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

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

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Answer 1 · 2017-01-09T22:02:40

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

Explanation:

$I = \int_{π / 2}^{π} \frac{sin x}{1 + {cos}^{2} x} d x$ $I=int_(pi//2)^pisinx/(1+cos^2x)dx$

We will make the substitution $cos x = tan θ$ $cosx=tantheta$ . Differentiating this shows that $- sin x . d x = {sec}^{2} θ . d θ$ $-sinxcolor(white).dx=sec^2thetacolor(white).d theta$ .

When making this substitution from $x$ $x$ to $θ$ $theta$ , we also need to change the bounds.

$x = \frac{π}{2} \Rightarrow cos (x) = cos (\frac{π}{2}) = 0 = tan (θ) \Rightarrow θ = 0$ $x=pi/2=>cos(x)=cos(pi/2)=0=tan(theta)=>theta=0$
$x = π \Rightarrow cos (x) = cos (π) = - 1 = tan (θ) \Rightarrow θ = - \frac{π}{4}$ $x=pi=>cos(x)=cos(pi)=-1=tan(theta)=>theta=-pi/4$

So:

$I = - \int_{π / 2}^{π} \frac{- sin x . d x}{1 + {cos}^{2} x} = - \int_{0}^{- π / 4} \frac{{sec}^{2} θ . d θ}{1 + {tan}^{2} θ}$ $I=-int_(pi//2)^pi(-sinxcolor(white).dx)/(1+cos^2x)=-int_0^(-pi//4)(sec^2thetacolor(white).d theta)/(1+tan^2theta)$

Reversing the order of the bounds with the negative sign and using the identity ${tan}^{2} θ + 1 = {sec}^{2} θ$ $tan^2theta+1=sec^2theta$ :

$I = \int_{- π / 4}^{0} d θ = {[θ]}_{- π / 4}^{0} = 0 - (- \frac{π}{4}) = \frac{π}{4}$ $I=int_(-pi//4)^0d theta=[theta]_(-pi//4)^0=0-(-pi/4)=pi/4$

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

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