Expert: Ahmed Salami - 7/30/2009

Question
QUESTION: Can you tell me, can i use integration by part or u sub for 2secX/2+tanX, or both, i cant find secx in the integral tables anywhere. thanks for your recent answers.

Dave


ANSWER: Hi Dave,
First, secx = 1/cosx
2secx / (2 + tanx) = (2/cosx) / (2 + tanx)
         = 2/(2 + tanx)cosx  
         = 2/(2cosx + tanx.cosx)
         = 2/(2cosx + sinx)
To integrate this kind of expression, we use the substitution t = tan(x/2)
Combining with other trigonometric identities, we can show then that
cosx = (1-t²)/(1+t²)
sinx = 2t/(1+t²)

x/2 = arctan t
x = 2.arctan t
dx/dt = 2/(1+t²)
dx = 2.dt/(1+t²)
And so,
∫[2secx / (2 + tanx)]dx = ∫[2/(2cosx + sinx)]dx
         = 2∫[2.dt/(1+t²)] / [2(1-t²)/(1+t²)] + [2t/(1+t²)]
         = 2∫ dt / (1-t²) + 2t
         = 2∫ dt/ 1 + 2t - t²
Let me know how it goes, i'm just leading you on here.
This might help a bit;
http://2000clicks.com/MathHelp/GeometryTrigEquivTanHalfAngle.htm

Regards

---------- FOLLOW-UP ----------

QUESTION: Hi Ahmed, can you fault me here when applicable please.

Secx = 1/cos x, tan x =sinx/cosx

so for 2secx/2+tanx, can i say 1/cosx/sinx/cosx =(1/cosx)*(cosx/sinx) so (cosx/cosx)*(1/sinx) =(1+(1/sinx)).putting in the numerals we get 2*1/(1/sinx)+2, so is this not the integral of 2/(sinx)+2???? Which ends up as 2Ln(sinx)+2x??? I have consulted one of my books and was told that it is just the Ln of what ever the denominator is.Does the end product depend upon whether it is a definate or indefinate??? Thanks for your reply.

Answer
Hi Dave,
This is where you got it all mixed up;  2secx/2+tanx is not (1/cosx)/(sinx/cosx) because it is not just tanx as the denominator but '2 + tanx' and you cant isolate the '2' as you are adding and not multiplying.
And, the integral of a rational function is only equal to the ln of the denominator if the numerator is the derivative of the denominator e.g
∫(2x+3)/(x²+3x+2) dx = ln(x²+3x+2) + c
or
∫cosx/sinx dx = ln(sinx) + c
The difference between a definite and indefinite integral is that the definite integral is always evaluated between two limits to get a value while an indefinite integral is left as an expression with an arbitrary constant.

Regards