Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# How would i go about solving with trig sub?

Hi, ive been stumped by this integral as it does not contain a root, can anyone help?

=∫(-x^2+9) dx

thanks.

Relevance
• 1 month ago

∫(-x^2+9) dx

=- 1/3x^3 + 9x + C answer//

• ?
Lv 7
1 month ago

I=S(9-x^2)dx

=>

=>

I=27[sinA-sin^3(A)/3]+C

=>

I=27sinA[1-sin^2(A)/3]+C

=>

I=27sinAcos^2(A)/3+C

If you want the result in terms of x, you

should integrate directly with respect

to x getting I=9x-(x^3)/3+C.

• 1 month ago

Hi, I'm sure you know how to solve it by the easiest method, although it is likely that you were asked to solve it with trig sub.

So to solve it by trig sub in this case you can use a mathematical artifice. you artificially add a root and thus use trig sub normally.

• alex
Lv 7
1 month ago

∫ax^n dx = (a/(n+1))x^(n+1)+C , where n≠ -1

• ?
Lv 7
1 month ago

So easy it's difficult!

Integral of -x^2 dx is -x^3/3,

integral of 9 dx is 9x,

Don't do "trig sub" !!!

• ?
Lv 7
1 month ago

You would directly integrate this. You would not use trig sub to do this. Although you can, you will just really really overcomplicate matters.

• 1 month ago

= - ∫(x^2-9) dx

= -∫x^2dx + ∫9 dx

= -x^3/3 + 9x +C

stop looking for complications!

• 1 month ago

integral(-x^2 + 9) dx = 9 x - x^3/3 + constant