Show that the integral ∫(1-2 sin^2x)/(1+2sinxcosx) dx = (1/2) ln2 between the limits π/4 and 0. [5 marks]
First, we use the trig identities: cos2x=cos^2x-sin^2x, cos^2x+sin^2x=1 and sin2x=2sinxcosx to transform the integral to ∫(cos2x)/(1+sin2x)dx.
We know that ∫f'(x)/f(x)dx = ln|f(x)|+c, so we let f(x)=1+sin2x. f'(x) can then be found by differentiating f(x), so f'(x) = 2cos2x.
So we multiply top and bottom of the fraction by 2, and take out a factor of 1/2 to express the integral as: (1/2)∫(2cos2x)/(1+sin2x)dx, which can be integrated using the above rule to get (1/2)ln|1+sin2x|+ c.
We then substitute the limits given, to get (1/2)ln(1+1)-(1/2)ln(1+0) = (1/2)ln(2).
