Possible webpage title: Solving Integrals Using the Substitution Method

roam
Messages
1,265
Reaction score
12
Evaluate the following integral using integration by substitution: http://img254.imageshack.us/img254/750/44900023cm4.png [/URL]




Here is my attempt:
Let x = sinu, then dx/du = cosu
Substituting gives, ∫1/(1-sin2u)×cosu du
= ∫1/(1-sin2u)×cosu du
= ∫cosu/√cos2u du
= ∫cosu/cosu du
= ∫1 du = u + c = sin-1x + c

Am I right? Did I get the right solution?

Regards,

 
Last edited by a moderator:
Physics news on Phys.org
You don't need Trig sub. for this.

\int\frac{xdx}{\sqrt{1-x^2}}

But we'll go with it!

x=\sin u
dx=\cos udu

\int\frac{\sin u \cos u du}{\sqrt{1-\sin^2 u}}
 
Try u=1-x^2 instead... Might be a bit easier.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top