(I tried numbering the lines for reference but it may be even more confusing)

A+B-D=0\\

-4A+4B+C=0\\

4A+4B+D=1\)

Third equation edited

This is where I got stuck. I keep trying to solve this system of equations and I keep getting different answers.

- I was given the following problem:
- \(\displaystyle \int\frac{1}{(\cos(x))\left(\sin^2(x)+4\right)}\ dx\)
- I am not quite sure how to go about solving it. This is what I've done:
- \(\displaystyle \int\frac{1}{(\cos(x))\left(\sin^2(x)+4\right)}\ dx \\ =\int\frac{\cos(x)}{(1-\sin^2(x))\left(\sin^2(x)+4\right)}\ dx\\u=\sin(x)\ \ \ \ \ \ \frac{du}{dx}=\cos(x)\ \ \ \ \ \ du=\cos(x)dx\)
- Plugging in the substitution:
- \(\displaystyle \int\frac{1}{(1+u)(1-u)(u^2+4)}\ dx\)
- This looks like something for partial fractions: \(\displaystyle \frac{A}{1+u}+\frac{B}{1-u}+\frac{Cu+D}{u^2+4}\)
- \(\displaystyle \frac{(A)(-u^3+u^2-4u+4)+B(u^3+u^2+4u+4)+(Cu+D)(-u^2+1)}{(1+u)(1-u)(u^2+4)}\\\frac{(-A+B-C)(u^3)+(A+B-D)(u^2)+(-4A+4B+C)(u)+4A+4B+D}{(1+u)(1-u)(u^2+4)}\)
- This now gives me:

A+B-D=0\\

-4A+4B+C=0\\

4A+4B+D=1\)

Third equation edited

This is where I got stuck. I keep trying to solve this system of equations and I keep getting different answers.

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