Question #39670

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,

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

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