#20190218
Solution:
(a)
∫1sinx+cosxdx
=√22∫1√22sinx+√22cosxdx
=√22∫1sin(x+π4)dx
=√22∫csc(x+π4)dx
=√22∫csc(x+π4)(csc(x+π4)+cot(x+π4))csc(x+π4)+cot(x+π4)dx
=−√22ln|csc(x+π4)+cot(x+π4)|+C
(b)
∫1sinx+√3cosxdx
=12∫112sinx+√32cosxdx
=12∫1sin(x+π3)dx
=12∫csc(x+π3)dx
=12∫csc(x+π3)(csc(x+π3)+cot(x+π3))csc(x+π3)+cot(x+π3)dx
=−12ln|csc(x+π3)+cot(x+π3)|+C
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