$\displaystyle \int \csc(x)\left(\sin(x)+\cot(x)\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}\left(\sin(x)+\frac{\cos(x)}{\sin(x)}\right)dx\\\\\displaystyle \int \frac{1}{\sin(x)}*\sin(x)+\frac{1}{\sin(x)}*\frac{\cos(x)}{\sin(x)}dx\\\\\displaystyle \int 1+\frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle \int 1 dx+\int \frac{\cos(x)}{\sin^2(x)}dx\\\\\displaystyle x+C_1+\int \frac{1}{u^2}du \ \text{ ... where } u = \sin(x)\\\\\displaystyle x+C_1+\int u^{-2}du\\\\$

Part 2

$\displaystyle x+C_1+\frac{1}{1+(-2)}u^{-2+1}+C_2\\\\\displaystyle x+C_1+\frac{1}{-1}u^{-1}+C_2\\\\\displaystyle x+C_1-u^{-1}+C_2\\\\\displaystyle x+C_1-\frac{1}{u}+C_2\\\\\displaystyle x+C_1-\frac{1}{\sin(x)}+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C_1+C_2\\\\\displaystyle x-\frac{1}{\sin(x)}+C\\\\\displaystyle x-\csc(x)+C\\\\$

<h3>Answer: x - csc(x) + C</h3>

Don't forget about the plus C constant

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Problem 26

Fortunately, there aren't as many steps for this problem.

$\displaystyle \int \frac{dy}{\csc(y)}\\\\\displaystyle \int \frac{1}{\csc(y)}dy\\\\\displaystyle \int \sin(y)dy\\\\\displaystyle -\cos(y)+C\\\\$

<h3>Answer: -cos(y) + C</h3>

6 0

3 years ago