what is the area under the curve f(x)=tanx/(secx+tanx) and the x-axis, from x=0 to x=pi

Answer 1

Perhaps easier to see

Multiply sin x/(1+ sin x) by (1-sin x)/(1-sin x) to get

sec x tan x - tan²x which integrates to sec x - tan x + x.

Answer 2

This is a tough one!! Here goes! First begin by converting all the functions into sin(x) and cos(x) versions, or

f(x) = (sin(x)/cos(x)) / ( 1/cos(x) + sin(x)/cos(x))

This simplifies to:

sin(x) / (1 + sin(x))

Rewrite the numerator to : sin(x) + 1 - 1 (which has no effect on its value)

So now you have (sin(x) +1 - 1) / (sin(x) + 1) which then splits into 1 - (1/(sin(x) + 1)).

When you integrate the 1 you get an x. So move to integrating the second part: (1/(sin(x) + 1))

Now, multiply the function by (sin(x)-1)/(sin(x)-1) aka "1" and by the difference of squares you get

sin(x)-1/(sin²x - 1) = sin(x) - 1 (-cos²(x)) = (1-sin(x))/(cos²x) = sec²x - sin(x)/cos²(x)

so the integral of sec²x = tan(x). Now we have two pieces done - namely x - (tan(x) - "something else"). Let's do the something else:

to integrate sin(x)/sin(x)/cos²(x) - this is done by substitution. Let u = cosx. By the chain rule, du = -sin(x)dx. So we already have most of this in the numerator except the -1, so we multiply and divide by -1 and get - (integral sign) u^-2 du. By the power rule this integrates to -1/u = -1/cos(x) which is your something else.

So the anti-derivative is x - (tan(x) + 1/cos(x)) = 1 - tanx(x) - 1/cos(x). Just evalate at the end points.