[tex]Evaluate: \int \frac{ \sin(2x) }{1 + \sin^{2} (x) }dx \\ [/tex]
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physicsgeniuspi physicsgeniuspi · Answer 1 · 2021-10-25T03:31:30+02:00

Answer:

ln (1 + sin^2(x)) + C

Step-by-step explanation:

Note that sin(2x) = 2 sin(x) cos(x).

So, the integrand is rewritten as:

(2 sin(x) cos(x)) / (1 + sin^2(x) ).

Now, make the substitution u = sin(x), so that du = cos(x) dx. Thus, in terms of the variable u, the integrand is

(2 u )/(1 + u^2).

Make another substitution v = 1 + u^2, so that dv = 2 u du, and the integrand in terms of v is

1 / v

Now, we integrate, and then back substitute to write that the indefinite integral is

ln |v| + C = ln |1 + u^2| + C = ln |1 + sin^2(x)| + C + ln (1 + sin^2(x)) +C, since 1 + sin^2(x) + 1 is positive for all real x.

SalmonWorldTopper SalmonWorldTopper · Answer 2 · 2021-10-25T03:43:44+02:00

Step-by-step explanation:

[tex]\large\underline{\sf{Solution-}}[/tex]

We have to evaluate the given integral.

[tex] \displaystyle \rm = \int \dfrac{ \sin(2x) }{1 + \sin^{2} (x) } \: dx[/tex]

[tex] \displaystyle \rm = \int \dfrac{2 \sin(x) \cos(x) }{1 + \sin^{2} (x) } \: dx[/tex]

Now let us assume that:

[tex] \rm \longmapsto u = 1 + { \sin}^{2}(x)[/tex]

[tex] \rm \longmapsto \dfrac{du}{dx} = 2 \sin(x) \cos(x) [/tex]

[tex] \rm \longmapsto dx = \dfrac{1}{2 \sin(x) \cos(x) } \: du[/tex]

Therefore, the integral becomes:

[tex] \displaystyle \rm = \int \dfrac{1}{u} \: du[/tex]

[tex] \displaystyle \rm = \ln(u) + C[/tex]

[tex] \displaystyle \rm = \ln(1 + sin^{2}x ) + C[/tex]

Therefore:

[tex] \displaystyle \rm \longmapsto \int \dfrac{ \sin(2x) }{1 + \sin^{2} (x) } \: dx = \ln(1 + sin^{2}x ) + C[/tex]