How do you integrate ∫x2√16−x2 by trigonometric substitution?
2 Answers
Explanation:
Let's use the substitution
x2=16sin2θ √16−x2=√16−16sin2θ=4√1−sin2θ=4cosθ dx=4cosθdθ
Now, let's substitute these back into our integral:
To solve this, I like to use the identity
We need to get our final answer back in terms of
Recall we started with
sinθ=x4 cosθ=√1−sin2θ=√1−x216=√16−x24 θ=sin−1(x4)
And finally:
Explanation:
Let
For
then:
Use now the trigonometric identity:
to have:
and based on the linearity of the integral:
and undoing the substitution: