A-Level Techniques of integration: worked solution

Question

Using x = cos theta, evaluate INT sin theta/(2 cos 2theta + 1) dtheta.

Step 1: prepare the substitution x = cos theta. Differentiating, dx/dtheta = -sin theta, so sin theta dtheta = -dx. That is convenient, because sin theta dtheta is exactly what sits on top of the integral. Step 2: rewrite the denominator in terms of x. Use the double angle formula cos 2theta = 2cos² theta - 1, so with x = cos theta, cos 2theta = 2x² - 1. Then 2 cos 2theta + 1 = 2(2x² - 1) + 1 = 4x² - 2 + 1 = 4x² - 1. Step 3: change the integral over. INT sin theta/(2 cos 2theta + 1) dtheta = INT [1/(4x² - 1)] sin theta dtheta = INT [1/(4x² - 1)] (-dx) = - INT dx/(4x² - 1). Step 4: split 1/(4x² - 1) using the difference of two squares and partial fractions. 4x² - 1 = (2x - 1)(2x + 1), so write 1/((2x - 1)(2x + 1)) = A/(2x - 1) + B/(2x + 1) 1 = A(2x + 1) + B(2x - 1) Put x = 1/2: 1 = A(2) so A = 1/2. Put x = -1/2: 1 = B(-2) so B = -1/2. Step 5: integrate term by term (INT dx/(2x + k) = (1/2)ln|2x + k|). - INT [ (1/2)/(2x - 1) - (1/2)/(2x + 1) ] dx = -(1/2)(1/2)ln|2x - 1| + (1/2)(1/2)ln|2x + 1| + C = (1/4)ln|2x + 1| - (1/4)ln|2x - 1| + C = (1/4) ln|(2x + 1)/(2x - 1)| + C Step 6: put x = cos theta back in. = (1/4) ln|(2 cos theta + 1)/(2 cos theta - 1)| + C Since |2 cos theta - 1| = |1 - 2 cos theta|, this is the same as (1/4) ln|(1 + 2 cos theta)/(1 - 2 cos theta)| + C.