Solve Trig Derivative: -8\int sec^3(\theta)+2sec^2(\theta)+sec(\theta)d\theta

[tangur]

$$\int \frac {x^2}{\sqrt{4x-x^2}}dx$$

I just want to be sure I'm right on this, complete the square first of all so you get $$-\int \frac {x^2}{\sqrt{(x-2)^2-4}}dx$$ let $$u=x-2$$ thus $$-\int \frac {(x+2)^2}{\sqrt{u^2-4}}dx$$then let$$u=2sec(\theta)$$
hence integral becomes $$-8\int sec^3(\theta)+2sec^2(\theta)+sec(\theta)d\theta$$

and then solve.

Thanks

 

[Parth Dave]

I got -4 as the constant

 

[ReyChiquito]

the first integral is always positive, the second one doesnt. You can't just invert the sign on the radical and take one minus out... the sign on the radical should stay, wich, if you think about it, might make things easier

 

[tangur]

true, since its under a sqrt root, you keep it under, but you put the whole expression under square root in parenthesis and take out the minus one, and then go x^2-4x. makes sense

thx guys