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Integral of 7/(1+x^2) dx?
Can't figure out how to do this problem. Plugged it in my calculator to get the answer, but I'd like to know how to do it without a calculator...
Integral from 1 to 2 of 7/(1+x^2) dx
Ahh I totally forgot about arctan. Thanks guys!
3 Answers
- 8 years agoFavorite Answer
Integral of 1/(1+x^2) would be inverse tangent(x).
so the integral of 7/(1+x^2) would be 7*(inverse tangent(x)).
from 1 to 2:
7*(inverse tan(2) - inverse tan(1)) = 7*(63-45) = 7*18 = 126 Degree = 0.7pi
- θ βяιαη θLv 78 years ago
â« 1/(1 + x^2) dx = arctan(x) + C is a standard integral (you can derive this via the trigonometric substitution u = tan(x) ==> du = sec^2(x) dx if needed), so we have:
â« 7/(1 + x^2) dx (from x=1 to 2) = 7 â« 1/(1 + x^2) dx (from x=1 to 2)
= 7arctan(x) (evaluated from x=1 to 2), by the FTC part 2 and the above integral
= 7[arctan(2) - arctan(1)]
= 7arctan(2) - 7Ï/4, by expanding (note: tan(Ï/4) = 1 ==> arctan(1) = Ï/4).
I hope this helps!
- Shaun DizzleLv 78 years ago
Trig substitution.
x = tanø
dx = sec^2ø dø
7 ⫠(sec^2ø)dø / (1 + tan^2ø)
Note that 1 + tan^2ø = sec^2ø by the pythagorean identity:
7 ⫠sec^2ø dø / sec^2ø
7 ⫠1 dø
7ø
But we want it in terms of x.
So look back at the original substitution:
x = tanø
That means that ø = tan^-1(x)
Therefore:
7ø
is the same as:
7tan^-1(x)
Now evaluate from 1 to 2:
7[tan^-1(2) - tan^-1(1)]
