How do you evaluate $\int \frac{cos x}{1 + {sin}^{2} x}$ $int cosx/(1+sin^2x)$ from $[0, \frac{π}{2}]$ $[0, pi/2]$ ?

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How do you evaluate $\int \frac{cos x}{1 + {sin}^{2} x}$ $int cosx/(1+sin^2x)$ from $[0, \frac{π}{2}]$ $[0, pi/2]$ ?

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Answer 1 · 2016-11-09T05:34:02

$\int \frac{cos x}{1 + {sin}^{2} x} d x = \frac{π}{4}$ $intcosx/(1+sin^2x)dx=pi/4$

Explanation:

First examining without the bounds:

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

Let $sin x = tan θ$ $sinx=tantheta$ . This may look like a wild substitution, but the goal is to get the denominator of the fraction to $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta=sec^2theta$ . Note that differentiating $sin x = tan θ$ $sinx=tantheta$ gives $cos x d x = {sec}^{2} θ d θ$ $cosxdx=sec^2thetad theta$ . It's also helpful that $cos x d x$ $cosxdx$ is already present in the integrand. Substituting these in:

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

Since $sin x = tan θ$ $sinx=tantheta$ , we see that $θ = arctan (sin x)$ $theta=arctan(sinx)$ :

$I = arctan (sin x) + C$ $I=arctan(sinx)+C$

Adding the bounds:

$I_{B} = \int_{0}^{\frac{π}{2}} \frac{cos x}{1 + {sin}^{2} x} d x = {[arctan (sin x)]}_{0}^{\frac{π}{2}}$ $I_B=int_0^(pi/2)cosx/(1+sin^2x)dx=[arctan(sinx)]_0^(pi/2)$

$I_{B} = arctan (sin (\frac{π}{2})) - arctan (sin (0))$ $color(white)(I_B)=arctan(sin(pi/2))-arctan(sin(0))$

$I_{B} = arctan (1) - arctan (0)$ $color(white)(I_B)=arctan(1)-arctan(0)$

$I_{B} = \frac{π}{4} - 0$ $color(white)(I_B)=pi/4-0$

$I_{B} = \frac{π}{4}$ $color(white)(I_B)=pi/4$

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How do you evaluate $\int \frac{cos x}{1 + {sin}^{2} x}$ $int cosx/(1+sin^2x)$ from $[0, \frac{π}{2}]$ $[0, pi/2]$ ?

1 Answer

Explanation:

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How do you evaluate $\int \frac{cos x}{1 + {sin}^{2} x}$ $int cosx/(1+sin^2x)$ from $[0, \frac{π}{2}]$ $[0, pi/2]$ ?