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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.

### 8 Answers

Relevance
• ∫(-x^2+9) dx

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

• I=S(9-x^2)dx

Let x=3sinA, then dx=3cosAdA

I=9S[1-sin^2(A)](3cosAdA)

=>

I=27[S cosAdA-S sin^2(A)d(sinA)]

=>

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.

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

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• So easy it's difficult!

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

integral of 9 dx is 9x,

answer to your problem is 9x - (1/3)x^3 + C.

Don't do "trig sub" !!!

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

• = - ∫(x^2-9) dx

= -∫x^2dx + ∫9 dx

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

stop looking for complications!

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

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