$ \int e^{-\sin t} \cos t~ dt - \int e^{-\sin t} \cos t \sin t~ dt $

Question

$ \int e^{-\sin t} \cos t~ dt - \int e^{-\sin t} \cos t \sin t~ dt $

According to the answer,

It substituted $x=\sin t$

$ \int e^{-x}~ dx- \int e^{-x}(x)~ dx $

Why did the answer remove $\cos t$ ?

Answer 1

$$I=\int e^{-\sin t} \cos t dt - \int e^{-\sin t} \cos t \sin t dt$$

You are substituting $x=\sin t$ Differentiate both sides w.r.t $t$

Thus $$\frac{dx}{dt}=\cos t$$ $${dx}=\cos(t) {dt}$$ Thus $$I=\int e^{-\sin t} \cdot dx - \int e^{-\sin t} \sin t \cdot dx$$ $$I=\int e^{-x} \cdot dx - \int e^{-x} \cdot x \cdot dx$$

Answer 2

[\begin{array}{l}\int {{e^{ - \sin t}}} \cos tdt - \int {{e^{ - \sin t}}} \sin t\cos tdt\\There\,are\,two\,parts\,of\,this\,question.\\Use\,the\,formula\int {{U^n}} du = \frac{{{u^{n + 1}}}}{{n + 1}} + c\,to\,solve\,the\,\sec ond\,part.\\You\,donot\,need\,to\,solve\,the\,first\,part\,it\,will\,cancel\,when\,you\,{\mathop{\rm int}} egrate\,the\,\sec ond\,part.\end{array}]