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Question Number 201027 by sonukgindia last updated on 28/Nov/23

Answered by Atomist last updated on 28/Nov/23

$$sechx=\frac{1}{coshx}$$$$coshx=\frac{e^x+e^{-x}}{2}$$$$\int sechxdx=\int \frac{2}{e^x+e^{-x}}dx=$$$$=\int \frac{2}{e^x+e^{-x}}\times \frac{e^x}{e^x}dx=$$$$\int \frac{2e^x}{e^{2x}+1}dx=$$$$u=e^x$$$$2dv=2e^{x}dx$$$$=\int \frac{2u}{1+u^2}du=2\int \frac{du}{1+u^2}=$$$$\int \frac{dx}{a^2+x^2}=\frac{1}{a}arctg\frac{x}{a}+C$$$$=2arctg\left(e^x\right)+C$$$$2(\frac{\pi }{2}-\frac{\pi }{4})=\frac{\pi }{2}$$$$Answer.\frac{\pi }{2}$$

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