## University Calculus: Early Transcendentals (3rd Edition)

$$\int e^x\sec^3(e^x)dx=\frac{1}{2}\sec e^x\tan e^x+\frac{1}{2}\ln|\sec e^x+\tan e^x|+C$$
$$A=\int e^x\sec^3(e^x)dx=\int\sec^3(e^x)d(e^x)$$ Set $a=e^x$, we have $$A=\int\sec^3ada$$ Take $u=\sec a$ and $dv=\sec^2ada$ That makes $du=\sec a\tan ada$ and $v=\tan a$ According to Integration by Parts rule, we have $$A=\sec a\tan a-\int\sec a\tan^2ada$$ $$A=\sec a\tan a-\int\sec a(\sec^2a-1)da$$ $$A=\sec a\tan a-\int\sec^3ada+\int\sec ada$$ $$A=\sec a\tan a+\ln|\sec a+\tan a|-\int\sec^3ada$$ The integral $\int\sec^3ada$ equals $A$, therefore, $$A=\sec a\tan a+\ln|\sec a+\tan a|-A$$ $$2A=\sec a\tan a+\ln|\sec a+\tan a|+C$$ $$A=\frac{1}{2}\sec a\tan a+\frac{1}{2}\ln|\sec a+\tan a|+C$$ $$A=\frac{1}{2}\sec e^x\tan e^x+\frac{1}{2}\ln|\sec e^x+\tan e^x|+C$$