Question

∫(cos2θ–1/cos2θ+1) dθ ​

Answer 1

Formulas Used 》

$☆》1 - cos 2 \theta = 2 {sin}^{2} \theta \\ \\☆》1 + cos2\theta = 2 {cos}^{2 } \theta \\ \\ ☆》\frac{sin \theta }{cos \theta } = tan\theta$

$☆》{tan}^{2} \theta = {sec}^{2} \theta - 1$

Solution 》

$I = \int \dfrac{cos 2\theta - 1}{cos 2\theta + 1} \: d \theta \\ \\ = - \int \frac{1 - cos 2\theta}{1 + cos2 \theta} \: d \theta \\ \\ = - \int \frac{2 {sin}^{2} \theta}{2 {cos}^{2} \theta} \: d \theta \\ \\ = - \int {tan}^{2} \theta \: d \theta \\ \\ = - \int( {sec}^{2} \theta - 1) \:d \theta \\ \\ = - \int {sec}^{2} \theta \: d \theta + \int1 \: d \theta \\ \\ = - tan \theta + \theta + C \\ \\ = \theta - tan \theta + C$

Which is the required Answer.

Answer 2

$I = \int \dfrac{cos 2\theta - 1}{cos 2\theta + 1} \: d \theta \\ \\ = - \int \frac{1 - cos 2\theta}{1 + cos2 \theta} \: d \theta \\ \\ = - \int \frac{2 {sin}^{2} \theta}{2 {cos}^{2} \theta} \: d \theta \\ \\ = - \int {tan}^{2} \theta \: d \theta \\ \\ = - \int( {sec}^{2} \theta - 1) \:d \theta \\ \\ = - \int {sec}^{2} \theta \: d \theta + \int1 \: d \theta \\ \\ = - tan \theta + \theta + C \\ \\ = \theta - tan \theta + C$