Question

$\int\limits { e^{arcsin(x) } } \, dx$

Answer 1

[tex] \int\limits { e^{arcsin(x) } } \, dx u \rightarrow e^{arcsin(x)} \rightarrow dx = \sqrt{1-x^2} du sin^2(x)= 1 - cos^2 (x) x^2 = sin^2(u) \int e^u cos(u) du f= cos(u); f'= -sin(u) g= e^u ; g'= e^u = \mathrm{e}^u\cos\left(u\right)-\left(-\mathrm{e}^u\sin\left(u\right)-{\displaystyle\int}-\mathrm{e}^u\cos\left(u\right)\,\mathrm{d}u\right) =\dfrac{\mathrm{e}^u\sin\left(u\right)+\mathrm{e}^u\cos\left(u\right)}{2} [/tex]

[tex]=\dfrac{\sqrt{1-x^2}\mathrm{e}^{\arcsin\left(x\right)}+x\mathrm{e}^{\arcsin\left(x\right)}}{2}+C Simplificat/rescris: \dfrac{\left(\sqrt{1-x^2}+x\right)\mathrm{e}^{\arcsin\left(x\right)}}{2}+C[/tex]


[tex] Aici \dfrac{\mathrm{e}^u\sin\left(u\right)+\mathrm{e}^u\cos\left(u\right)}{2} Uitam de vechea substitutie u= arcsin(x) Folosim: sin(arcsin(x))=x si cos(arcsin(x))= \sqrt{1-x^2} [/tex]