# Integrate secx/(1+cosecx)dx ?

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Feb 12, 2018

$\int \sec \frac{x}{1 + \csc x} \cdot \mathrm{dx}$

=$\frac{1}{4} L n \left(\frac{1 + \sin x}{1 - \sin x}\right) + \frac{1}{2} \cdot {\left(1 + \sin x\right)}^{- 1} + C$

#### Explanation:

$\int \sec \frac{x}{1 + \csc x} \cdot \mathrm{dx}$

=$\int \frac{\frac{1}{\cos} x}{1 + \frac{1}{\sin} x} \cdot \mathrm{dx}$

=$\int \sin \frac{x}{\cos x \cdot \left(1 + \sin x\right)} \cdot \mathrm{dx}$

=$\int \frac{\sin x \cdot \cos x}{{\left(\cos x\right)}^{2} \cdot \left(1 + \sin x\right)} \cdot \mathrm{dx}$

=$\int \frac{\sin x \cdot \cos x \cdot \mathrm{dx}}{\left(1 - {\left(\sin x\right)}^{2}\right) \cdot \left(1 + \sin x\right)}$

=$\int \frac{\sin x \cdot \cos x \cdot \mathrm{dx}}{\left(1 - \sin x\right) \cdot {\left(1 + \sin x\right)}^{2}}$

After using $y = \sin u$ and $\mathrm{dy} = \cos u \cdot \mathrm{du}$ transforms, this integral became

$\int \frac{y \cdot \mathrm{dy}}{\left(1 - y\right) \cdot {\left(1 + y\right)}^{2}}$

Now, I decomposed integrand into basic fractions,

$\frac{y}{\left(1 + y\right) \cdot {\left(1 - y\right)}^{2}} = \frac{A}{1 - y} + \frac{B}{1 + y} + \frac{C}{1 + y} ^ 2$

After expanding denominator,

$A \cdot {\left(1 + y\right)}^{2} + B \cdot \left(1 - {y}^{2}\right) + C \cdot \left(1 - y\right) = y$

Set $x = - 1$, $2 C = - 1$, so $C = - \frac{1}{2}$

Set $x = 1$, $4 A = 1$, so $A = \frac{1}{4}$

Set $x = 0$, $A + B + C = 0$, so $B = \frac{1}{4}$

Hence,

$\int \frac{y \cdot \mathrm{dy}}{\left(1 - y\right) \cdot {\left(1 + y\right)}^{2}}$

=$\frac{1}{4} \int \frac{\mathrm{dy}}{1 - y} + \frac{1}{4} \int \frac{\mathrm{dy}}{1 + y} - \frac{1}{2} \int \frac{\mathrm{dy}}{1 + y} ^ 2$

=$\frac{1}{4} L n \left(1 + y\right) - \frac{1}{4} L n \left(1 - y\right) + \frac{1}{2} \cdot \frac{1}{1 + y} + C$

=$\frac{1}{4} L n \left(\frac{1 + y}{1 - y}\right) + \frac{1}{2} \cdot {\left(1 + y\right)}^{- 1} + C$

Thus,

$\int \sec \frac{x}{1 + \csc x} \cdot \mathrm{dx}$

=$\frac{1}{4} L n \left(\frac{1 + \sin x}{1 - \sin x}\right) + \frac{1}{2} \cdot {\left(1 + \sin x\right)}^{- 1} + C$

Then teach the underlying concepts
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#### Explanation

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#### Explanation:

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Dec 22, 2017

$\int \sec \frac{x}{1 + \csc x} \mathrm{dx} = \frac{1}{2} \ln | \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x} | + \frac{1}{2} \frac{1}{1 + \sin x} + C$

#### Explanation:

Let

$I = \int \sec \frac{x}{1 + \csc x} \mathrm{dx}$

Apply the substitution $x = 2 \theta$:

$I = 2 \int \frac{\sec 2 \theta}{1 + \csc 2 \theta} d \theta$

Apply the appropriate double-angle Trigonometric identities:

$I = 2 \int \frac{\left(\frac{1 + {\tan}^{2} \theta}{1 - {\tan}^{2} \theta}\right)}{1 + \left(\frac{1 + {\tan}^{2} \theta}{2 \tan \theta}\right)} d \theta$

Rearrange:

$I = 4 \int \left(\tan \frac{\theta}{\left(1 - \tan \theta\right) {\left(1 + \tan \theta\right)}^{3}}\right) {\sec}^{2} \theta d \theta$

Apply the substitution $u = \tan \theta$:

$I = 4 \int \frac{u}{\left(1 - u\right) {\left(1 + u\right)}^{3}} \mathrm{du}$

Apply partial fraction decomposition:

$I = \frac{1}{2} \int \left(\frac{1}{1 - u} + \frac{1}{1 + u} + \frac{2}{1 + u} ^ 2 - \frac{4}{1 + u} ^ 3\right) \mathrm{du}$

Integrate directly:

$I = \frac{1}{2} \left\{- \ln | 1 - u | + \ln | 1 + u | - \frac{2}{1 + u} + \frac{2}{1 + u} ^ 2\right\} + C$

Rearrange:

$I = \frac{1}{2} \ln | \frac{1 + u}{1 - u} | - \frac{u}{1 + u} ^ 2 + C$

Reverse the substitutions:

$I = \frac{1}{2} \ln | \frac{1 + \tan \left(\frac{x}{2}\right)}{1 - \tan \left(\frac{x}{2}\right)} | - \tan \frac{\frac{x}{2}}{1 + \tan \left(\frac{x}{2}\right)} ^ 2 + C$

Apply the appropriate Trigonometric identities:

$I = \frac{1}{2} \ln | \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x} | - \frac{\sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{\cos \left(\frac{x}{2}\right) + \sin \left(\frac{x}{2}\right)} ^ 2 + C$

Rearrange:

$I = \frac{1}{2} \ln | \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x} | - \frac{1}{2} \frac{2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)}{1 + 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)} + C$

Apply the appropriate double-angle Trigonometric identity:

$I = \frac{1}{2} \ln | \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x} | - \frac{1}{2} \sin \frac{x}{1 + \sin x} + C$

Rearrange:

$I = \frac{1}{2} \ln | \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x} | + \frac{1}{2} \frac{1}{1 + \sin x} - \frac{1}{2} + C$

Redefine $C$ to remove the constant:

$I = \frac{1}{2} \ln | \frac{1 + \cos x + \sin x}{1 + \cos x - \sin x} | + \frac{1}{2} \frac{1}{1 + \sin x} + C$

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