9}{u^{9/5}} + C \cr & {\text{write in terms of }}x \cr & = \frac{5}{9}{\left( {{x^2} + 7x + 3} \right)^{9/5}} + C \cr}

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