integrate 1/(sinx+cosx) - Maths - Integrals

Question

integrate 1/(sinx+cosx)?

Ajanta Trivedi · Accepted Answer

$I = \int \frac{1}{(\sin x + \cos x)} ⅆ x I = \int \frac{1}{2 \sin \frac{x}{2} . \cos \frac{x}{2} + 1 - 2 \sin^{2} \frac{x}{2}} ⅆ x$ since $\sin x = 2 \sin \frac{x}{2} . \cos \frac{x}{2}, \cos x = 1 - 2 \sin^{2} \frac{x}{2}$

now dividing numerator and denominator by $\cos^{2} \frac{x}{2}$

$I = \int \frac{\sec^{2} \frac{x}{2}}{2 \tan \frac{x}{2} + \sec^{2} \frac{x}{2} - 2 \tan^{2} \frac{x}{2}} ⅆ x I = \int \frac{\sec^{2} \frac{x}{2}}{2 \tan \frac{x}{2} + 1 - \tan^{2} \frac{x}{2}} ⅆ x ∵ \sec^{2} \frac{x}{2} - \tan^{2} \frac{x}{2} = 1$

let $\tan \frac{x}{2} = t, \Rightarrow \frac{1}{2} . \sec^{2} \frac{x}{2} d x = d t$

now substituting these in the above integral.

$I = \int \frac{2}{2 t + 1 - t^{2}} ⅆ t I = \int \frac{2}{1 - (t^{2} - 2 t + 1 - 1)} ⅆ t I = \int \frac{2}{2 - (t - 1)^{2}} ⅆ t I = \int \frac{2}{(\sqrt{2})^{2} - (t - 1)^{2}} ⅆ t$

since $\int \frac{1}{a^{2} - x^{2}} ⅆ x = \frac{1}{2 a} \log | \frac{a + x}{a - x} | + c$ where c is the integration constant.

therefore the above integral becomes

$I = 2 * \frac{1}{2 \sqrt{2}} \log | \frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1} | + c I = \frac{1}{\sqrt{2}} \log | \frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1} | + c$

now substituting the value of t.

$I = \frac{1}{\sqrt{2}} \log | \frac{\sqrt{2} - 1 + \tan \frac{x}{2}}{\sqrt{2} + 1 - \tan \frac{x}{2}} | + c$