# 求解$\ dfrac {dy} {dx} = \ cos（x + y）+ \ sin（x + y）$

ss1729 06/18/2018. 2 answers, 86 views

My Attempt $$\ frac {dy} {dx} = \ sqrt {2} \ cos（x + y- \ tfrac {\ pi} {4}）$$设置$t = x + y- \ tfrac {\ pi } {4} \暗示y = t-x + \ tfrac {\ pi} {4}$ $$\ frac {dy} {dx} = \ frac {dt} {dx} -1 = \ sqrt {2} \ cos t \\ \ frac {dt} {dx} = \ sqrt {2} \ cos（t）+1 \\ \ int \ frac {dt} {\ sqrt {2} \ cos（t）+1} = \ int dx$$如何进一步查找常规解决方案？

Cesareo 06/18/2018.

$$\ frac {du} {dx} = \ sin u + \ cos u +1$$

$$\ frac {du} {\ sin u + \ cos u + 1} = dx$$

$$\ sin u + \ cos u = \ sqrt 2 \ sin（u + \ frac {\ pi} {4}）$$

$$\ cos u = 2 \ cos ^ 2（\ frac u2）-1 \\ \ sin u = 2 \ cos（\ frac u2）\ sin（\ frac u2）$$

$$\ frac {1} {\ cos u + \ sin u + 1} = \ frac {1} {2 \ cos（\ frac u2）（\ cos（\ frac u2）+ \ sin（\ frac u2））} = \ frac 12 \ left（\ frac {\ sin（\ frac u2）} {\ cos（\ frac u2）} + \ frac { - \ sin（\ frac u2）+ \ cos（\ frac u2）} {\ cos（\ frac u2）+ \ sin（\ frac u2）} \ right）$$

ss1729 06/19/2018.

\ begin {align}＆\ int dx = \ int \ frac {dt} {\ sqrt {2} \ cos（t）+1} \\ x＆= \ frac {1} {\ sqrt {2}} \ int \ frac {dt} {\ cos t + \ frac {1} {\ sqrt {2}}} = \ frac {1} {\ sqrt {2}} \ int \ frac {dt} {\ cos t + \ cos \压裂{\ PI} {4}} = \压裂{1} {\ SQRT {2}} \ INT \压裂{DT} {2个\ COS \大（\压裂{吨} {2} + \压裂{\ PI } {8} \ big）\ cos \ big（\ frac {t} {2} - \ frac {\ pi} {8} \ big）} \\＆= \ frac {\ sqrt {2}} {2 \ SQRT {2}} \ INT \压裂{\罪\大[\大（\压裂{吨} {2} + \压裂{\ PI} {8} \大） - \大（\压裂{吨} {2 } - \压裂{\ PI} {8} \大）\大]} {\ COS \大（\压裂{吨} {2} + \压裂{\ PI} {8} \大）\ COS \大（ \ frac {t} {2} - \ frac {\ pi} {8} \ big）} dt \\＆= \ frac {1} {2} \ bigg [\ int \ tan \ Big（\ frac {t} {2} + \压裂{\ PI} {8} \大）DT- \ INT \黄褐色\大（\压裂{吨} {2} - \压裂{\ PI} {8} \大）DT \比格] \\＆\ text {Set} u = \ frac {t} {2} + \ frac {\ pi} {8} \暗示dt = 2du \ quad \＆\ quad v = \ frac {t} {2} - \ frac {\ pi} {8} \暗示dt = 2dv \\ x＆= \ int \ tan u.du- \ int \ tan v.dv = \ log | \ sec u | - \ log | \ sec v | = \ log | \ frac {\ sec u} {\ sec v} | + C \\＆= \ log | \ frac {\ cos v} {\ cos u} | + C = \ log \ bigg | \ frac {\ COS \大（\ tfrac {吨} {2} - \ tfrac {\ PI} {8} \大）} {\ COS \大（\ tfrac {吨} {2} + \ tfrac {\ PI} {8} \ big）} \ bigg | + C \\＆= \ log \ bigg | \ frac {\ cos \ big（\ tff 外消旋{X + Y} {2} - \ tfrac {\ PI} {8} - \ tfrac {\ PI} {8} \大）} {\ COS \大（\ tfrac {X + Y} {2} - \ tfrac {\ PI} {8} + \ tfrac {\ PI} {8} \大）} \比格| + C = \ LOG \比格| \压裂{\ COS \大（\ tfrac {X + Y} { 2} - \ tfrac {\ pi} {4} \ big）} {\ cos \ big（\ tfrac {x + y} {2} \ big）} \ bigg | + C \\＆= \ log \ bigg | \压裂{1} {\ SQRT {2}} + \压裂{1} {\ SQRT {2}} \黄褐色\大（\ tfrac {X + Y} {2} \大）\比格| + C \\ ＆\ color {red} {x = \ log \ bigg | 1+ \ tan \ big（\ tfrac {x + y} {2} \ big）\ bigg | + K} \ end {align}