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# Week 2 Thursday Problems
## Reading.
Chapter 7.2. And preview chapter 7.3, and more practice in 7.5.
Keep going keep going ! ! ! You can do it !
## Problems.
1. Use the methods described today to deal with the following integrals involving trigonometric functions.
1. $\displaystyle\int \sin^3(x)\cos^2(x)dx$
2. $\displaystyle\int \cos^9(x)\sin^5(x)dx$
3. $\displaystyle\int \cos^2(x)dx$
4. $\displaystyle\int \sin^2(x)\cos^2(x)dx$
5. $\displaystyle\int\tan(x)\sec^3(x)dx$
6. $\displaystyle\int \sin(x)\sec^5(x)dx$ .... do you see a tangent here?
7. $\displaystyle\int \sin(5x)\sin(2x)dx$ ... Hint: Use the product-to-sum trigonometric identities.
8. $\displaystyle\int\sin(8x)\cos(5x)dx$
2. Use Weierstrass substitution to find the following:
1. $\displaystyle\int \frac{dx}{1-\cos(x)}$
2. $\displaystyle\int \frac{dx}{3\sin(x)-4\cos(x)}$
3. Try to derive $\displaystyle\int \sec(x)dx$ in the following way: (as outlined in text):
- Multiply the integrand by $1=\frac{\tan(x)+\sec(x)}{\tan(x)+\sec(x)}$
- Then do a substitution $u=\tan(x)+\sec(x)$
4. Similarly, try to derive $\displaystyle\int \csc(x)dx$ in the following way::
- Multiply the integrand by $1=\frac{\csc(x)+\cot(x)}{\csc(x)+\cot(x)}$
- Then do a substitution $u=\csc(x)+\cot(x)$
5. (Optional but good for your development) Let us try to figure out why if we set $t=\tan(\frac{x}{2})$ in Weierstrass substitution, we get $\cos(x)= \frac{t^2-1}{t^2+1}$, $\sin(x)=\frac{2t}{t^2+1}$, and $dx= \frac{2}{1+x^2} dt$.
1. First, draw a right triangle where one of the angle is $\frac{x}{2}$. If $t=\tan(\frac{x}{2})$, label the sides of the triangle in terms of $t$.
2. Using your triangle, show that $\cos\left( \frac{x}{2} \right)= \frac{1}{\sqrt{1+t^2}}$ and $\sin\left( \frac{x}{2} \right)=\frac{t}{\sqrt{1+t^2}}$.
3. Use double angle formula and/or Pythagorean identity to show $\cos(x) = \frac{t^2-1}{t^2+1}$ and $\sin(x) = \frac{2t}{t^2+1}$.
4. Finally, if $t=\tan(\frac{x}{2})$, show $dx = \frac{2}{1+t^2}dt$.
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