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Question #e4efb

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Oct 12, 2017

#int sinx/sin(x+a) dx = (x+a)cosa-log(sin(x+a))sina + C#

Explanation:

Making #y = x+a# and #dy = dx# we have

#int sinx/sin(x+a) dx = int sin(y-a)/siny dy = int (cosa - cos y/siny sina)dy = y cosa-log(siny) sina + C#

and finally

#int sinx/sin(x+a) dx = (x+a)cosa-log(sin(x+a))sina + C#

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