Question

i need help badly, these are trigonometric substitutions. the question is integral dx/8+2x^2 from 0 to 2

Answer 1

\[\int_0^2\frac{dx}{8+2x^2}\]
Setting \(x=2\tan t\), we have \(dx=2\sec^2t\,dt\).
\[\int_0^{\pi/4}\frac{2\sec^2t}{8+2(2\tan t)^2}\,dt=\frac{1}{4}\int_0^{\pi/4}\frac{\sec^2t}{1+\tan^2t}\,dt=\frac{1}{4}\int_0^{\pi/4}dt\]

Answer 2

how did you get 0 and pi/4

Answer 3

It comes with the substitution. Setting \(x=2\tan t\), we also have \(t=\arctan\dfrac{x}{2}\).

When the lower limit is \(x=0\), you have \(t=\arctan\dfrac{0}{2}=\arctan0=0\).

When the upper limit is \(x=2\), you have \(t=\arctan\dfrac{2}{2}=\arctan1=\dfrac{\pi}{4}\).

Answer 4

so then a does not equal 8, instead a =
\[2x ^{2}\]

Answer 5

What is \(a\)? Are you referring to a table or something?

Answer 6

its how my teacher explained it, a=2x^2, then x=2tant, then dx would be 2sec^2tdt, then the sqaure root of X^2-3 would = square root of 8

Answer 7

Let's consider a more general case. Suppose you're given the (indefinite) integral
\[\int\frac{dx}{a+bx^2}\]
(with \(a,b\not=0\))

The substitution we would use here would take on the form \(x=\sqrt{\dfrac{a}{b}}\tan t\). The reason for this is made clear when we replace accordingly:
\[\int\frac{dx}{a+b\left(\sqrt{\frac{a}{b}}\tan t\right)^2}=\int\frac{dx}{a+b\left(\frac{a}{b}\tan^2 t\right)}=\int\frac{dx}{a+a\tan^2 t}=\frac{1}{a}\int\frac{dx}{1+\tan^2t}\]
We also have to replace the differential, which would take the form \(dx=\sqrt{\dfrac{a}{b}}\sec^2t\,dt\), so the integral is
\[\frac{1}{a}\int\frac{\sqrt{\frac{a}{b}}\sec^2t}{1+\tan^2t}\,dt=\frac{1}{\sqrt {ab}}\int\frac{\sec^2t}{1+\tan^2t}\,dt=\frac{1}{\sqrt {ab}}\int dt\]
The point is that your substitution should reflect the relationship needed for you to be able to remove the constant coefficients in the denominator. We want to change \(b\) in \(bx^2\) to \(a\) in \(a\tan^2t\), and so we replace \(x\) with \(\sqrt{\dfrac{a}{b}}\tan t\).