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Question Number 24651 by ajitkumarrajak99@gmail.com last updated on 23/Nov/17

$$\int \sin x+\cos ydx$$

Answered by jota+ last updated on 23/Nov/17

$$Seay=ax+b$$$$\Rightarrow I=-cosx+\frac{1}{a}siny+C.$$$$$$

Answered by ajfour last updated on 24/Nov/17

$$=-\cos x+\int \left(\frac{dx}{dy}\right)\cos ydy+c_1$$$$=-\cos x+\left(\frac{dx}{dy}\right)\sin y-\int \left(\frac{d^{2}x}{{dy}^2}\right)\sin ydy+c_2$$$$=-\cos x+\left(\frac{dx}{dy}\right)\sin y+\left(\frac{d^{2}x}{{dy}^2}\right)\cos y$$$$+\int \left(\frac{d^{3}x}{{dy}^3}\right)\cos ydy+c_3$$$$=-\cos x+\underset{r=1}{\sum ^{?}}[\left(\frac{d^{r}x}{{dy}^r}\right)\sin y+\left(\frac{d^{r+1}x}{{dy}^{r+1}}\right)\cos y]+c$$

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