∫ cosx / √(1+sin²x) dx : am I doing something wrong?

Question

∫ cosx / √(1+sin²x) dx

let

t = sinx

dt = cosx dx   →   dx = dt/cosx   →   cosx = dt/dx

∫ cosx / √(1+sin²x) dx = ∫ [1/√(1+sin²x)] * cosx dx = ∫ [1/√(1+t²)] dt/dx dx = ∫ 1/√(1+t²) dt

Ok, my answer has to contain [ln]

ln(1+t²)/2 + c = ln(1+sin²x)/2 + c = F(x)

---------

pi/2

∫ cosx / √(1+sin²x) dx = F(pi/2) - F(0)

0

= ln(1+sin²[pi/2])/2 - ln(1+sin²[0])/2 + c - c

= ln(1+1)/2 - ln(1+0)/2

= ln(2)/2 - 0

= ln(2)/2   {answer}

--------

HP50g GraphCalc evaluation: -ln(√2 - 1)

---------

alternatively?

∫ 1/√(1+t²) dt = arctan(t) + c = arctan(sinx) + c

over [0,pi/2] arctan(sinx) = pi/4

=========

integrate by parts a bit complex here

Where is my error? please show steps

Answer 1

Your error is in the last integration step.

∫ 1/√(1+t²) dt

= ln [t + √(1+t²)]

limit of x changes from 0 to π/2 to 0 to 1

=> definite integration

= ln (1 + √2)

This matches with the given answer - ln (√2 - 1)which is same as

ln [1/(√2 - 1)]

= ln[(√2 + 1)/(√2 - 1)*(√2 + 1)]

= ln (√2 + 1).

Answer 2

∫ [cosx /√(1+sin²x)] dx

let

t = sinx

dt = cosx dx

∫ cosx dx/√(1+sin²x)= ∫ [1/√(1+t²)] dt=J ==> It is OK up to here

J=ln |t+√(1+t²)|+C=

ln |sinx+√(1+sin²x)|+C=F(x)

pi/2

∫ [cosx/√(1+sin²x)] dx = F(pi/2) - F(0)=

0

ln |1+√(1+1)| - ln |0+√(1+0)|=ln(1+√2) - ln1=ln(1+√2)

Edit:

Usually, when solving an integral by substitution, it's easier and faster to change the limits of integration too and not to go back to the original argument.

In your integral the limits are from x=0 to π/2 and the substitution is sinx=t ==>

the limits of t are: t=sin0 to sin(π/2) or t=0 to 1. Then

π/2

∫ [cosx/√(1+sin²x)] dx =

0

1

∫ [1/√(1+t²)] dt =ln |t+√(1+t²)| [t=0 to 1]=

0

ln |1+√(1+1)| - ln |0+√(1+0)|=ln(1+√2) ==>

no need to go back to x

Edit:

Actually ∫[1/√(1+t²)]dt is a standard integral, see here:

http://en.wikipedia.org/wiki/Hyperbolic_...

so,  ÃƒÂ¢Ã‚ˆÂ«[1/√(1+t²)]dt=sinh⁻¹(t)+C, where

sinh⁻¹(t)=arsinh(t)=ln |t+√(1+t²)|, which you can find here:

http://en.wikipedia.org/wiki/Inverse_hyp...

And finally that's the simplest way to solve ∫[1/√(1+t²)]dt = ln |t+√(1+t²)|+C.

Isn't it?

*** But the solution of Kaksi_guy is the nicest.

And without  using any artificial multiplying and dividing.

Answer 3

Don't forget the root in the denominator.   ∫1/√(1+t²) dt = arctan(t) + c is wrong because of the root present in the denominator of the fraction being integrated.  Also, pay attention to the change of the limits of integration when the variable changes.

Answer 4

∫ [cosx / √(1 + sin²x)] dx =

the integrand includes both the function sinx (even if powered) and its

derivative cosx, thus you properly substituted sinx = u →

d(sinx) = du →

cosx dx = du

and then substituting,

∫ (cosx dx)/ √(1 + sin²x) = ∫ du / √(1 + u²) =

let's multiply and divide the integrand by √(1 + u²) + u:

∫ {[√(1 + u²) + u] / √(1 + u²)} {1 /[√(1 + u²) + u]} du =

distributing the first factor:

∫ {[√(1 + u²)/ √(1 + u²)] + [u / √(1 + u²)]} {1 /[√(1 + u²) + u]} du =

simplifying and rearranging the integrand:

∫ { {1 + [u / √(1 + u²)]} / [√(1 + u²) + u] } du =

∫ { {1 + [u / √(1 + u²)]} / [u + √(1 + u²)] } du =

note, that, due to the rearrangements, the numerator is now exactly

the derivative of the denominator (in fact d[√(1 + u²)] / du =

(1/2)(2u)(1 + u²)^[(1/2) -1] = u (1 + u²)^(-1/2) = [u / √(1 + u²)] ), thus:

∫ {d[u + √(1 + u²)]} / [u + √(1 + u²)] = ln [u + √(1 + u²)] + C (abs value is not needed

in that ln argument is positive)

thus ln [u + √(1 + u²)] + C is the antiderivative in terms of u;

if you want the antiderivative in terms of x, you naturally have to substitute back u = sinx, yielding:

ln [sinx + √(1 + sin²x)] + C

well, these functions, ln [u + √(1 + u²)] and ln [sinx + √(1 + sin²x)], are not the same..! Though both representing the antiderivative, they are quite different from each other, aren't they?

Therefore an interval referred to x variable, can't remain unchanged as switching to u variable;

in your case, 0 and π/2 limits are clearly referred to x, thus, in order to avoid trouble, I suggest you to plug them straight into ln [sinx + √(1 + sin²x)], yielding:

ln {sin(π/2) + √[1 + sin²(π/2)]} - ln {sin(0) + √[1 + sin²(0)]} =

ln [1 + √(1 + 1²)] - ln (1) = ln (1 + √2) - 0 = ln (1 + √2)

thus, to sum up, in order to avoid problems, FIRST substitute back in your antiderivative, and AFTERWARDS evaluate the definite integral using the given integration limits without changing them

I hope my explanation is clear...(if not, let me know)

Bye and good luck!

Answer 5

substitution: sh t = sin x, dt * ch t = dx * cos x; limits:

1)if x=0, then t=0;

2)if x=pi/2, then sh t=1; or; (exp(t) –exp(-t))/2 =1;

or; exp(2t) –2exp(t) –1 =0;

solving quadratic eqn we get:

exp(t) = 1 +√2 > 0, hence t=ln(1 +√2);

3) Y = ∫ dt* ch t / √(1 +(sh t)^2) {t=0 to ln(1 +√2)} =  

= ∫ dt = ln(1 +√2);