How do you find the integral of $\frac{1}{1 + {sin}^{2} x}$ $1 / (1 + sin^2 x)$ ?

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How do you find the integral of $\frac{1}{1 + {sin}^{2} x}$ $1 / (1 + sin^2 x)$ ?

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Answer 1 · 2016-10-20T20:53:04

$\frac{arctan (\sqrt{2} tan x)}{\sqrt{2}} + C$ $arctan(sqrt2tanx)/sqrt2+C$

Explanation:

$I = \int \frac{1}{1 + {sin}^{2} x} d x$ $I=int1/(1+sin^2x)dx$

Notice that dividing by ${cos}^{2} (x)$ $cos^2(x)$ will get us working with $tan x$ $tanx$ and $sec x$ $secx$ functions, which are derivatives of themselves:

$I = \int \frac{\frac{1}{{cos}^{2} x}}{\frac{1 + {sin}^{2} x}{{cos}^{2} x}} d x = \int \frac{{sec}^{2} x}{{sec}^{2} x + {tan}^{2} x} d x$ $I=int(1/cos^2x)/((1+sin^2x)/cos^2x)dx=intsec^2x/(sec^2x+tan^2x)dx$

Recall that ${sec}^{2} x = 1 + {tan}^{2} x$ $sec^2x=1+tan^2x$ :

$I = \int \frac{{sec}^{2} x}{(1 + {tan}^{2} x) + {tan}^{2} x} d x = \int \frac{{sec}^{2} x}{1 + 2 {tan}^{2} x} d x$ $I=intsec^2x/((1+tan^2x)+tan^2x)dx=intsec^2x/(1+2tan^2x)dx$

We will use the substitution $tan x = \frac{1}{\sqrt{2}} tan θ$ $tanx=1/sqrt2tantheta$ . This implies that ${sec}^{2} x d x = \frac{1}{\sqrt{2}} {sec}^{2} θ d θ$ $sec^2xdx=1/sqrt2sec^2thetad theta$ . Substituting:

$I = \int \frac{\frac{1}{\sqrt{2}} {sec}^{2} θ}{1 + 2 {(\frac{1}{\sqrt{2}} tan θ)}^{2}} d θ = \frac{1}{\sqrt{2}} \int \frac{{sec}^{2} θ}{1 + 2 (\frac{1}{2} {tan}^{2} θ)} d θ$ $I=int(1/sqrt2sec^2theta)/(1+2(1/sqrt2tantheta)^2)d theta=1/sqrt2int(sec^2theta)/(1+2(1/2tan^2theta))d theta$

$I = \frac{1}{\sqrt{2}} \int \frac{{sec}^{2} θ}{1 + {tan}^{2} θ} d θ$ $I=1/sqrt2int(sec^2theta)/(1+tan^2theta)d theta$

Using the same trigonometric identity:

$I = \frac{1}{\sqrt{2}} \int \frac{{sec}^{2} θ}{{sec}^{2} θ} d θ = \frac{1}{\sqrt{2}} \int d θ = \frac{1}{\sqrt{2}} θ + C$ $I=1/sqrt2intsec^2theta/sec^2thetad theta=1/sqrt2intd theta=1/sqrt2theta+C$

From $tan x = \frac{1}{\sqrt{2}} tan θ$ $tanx=1/sqrt2tantheta$ we see that $\sqrt{2} tan x = tan θ$ $sqrt2tanx=tantheta$ and $θ = arctan (\sqrt{2} tan x)$ $theta=arctan(sqrt2tanx)$ . Thus:

$I = \frac{arctan (\sqrt{2} tan x)}{\sqrt{2}} + C$ $I=arctan(sqrt2tanx)/sqrt2+C$

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How do you find the integral of $\frac{1}{1 + {sin}^{2} x}$ $1 / (1 + sin^2 x)$ ?

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