Trig Substitution Problem w/ tan substitution

Burjam
Messages
52
Reaction score
1

Homework Statement



Under #3

Homework Equations



Trig identities

The Attempt at a Solution



The picture attached is my attempt. The square in the upper upper left is the problem and the one in the lower right is my solution. I'm seeing that I'm getting the wrong answer, but not how.

1479619739190-1775315430.jpg
 
Physics news on Phys.org
You should type in the problem at least.
The formula for csc (theta) is wrong, You miss a square root.
 
  • Like
Likes fresh_42
I miss a square root? Where does a square root come into play? None of the sides of the triangle should have a square root in them.
 
Burjam said:
I miss a square root? Where does a square root come into play? None of the sides of the triangle should have a square root in them.
Really? So the dimension of the sides of a right triangle is length, but that of the hypotenuse is length-squared?
Recall Pythagoras' Theorem.
 
Haha sorry I did this while really tired... my bad.
 
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