Trigonometric Integration, Definite integral.

Pinedas42
Messages
11
Reaction score
0

Homework Statement


Use 2 or more substitutions to find the following integrals
hint : begin with u=cosx


Homework Equations


Integral 0--->pi/2 (cosxsinx)/swrt(cos2x+49 dx


The Attempt at a Solution


I'm still a little fuzzy on using multiple substitutions. From what I've read on the text and previous easier equations, it just means that there are multiple u=(something) that can work. Is that right?

so I tried u=cosx
du=-sinxdx

giving me


-1 * Integral 0--->pi/2 u/sqrt(u2+49) du


it's here that I am brickwalling. I really want to know how it works, so if you wouldn't mind a step by step process, I'd appreciate greatly.
 
Physics news on Phys.org
Pinedas42 said:

Homework Statement


Use 2 or more substitutions to find the following integrals
hint : begin with u=cosx

Homework Equations


Integral 0--->pi/2 (cosxsinx)/swrt(cos2x+49 dx

The Attempt at a Solution


I'm still a little fuzzy on using multiple substitutions. From what I've read on the text and previous easier equations, it just means that there are multiple u=(something) that can work. Is that right?

so I tried u=cosx
du=-sinxdx

giving me-1 * Integral 0--->pi/2 u/sqrt(u2+49) duit's here that I am brickwalling. I really want to know how it works, so if you wouldn't mind a step by step process, I'd appreciate greatly.

You should either change the limits of integration, or do the corresponding indefinite integration.

You have \displaystyle \int \frac{u}{\sqrt{u^2+49}} du.

Can you see a substitution which might work with your result?

(I can see two, either of which looks helpful.)
 
So I beat at it until I solved it :D ( I don't give up dammit)

I put the integral into terms of u
so
u=cos(0)=1
u=cos(pi/2)=0so

integral 0-->1 u(u^2+49)^-1/2 du
I took the second sub of t=u^2+49
dt=2udu
to give
1/2 integral0-->1 (t)^-1/2 dt
1/2 * 2 (t)^1/2
giving the function
(u^2+49)^1/2 |0-->1
then using the fundamental theorem of calculus
[(1^2+49)^1/2]-[0^2+49]^1/2]
sqrt(50)-sqrt(49)
giving
-7+5sqrt(2)

:D

pretty stoked lol
 
Excellent! (and welcome to PF !)

Another (very nice) subst. would have been to let t = √(u2+49) .

Try it, you might like it.
 
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