How do you calculate (step by step) the integral of= " [ (sin x)^2] / [1+(sin x)^2] " ??

Question

Is there any way to manipulate algebraicly the integral to make it easier to calculate?

Answer 1

Integral ( sin^2(x) / [1 + sin^2(x)] dx )

Try substitution. But remember to solve for x when doing so.

Let u = sin(x). Then
arcsin(u) = x, so
1/sqrt(1 - u^2) du = dx

Integral ( u^2 / [1 + u^2] 1/sqrt(1 - u^2) du )

This simplifies as

Integral ( u^2 / [1 + u^2]^(3/2) du )

At this point, we have no choice but to use trigonometric substitution.

Let u = tan(t).
du = sec^2(t) dt

Integral ( tan^2(t) / [1 + tan^2(t)]^(3/2) sec^2(t) dt )

Use the trig identity 1 + tan^2(t) = sec^2(t)

Integral ( tan^2(t) / [sec^2(t)]^(3/2) sec^2(t) dt )

(z^2)^(3/2) simplifies into z^3, so

Integral ( tan^2(t) / sec^3(t) sec^2(t) dt )

Note the cancellation.

Integral ( tan^2(t) / sec(t) dt )

Use the identity tan^2(t) = sec^2(t) - 1

Integral ( [sec^2(t) - 1] / sec(t) dt )

Split into two fractions.

Integral ( { [sec^2(t)]/sec(t) - 1/sec(t) } dt )

Integral ( { sec(t) - cos(t) } dt )

It's worth memorizing the integral of secant. The integral of
sec(t) is ln|sec(t) + tan(t)|. Integrating, we get

ln|sec(t) + tan(t)| - sin(t) + C

To convert back in terms of u, use SOHCAHTOA and right angle triangles.

Since u = tan(t), then
tan(t) = u/1 = opp/adj
opp = u
adj = 1, so by Pythagoras,
hyp = sqrt(u^2 + 1)

Since cos(t) = adj/hyp, sec(t) = hyp/adj.
sec(t) = sqrt(u^2 + 1)/1 = sqrt(u^2 + 1)
tan(t) = opp/adj = u/1 = u
sin(t) = opp/hyp = u/sqrt(u^2 + 1)

ln|sec(t) + tan(t)| - sin(t) + C becomes

ln|sqrt(u^2 + 1) + u| - u/sqrt(u^2 + 1) + C

Back-substituting x, since u = sin(x), our final answer is

ln|sqrt(sin^2(x) + 1) + sin(x)| - sin(x)/sqrt(sin^2(x) + 1) + C

Answer 2

I like Mr Pisces' approach, but the integral of
1/ (1+ sin ² x) presents a problem. If we let
u = 1 + sin² x, du = 2 sin x cos x = sin 2x,
and the integral is not logarithmic!
Let's try u = sin x, x = arcsin u, dx = du/√(1-u²)
Then we get
∫ du/ √(1-u²)(1 + u²).
= ∫ √(1-u²) du/ (1-u^4),
But this doesn't seem to lead anywhere either.

I finally got it! The method that finally works is to
let u = tan x, x = arctan u, dx = du/(1+u²)
Then
1 + u² = sec² x,
1/(1+u²) = cos² x
u/(1+ u²) = sin² x
Making the substitutions and simplifying, we get
∫ du/ (1+u²) * (1 + [u²/(1+u²] ),
which simplifies to
∫ du/ (1 + 2u²) = ∫ du/ (1 + [√2u]²).
Finally, letting t = √2 u, u = t /√2, du = dt/ √2,
we get
1/√2 * ∫ dt/ (1+ t²) = 1/√2 arctan t
= 1/√2 arctan (√2 u) = 1/√2 arctan(√2 tan x) + C.

Now we can use Mr. Pisces' approach to evaluate
the original integral:
∫ sin² x dx / (1 + sin² x) dx = ∫ (1 + sin² x -1) dx /(1 + sin² x)
= ∫ dx - ∫ dx/ (1+sin² x)
= x - 1/√2 arctan(√2 tan x) + C.

Answer 3

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