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# Evaluate the integrals. Remember to include a constant of integration with the indefinite integrals. Your answers may appear different from those in the Answers sec...

## Question

###### Evaluate the integrals. Remember to include a constant of integration with the indefinite integrals. Your answers may appear different from those in the Answers section but may still be correct. For example, evaluating $I=\int \sin x \cos x \, d x$ using the substitution $u=\sin x$ leads to the answer $I=\frac{1}{2} \sin ^{2} x+C ;$ using $u=\cos x$ leads to $I=-\frac{1}{2} \cos ^{2} x+C$ and rewriting $I=\frac{1}{2} \int \sin (2 x) d x$ leads to \,$I=-\frac{1}{4} \cos (2 x)+C$ These answers are

Evaluate the integrals. Remember to include a constant of integration with the indefinite integrals. Your answers may appear different from those in the Answers section but may still be correct. For example, evaluating $I=\int \sin x \cos x \, d x$ using the substitution $u=\sin x$ leads to the answer $I=\frac{1}{2} \sin ^{2} x+C ;$ using $u=\cos x$ leads to $I=-\frac{1}{2} \cos ^{2} x+C$ and rewriting $I=\frac{1}{2} \int \sin (2 x) d x$ leads to \,$I=-\frac{1}{4} \cos (2 x)+C$ These answers are all equal except for different choices for the constant of integration $C$ : $\sin ^{2} x=-\frac{1}{2} \cos ^{2}+\frac{1}{2}=-\frac{1}{4} \cos (2 x)+\frac{1}{4}$ $\sum_{i=1}^{100}$ You can always check your own answer to an indefinite integral by differentiating it to get back to the integrand. This is often easier than comparing your answer with the answer in the back of the book. You may find integrals that you can't do, but you should not make mistakes in those you can do because the answer is so easily checked. (This is a good thing to remember during tests and exams.) $$\int \frac{\sin ^{2} x}{\cos ^{4} x} d x$$

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