Question

how do you solve the integral of sqrt(sec^4x)dx from -pi/4 to pi/4?

Answer 1

\[I=\int\limits_{-\frac{ \pi }{ 4 }}^{\frac{ \pi }{4 }}\sec ^4x~dx\]
\[\left\{ \sec \left( -x \right) \right\}^4=\sec ^4x,so~\it ~is~an~even~function.\]
\[I=2\int\limits_{0}^{\frac{ \pi }{ 4 }}\sec ^4x~dx=\int\limits_{0}^{\frac{ \pi }{ 4 }}\sec ^2x~\sec ^2x~dx\]
\[=2\int\limits_{0}^{\frac{ \pi }{ 4 }}\left( 1+\tan ^2x \right)\sec ^2x~dx\]
\[=2\left[ \int\limits_{0}^{\frac{ \pi }{ 4 }}\sec ^2x~dx-\int\limits_{0}^{\frac{ \pi }{ 4 }}\tan ^2x~\sec ^2x~dx \right]+c\]
\[\left[ \int\limits \sec ^2x~dx=\tan x,\int\limits f^n \left( x \right)f'\left( x \right)dx=\frac{ f ^{n+1}\left( x \right) }{n+1 } \right]\]
complete it.

Answer 2

@surjithayer does it work that way with the square root of sec^4x?

Answer 3

now i have read \[\sqrt{\sec ^4x}dx=\sec ^2x,\it~ is~ an~ even~ function.\]
\[I=2\int\limits_{0}^{\frac{ \pi }{4 }}\sec ^2x~dx\]
it is much easier and i have already solved it.

Answer 4

@surjithayer thank you so much! i completely overlooked that, makes me feel kinda stupid. lol

Answer 5

NP