Question

what is the integration of 1/[{sinx}^4 + {cosx}^4]

Answer 1

Answer on Question #51630 – Math – Integral Calculus

Question

What is the integration of $1 / [\{\sin x\} ^4 + \{\cos x\} ^4]$?

Solution

Method 1

$$\begin{aligned} & \int \frac {1}{\sin^ {4} x + \cos^ {4} x} d x = \int \frac {1}{\sin^ {4} x + \cos^ {4} x} d x = \int \frac {1}{\cos^ {4} x \left(\frac {\sin^ {4} x}{\cos^ {4} x} + 1\right)} d x = \int \frac {1}{\cos^ {2} x (\tan^ {4} x + 1)} \frac {d x}{\cos^ {2} x} = \\ & = \int \frac {1}{\cos^ {2} x (\tan^ {4} x + 1)} \frac {d x}{\cos^ {2} x} = \int (\tan^ {2} x + 1) \frac {1}{\tan^ {4} x + 1} d \tan x = | t = \tan x | = \int \frac {(t ^ {2} + 1)}{t ^ {4} + 1} d t = \\ & = \int \frac {(t ^ {2} + 1)}{t ^ {4} + 1} d t = \int \frac {t ^ {2} + 1}{(t ^ {4} + 2 t ^ {2} + 1 - 2 t ^ {2})} d t = \int \frac {t ^ {2} + 1}{(t ^ {2} + 1 - t \sqrt {2}) (t ^ {2} + 1 + t \sqrt {2})} d t \end{aligned}$$ $$\begin{aligned} & \frac {t ^ {2} + 1}{(t ^ {2} - t \sqrt {2} + 1) (t ^ {2} + t \sqrt {2} + 1)} = \frac {A t + B}{t ^ {2} - t \sqrt {2} + 1} + \frac {C t + D}{t ^ {2} + t \sqrt {2} + 1} = \frac {A t ^ {3} + B t ^ {2} + A t ^ {2} \sqrt {2} + B t \sqrt {2} + A t + B}{(t ^ {2} - t \sqrt {2} + 1) (t ^ {2} + t \sqrt {2} + 1)} + \\ & \quad + \frac {C t ^ {3} + D t ^ {2} - C t ^ {2} \sqrt {2} - D t \sqrt {2} + C t + D}{(t ^ {2} - t \sqrt {2} + 1) (t ^ {2} + t \sqrt {2} + 1)} = \\ & = \frac {(A + C) t ^ {3} + (B + D) t ^ {2} + (A - C) t ^ {2} \sqrt {2} + (B - D) t \sqrt {2} + (A + C) t + B + D}{(t ^ {2} - t \sqrt {2} + 1) (t ^ {2} + t \sqrt {2} + 1)}, \end{aligned}$$ $$A + C = 0, B + D = 1, A - C = 0, B - D = 0, \text{ hence}$$ $$\begin{cases} A + C = 0 \\ A - C = 0 \end{cases} \text{which gives } \begin{cases} 2A = 0 \\ A = C \end{cases} \text{, so } A = C = 0$$ $$\begin{cases} B + D = 1 \\ B = D \end{cases} \text{, which gives } 2B = 1, B = D, \text{ so } B = D = \frac{1}{2}$$

obtain

$$\begin{aligned} & \frac {t ^ {2} + 1}{(t ^ {2} - t \sqrt {2} + 1) (t ^ {2} + t \sqrt {2} + 1)} = \frac {1}{2 (t ^ {2} - t \sqrt {2} + 1)} + \frac {1}{2 (t ^ {2} + t \sqrt {2} + 1)} = \\ & = \frac {1}{2 \left( \left( t - \frac {\sqrt {2}}{2} \right) ^ {2} + \frac {1}{2} \right)} + \frac {1}{2 \left( \left( t + \frac {\sqrt {2}}{2} \right) ^ {2} + \frac {1}{2} \right)} \\ \end{aligned}$$ $$\begin{array}{l} \int \frac {t ^ {2} + 1}{(t ^ {2} + 1 - t \sqrt {2}) (t ^ {2} + 1 + t \sqrt {2})} d t = \frac {1}{2} \int \frac {d t}{\left(\left(t - \frac {\sqrt {2}}{2}\right) ^ {2} + \frac {1}{2}\right)} + \frac {1}{2} \int \frac {d t}{\left(\left(t + \frac {\sqrt {2}}{2}\right) ^ {2} + \frac {1}{2}\right)} = \\ = \frac {\sqrt {2}}{2} \arctan \frac {t - \frac {\sqrt {2}}{2}}{\frac {1}{\sqrt {2}}} + \frac {\sqrt {2}}{2} \arctan \frac {t + \frac {\sqrt {2}}{2}}{\frac {1}{\sqrt {2}}} + c = \frac {\sqrt {2}}{2} \arctan (t \sqrt {2} - 1) + \frac {\sqrt {2}}{2} \arctan (t \sqrt {2} + 1) + c = \\ = \frac {\sqrt {2}}{2} \arctan \left(\sqrt {2} \tan x - 1\right) + \frac {\sqrt {2}}{2} \arctan \left(\sqrt {2} \tan x + 1\right) + c, \end{array}$$

