Why is the general indefinite integral of sinx/1-(sinx)^2 equal to secx?

[fk378]

Homework Statement

Find the general indefinite integral of sinx/1-(sinx)^2

The Attempt at a Solution

I arrived at tanxsecx(dx), rewrote it as sinx(1/cos^2 x) = cosxtanx = sinx

However, I know that the correct answer is secx. WHY? Can anyone explain why using trig functions (ie, without using u as a substitution)?

 

[Dick]

sin(x)/cos(x)^2 isn't equal to cos(x)*tan(x).

 

[dynamicsolo]

I arrived at tanxsecx(dx), rewrote it as sinx(1/cos^2 x) = cosxtanx = sinx

However, I know that the correct answer is secx. WHY? Can anyone explain why using trig functions (ie, without using u as a substitution)?

Don't simplify it: just leave it as (sin x) / (cos^2 x) . Now, could you use a u-substitution on

\int \frac{sin x}{cos^{2} x} dx?

 

[fk378]

sin(x)/cos(x)^2 isn't equal to cos(x)*tan(x).

Why can't it equal if you substitute?
sinx/(cosx)(cosx)
(sinx)(1/(cosx)(cosx)
the antiderivative of that gives...
(cosx)(secx)(secx)

 

[fk378]

Don't simplify it: just leave it as (sin x) / (cos^2 x) . Now, could you use a u-substitution on

\int \frac{sin x}{cos^{2} x} dx?

Is there any way to get to the answer without substituting? What is making my method invalid?

 

[HallsofIvy]

Because "tan(x) cos(x)" is equal to (sin(x)/cos(x))(cos(x))= sin(x). You have sin(x)/cos^2(x). The "second" cosine is in the denominator, not the numerator. That is (sin(x)/cos(x))(1/cos(x)= tan(x)sec(x). The integral of that is sec(x)+ C= 1/cos(x) + C which is exactly what you get if you leave it as sin(x)/cos^2(x) and make the substitution u= cos(x).