Question

\[\int\frac{1}{\sec x + cosec x}\text{  dx }\]

Answer 1

\[\text{We have}, \]
\[I = \int\frac{1}{\sec x + \text{ cosec x}} \text{ dx}\]
\[I = \int\frac{1}{\frac{1}{\cos x} + \frac{1}{\sin x}} \text{ dx}\]
\[I = \frac{1}{2}\int\frac{2\sin x \cos x}{\sin x + \cos x} \text{ dx}\]
\[I = \frac{1}{2}\int\frac{1 + 2\sin x \cos x - 1}{\sin x + \cos x} \text{ dx}\]
\[I = \frac{1}{2}\int\frac{\sin^2 x + \cos^2 x + 2\sin x \cos x - 1}{\sin x + \cos x} \text{ dx}\]
\[I = \frac{1}{2}\int\frac{\left( \sin x + \cos x \right)^2 - 1}{\sin x + \cos x} \text{ dx}\]
\[I = \frac{1}{2}\int\frac{\left( \sin x + \cos x \right)^2}{\sin x + \cos x} \text{ dx} - \frac{1}{2}\int\frac{1}{\sin x + \cos x} \text{ dx}\]
\[I = \frac{1}{2}\int\left( \sin x + \cos x \right) \text{ dx}- \frac{1}{2}\int\frac{1}{\sin x + \cos x} \text{ dx}\]
\[I = \frac{1}{2}\left( - \cos x + \sin x \right) + C_1 - \frac{1}{2\sqrt{2}}\int\frac{1}{\frac{1}{\sqrt{2}}\left( \sin x + \cos x \right)} \text{ dx}\]
\[I = \frac{1}{2}\left( - \cos x + \sin x \right) + C_1 - \frac{1}{2\sqrt{2}}\int\frac{1}{\sin x \cos\frac{\pi}{4} + \cos x \sin\frac{\pi}{4}} \text{ dx}\]
\[I = \frac{1}{2}\left( - \cos x + \sin x \right) + C_1 - \frac{1}{2\sqrt{2}}\int\frac{1}{\sin\left( x + \frac{\pi}{4} \right)} \text{ dx}\]
\[I = \frac{1}{2}\left( - \cos x + \sin x \right) + C_1 - \frac{1}{2\sqrt{2}}\int cosec\left( x + \frac{\pi}{4} \right) \text{ dx}\]
\[I = \frac{1}{2}\left( - \cos x + \sin x \right) - \frac{1}{2\sqrt{2}}\text{ log}\left| \tan\left( \frac{x}{2} + \frac{\pi}{8} \right) \right| + C\]