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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

Answer 1

Christopher Allers

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

#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/sqrt (sec^2x)+C#

# = -1/secx+C#

# = -cosx+C#

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Answer 2

Cameron Alexander

Here is another substitution that will work.

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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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

#int tanx/secx#

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

#int senx#

-cos x + c