Question

integrate sec^4x.tanx.​

Answer 1

Answer:

$\int \sec^{4} x \tan(x) dx \\ = > \int\sec ^{3} (x) \sec(x) \tan(x) dx$

Now use substitution method

Let secx = t

=> secx tanx dx =dt

now put it in I

We write sec^3x =t^3

and

secx tanx dx = dt

$\rightarrow \: \int {t}^{3} dt$

$\rightarrow \frac{ {t}^{4} }{4} + c \\ \rightarrow \: \frac{ \sec ^{4} (x) }{4} + c$

This is the required answer .

formula used

$\frac{d}{dx} ( \sec(x) ) = \sec(x) \tan(x) \\ and \\ \int \: {x}^{n } = \frac{ {x}^{n + 1} }{n + 1}$

Answer 2

write sec^4xtanx as

int sec^3xsecx tanx dx

let secx =. t

secx tanx dx =dt

int t^3dt

= t^4/4 +c

put t = secx

= sec^4x/ 4 +. c