Is it possible to solve this integrate by integrate substitution method?

#int tanx/secx#

Not this away: #int [(senx)/(cosx)]##cosx/1#

#int senx#

-cos x + c

2 Answers
Jim H
Apr 2, 2018

Here is one way. (I think it is harder than what you did.)

Explanation:

#int tanx/secx dx#

Let #u = tanx#. Then #du = sec^2x dx# and

#secx = sqrt(tan^2x+1) = sqrt(u^2+1)# So that #du = (u^2+1) dx# and #dx = 1/(u^2+1) du#

#int tanx/secx dx = int u/sqrt(u^2+1) * 1/(u^2 +1) du#

# = int u (u^2+1)^(-3/2) du#

# = 1/2 int 2u (u^2+1)^(-3/2) du#

# = 1/2[-2/1 (u^2+1)^(-1/2)]+C#

# = -1/(u^2+1)^(1/2)+C#

# = -1/(u^2+1)^(1/2)+C#

# = -1/sqrt (sec^2x)+C#

# = -1/secx+C#

# = -cosx+C#

Jim H
Apr 3, 2018

Here is another substitution that will work.

Explanation:

Let #u = secx# so #du = secxtanxdx# and

#int tanx/secx dx = int 1/sec^2x secx tanx dx#

# = int 1/u^2 du#

# = -u^(-1)+C = -1/u +C#

# = -1/secx +C#

# = -cosx +C#

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