How do you compute the antiderivative #sin x/cos (2x)#?

1 Answer
Mar 6, 2015

The required integral is

#I = intsinx/cos(2x)dx#

The first part of the solution requires the use of the trigonometric identity #cos(2x) = 2cos^2x - 1#

This leaves you with,

#I = intsinx/(2cos^2x - 1)dx#

Let #cosx = t#

Differentiating both sides with respect to t,

#-sinx dx/dt = 1#
#=> sinxdx=-dt#

Thus,

#I = -int1/(2t^2-1)dt#

#=> I = -1/2int1/(t^2-1/2)dt#

This sort of integral has a standard result, which is

#int1/(x^2-a^2)dx = 1/(2a)ln|(x-a)/(x+a)| + C#

which I will derive once we have the answer to your question, which is

#I = -1/2[1/(2(1/sqrt(2)))ln|(t-1/sqrt2)/(t+1/sqrt2)|] + C#

#=> I = -sqrt2/4ln|(t-1/sqrt2)/(t+1/sqrt2)| + C#

Finally, replacing #t=cosx#, we get,

#intsinx/cos(2x)dx = -sqrt2/4ln|(cosx-1/sqrt2)/(cosx+1/sqrt2)|#+C

Now, for the derivation of #int1/(x^2-a^2)dx = 1/(2a)ln|(x-a)/(x+a)| + C# here's a screenshot from a textbook.

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