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sin-x-cos-y-dx-




Question Number 24651 by ajitkumarrajak99@gmail.com last updated on 23/Nov/17
∫sin x+cos y dx
$$\int\mathrm{sin}\:{x}+\mathrm{cos}\:{y}\:{dx} \\ $$
Answered by jota+ last updated on 23/Nov/17
Sea y=ax+b  ⇒I=−cosx+(1/a)siny+C.
$${Sea}\:{y}={ax}+{b} \\ $$$$\Rightarrow{I}=−{cosx}+\frac{\mathrm{1}}{{a}}{siny}+{C}. \\ $$$$ \\ $$
Answered by ajfour last updated on 24/Nov/17
=−cos x+∫((dx/dy))cos y dy+c_1   =−cos x+((dx/dy))sin y−∫((d^2 x/dy^2 ))sin ydy+c_2   =−cos x+((dx/dy))sin y+((d^2 x/dy^2 ))cos y                        +∫((d^( 3) x/dy^3 ))cos ydy+c_3   =−cos x+Σ_(r=1) ^? [((d^r x/dy^r ))sin y+((d^(r+1) x/dy^(r+1) ))cos y]+c
$$=−\mathrm{cos}\:{x}+\int\left(\frac{{dx}}{{dy}}\right)\mathrm{cos}\:{y}\:{dy}+{c}_{\mathrm{1}} \\ $$$$=−\mathrm{cos}\:{x}+\left(\frac{{dx}}{{dy}}\right)\mathrm{sin}\:{y}−\int\left(\frac{{d}^{\mathrm{2}} {x}}{{dy}^{\mathrm{2}} }\right)\mathrm{sin}\:{ydy}+{c}_{\mathrm{2}} \\ $$$$=−\mathrm{cos}\:{x}+\left(\frac{{dx}}{{dy}}\right)\mathrm{sin}\:{y}+\left(\frac{{d}^{\mathrm{2}} {x}}{{dy}^{\mathrm{2}} }\right)\mathrm{cos}\:{y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\int\left(\frac{{d}^{\:\mathrm{3}} {x}}{{dy}^{\mathrm{3}} }\right)\mathrm{cos}\:{ydy}+{c}_{\mathrm{3}} \\ $$$$=−\mathrm{cos}\:{x}+\underset{{r}=\mathrm{1}} {\overset{?} {\sum}}\left[\left(\frac{{d}^{{r}} {x}}{{dy}^{{r}} }\right)\mathrm{sin}\:{y}+\left(\frac{{d}^{{r}+\mathrm{1}} {x}}{{dy}^{{r}+\mathrm{1}} }\right)\mathrm{cos}\:{y}\right]+{c} \\ $$

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