1. The problem statement, all variables and given/known data I'm just trying to solve this [tex] \int {\sec^3{x} dx [/tex] 2. Relevant equations [tex] sec^2{x} = 1 + tan^2{x} [/tex] 3. The attempt at a solution well i was able to simplify it to this: [tex] \int {\sec{x}*\tan^2{x}} dx + \ln{|\sec{x} + \tan{x}|} [/tex] but I still was not able to find that new integral
INT[sec^3x]dx = INT[secx.sec^2x]dx By parts (differentiating secx and integrating sec^2x): = INT[secx.sec^2x]dx = secx.tanx - INT[secx.tan^2x]dx = secx.tanx - INT[secx(sec^2x - 1)]dx = secx.tanx - INT[sec^3x - secx]dx So we have: INT[sec^3x]dx = secx.tanx - INT[sec^3x - secx]dx INT[sec^3x]dx = secx.tanx - INT[sec^3x]dx + INT[secx]dx 2 INT[sec^3x]dx = secx.tanx + INT[secx]dx 2 INT[sec^3x]dx = secx.tanx + Ln|secx + tanx| INT[sec^3x]dx = 1/2 secx.tanx + 1/2 Ln|secx + tanx|
[tex]\int \frac{\cos x}{\left( 1-\sin ^{2}x\right) ^{2}}dx=\allowbreak \int \frac{1}{\left( 1-t^{2}\right) ^{2}}\,dt [/tex] for the last integral, use simple fractions. [itex] \sin x=t [/itex] has been used.
Another way: Since sec(x)= 1/cos(x), [tex]\int sec^3(x)dx= \int\frac{dx}{cos^3(x)}[/tex] which is an odd power of cos(x). Multiply numerator and denominator by cos(x): [tex]\int \frac{cos(x)dx}{cos^4(x)}= \int\frac{cos(x)dx}{(1- sin^2(x))^2}[/tex] Let u= sin(x) so du= cos(x)dx [tex]\int \frac{du}{(1-u^2)^2}= \int\frac{du}{(1-u)^2(1+u)^2}[/tex] and, again, use partial fractions.