Question

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

Answer 1

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.