Solving an immediate indefinite integral of a composite function

[greg_rack]

Homework Statement

$$\int (\frac{1}{cos^2x\cdot tan^3x})dx$$

Relevant Equations

none

That's my attempt:
$$\int (\frac{1}{cos^2x\cdot tan^3x})dx = \int (\frac{1}{cos^2x}\cdot tan^{-3}x) dx$$
Now, being $\frac{1}{cos^2x}$ the derivative of $tanx$, the integral gets:
$$-\frac{1}{2tan^2x}+c$$
But there is something wrong... what?

 

[PeroK]

I'm missing what's supposed to be wrong here!

 

[Mark44]

Homework Statement:: $$\int (\frac{1}{cos^2x\cdot tan^3x})dx$$
Relevant Equations:: none

That's my attempt:
$$\int (\frac{1}{cos^2x\cdot tan^3x})dx = \int (\frac{1}{cos^2x}\cdot tan^{-3}x) dx$$
Now, being $\frac{1}{cos^2x}$ the derivative of $tanx$, the integral gets:
$$-\frac{1}{2tan^2x}+c$$
But there is something wrong... what?

Convert the denominator into a fraction with only sine and cosine factors. IOW, convert the $\tan^3$ factor. The resulting expression becomes a reasonably simple problem that can be done by nothing more complicated than substitution.

 

[PeroK]

Convert the denominator into a fraction with only sine and cosine factors. IOW, convert the $\tan^3$ factor. The resulting expression becomes a reasonably simple problem that can be done by nothing more complicated than substitution.

Is the OP's answer wrong?

 

[greg_rack]

Is the OP's answer wrong?

The correct answer is $-\frac{1}{2sin^2x}+c$

 

[Mark44]

Is the OP's answer wrong?

I didn't go through his work very closely -- I just showed an easier way to go about it.

 

[PeroK]

The correct answer is $-\frac{1}{2sin^2x}+c$

That's what you got, isn't it?

 

[greg_rack]

Convert the denominator into a fraction with only sine and cosine factors. IOW, convert the $\tan^3$ factor. The resulting expression becomes a reasonably simple problem that can be done by nothing more complicated than substitution.

I did it, and it took me to the correct answer.
But then I found out the method explained in the OP which looks legit to me, but takes me to a wrong answer, and so I wanted to understand why

 

[greg_rack]

That's what you got, isn't it?

I got $-\frac{1}{2tan^2x}+c$

 

[PeroK]

I got $-\frac{1}{2tan^2x}+c$

Same thing. Don't forget the constant of integration!

 

[greg_rack]

Same thing. Don't forget the constant of integration!

What do you mean? How could $tan^2x=sin^2x$?

 

[PeroK]

What do you mean? How could $tan^2x=sin^2x$?

$$\frac{1}{tan^2x}= \frac{1}{sin^2x} + C$$ For some constant $C$.

 

[etotheipi]

@greg_rack it's because$$-\frac{1}{2\sin^2{x}} + c_1 = -\frac{1}{2} \csc^2 x + c_1 = -\frac{1}{2} (1+ \cot^2{x}) + c_1 = - \frac{1}{2\tan^2{x}} + c_2$$

 

[vela]

I got $-\frac{1}{2tan^2x}+c$

When you get two answers that don't appear to be equivalent, try plotting them both or better yet their difference.

https://www.wolframalpha.com/input/?i=Plot[{1/Sin[x]^2}+-+1/Tan[x]^2,+{x,+-Pi,+Pi}]

 

[greg_rack]

Thanks a lot guys!

 

[Mark44]

This is the same sort of situation you get with the integral $\int \sin(x)\cos(x)dx$.
You can do this in at least three different ways, of which I'll show two.
1. Let $u = \sin(x)$, so $du = \cos(x)dx$. The resulting antiderivative is $-\cos^2(x) + C$.
2. Let $u = \cos(x)$, so $du = -\sin(x)dx$. The resulting antiderivative is $\sin^2(x) + C$
These results look different, but because $\sin^2(x) + \cos^2(x) = 1$, $\sin^2(x)$ and $-\cos^2(x)$ differ only by a constant, 1.

The third way involves writing the integrand as $\sin(2x)$.

 

[PeroK]

Thanks a lot guys!

Everyone has to learn this lesson once! Usually it's:
$$\int \sin (2x) dx = -\frac 1 2 \cos(2x) + C$$ And
$$\int \sin (2x) dx = \int 2 \sin x \cos x dx = -\cos^2 x + C$$ Which are related by the trig identity $$\cos(2x) = 2\cos^2 x - 1$$

 

[etotheipi]

another non-trig one that can sometimes catch people out$$\frac{1}{a} \ln{ax} + c_1 = \int \frac{\mathrm{d}x}{ax} = \frac{1}{a} \int \frac{\mathrm{d}x}{x} = \frac{1}{a} \ln{x} + c_2$$

 

[Infrared]

Another "favorite": In the integral $\int\frac{1}{\sqrt{1-x^2}}dx$, depending on whether you substitute $x=\sin(u)$ or $x=\cos(u),$ you will get either $\arcsin(x)+c_1$ or $-\arccos(x)+c_2.$

 

[PeroK]

$$\int \frac{1}{\sqrt{x^2 + a^2}} \ dx = \sinh^{-1}(\frac x a) + C_1 = \ln(x + \sqrt{x^2 + a^2}) + C_2$$