where $\arctan$ is the inverse of $\tan$ function, $c$ is an arbitrary real constant.

Method 2

$$\begin{array}{l} \sin^ {4} x + \cos^ {4} x = (\sin^ {2} x + \cos^ {2} x) ^ {2} - 2 \sin^ {2} x \cos^ {2} x = 1 - 2 \sin^ {2} x \cos^ {2} x = 1 - \frac {1}{2} \sin^ {2} 2 x = \\ = (1 + \frac {\sqrt {2}}{2} \sin 2 x) (1 - \frac {\sqrt {2}}{2} \sin 2 x) \\ \end{array}$$ $$\int \frac {1}{\sin^ {4} x + \cos^ {4} x} d x = \int \frac {1}{(1 + \frac {\sqrt {2}}{2} \sin 2 x) (1 - \frac {\sqrt {2}}{2} \sin 2 x)} = \frac {1}{2} \left[ \int \frac {1}{1 + \frac {\sqrt {2}}{2} \sin 2 x} d x + \int \frac {1}{1 - \frac {\sqrt {2}}{2} \sin 2 x} d x \right]$$ $$\sin 2 x = 2 \sin x \cos x = \frac {2 \sin x}{\cos x} \cos^ {2} x = \frac {2 \tan x}{\frac {1}{\cos^ {2} x}} = \frac {2 \tan x}{1 + \tan^ {2} x} = [ t = \tan x ] = \frac {2 t}{1 + t ^ {2}}$$ $$d t = d (\tan x) = \frac {1}{\cos^ {2} x} d x = (1 + t ^ {2}) d x \Rightarrow d x = \frac {d t}{1 + t ^ {2}}$$ $$\begin{array}{l} \frac {1}{2} \left[ \int \frac {1}{1 + \frac {\sqrt {2}}{2} \sin 2 x} d x + \int \frac {1}{1 - \frac {\sqrt {2}}{2} \sin 2 x} d x \right] = \frac {1}{2} \left[ \int \frac {\frac {d t}{1 + t ^ {2}}}{1 + \frac {\sqrt {2}}{2} \left(\frac {2 t}{1 + t ^ {2}}\right)} + \int \frac {\frac {d t}{1 + t ^ {2}}}{1 - \frac {\sqrt {2}}{2} \left(\frac {2 t}{1 + t ^ {2}}\right)} \right] = \\ = \frac {1}{2} \left[ \int \frac {d t}{1 + t ^ {2} + \sqrt {2} t} + \int \frac {d t}{1 + t ^ {2} - \sqrt {2} t} \right] = \frac {1}{2} \left[ \int \frac {d t}{(t + \frac {1}{\sqrt {2}}) ^ {2} + \frac {1}{2}} + \int \frac {d t}{(t - \frac {1}{\sqrt {2}}) ^ {2} + \frac {1}{2}} \right] = \\ = \frac {1}{2} \left[ \int \frac {2 d t}{(\sqrt {2} t + 1) ^ {2} + 1} + \int \frac {2 d t}{(\sqrt {2} t - 1) ^ {2} + 1} \right] = \frac {1}{\sqrt {2}} (\arctan (\sqrt {2} \tan x + 1) + \arctan (\sqrt {2} \tan x - 1)) \\ \end{array}$$ $$\text{So} \int \frac {1}{\sin^ {4} x + \cos^ {4} x} d x = \frac {1}{\sqrt {2}} (\arctan (\sqrt {2} \tan x + 1) + \arctan (\sqrt {2} \tan x - 1)) + c,$$

where $\arctan$ is the inverse of $\tan$ function, $c$ is an arbitrary real constant.

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