1. Integrating Trigonometry

I tried doing 2 questions here which I fail to get the answer.

1) ∫ $tan^2 (1-3x) dx$

f(x) = tan(1-3x)
f'(x) = -3 sec^2 (1-3x)

By using the integration technique, that is ,
∫f(x)^n * f'(x) = f(x)^(n+1) /(n+1) + c

I'll get,
(1/ (-3sec^2 (1-3x)) * ( tan(1-3x)^3 / 3).
I know something is wrong somewhere, because the answer is $-1/3 tan(1-3x) -x + c$

2) The same goes to the second question, ∫ $sin^2 (2x) dx$.

2. Re: Integrating Trigonometry

1. Let $u = 1-3x$ so that $du = -3dx \text{ and } dx = -\dfrac{du}{3}$

Hence we have $\displaystyle -\dfrac{1}{3} \int \tan^2(u)\, du$ which may or may not be a standard integral. I'll outline the integration below:

We know from our trig identities that $\tan^2(u) = \sec^2(u)-1$ so let's sub that into the integral:

$\int \tan^2(u)\, du = \int (\sec^2(u) - 1)du$

We know that $\dfrac{d}{du} \tan(u) = \sec^2(u)$ so then the reverse must be true: $\int \sec^2(u) = \tan(u)$ (the constant will be added later)

Thus $\int (\sec^2(u)-1)du = \tan(u) - u + C$ where C is the constant of integration.

Back substitute to get your integral in terms of x

2. I'd go with a double angle identity:

$\cos(2\theta) = 1-2\sin^2(\theta)$ which can be rearranged to give $\sin^2(\theta) = \dfrac{1}{2} - \dfrac{\cos(2\theta)}{2}$

In your example $\theta = 2x$ and hence $\sin^2(2x) = \dfrac{1}{2} - \dfrac{\cos(4x)}{2}$.

The integral of cos(ax) is $\int \cos(ax) dx = \dfrac{1}{a} \sin(ax)$ for constant, non-zero a.

3. Re: Integrating Trigonometry

Thanks, I'm able to understand how to get the answer. So, should I substitute trigo identities when I come across similar questions like these?

4. Re: Integrating Trigonometry

Originally Posted by nav92
Thanks, I'm able to understand how to get the answer. So, should I substitute trigo identities when I come across similar questions like these?
If you can't solve them immediately then often an identity will make it easier to solve. The main one I come across is $\cos(2\theta) = 2\cos^2(\theta)-1 = 1-2\sin^2(\theta)$ because it allows you to change a squared term into a linear term.

5. Re: Integrating Trigonometry

You should know that $\displaystyle \int{\sec^2{x}\,dx} = \tan{x}$. You should also know that $\displaystyle 1 + \tan^2{x} \equiv \sec^2{x}$. So

\displaystyle \begin{align*}\int{\tan^2{(1 - 3x)}\,dx} &= \int{\sec^2{(1 - 3x)} - 1} \\ &= -\frac{1}{3}\int{-3\sec^2{(1 - 3x)}\,dx} - \int{1\,dx} \\ &= -\frac{1}{3}\int{\sec^2{(u)}\,du} - \int{1\,dx}\textrm{ after making the substitution }u = 1 - 3x \implies du = -3\,dx \\ &= -\frac{1}{3}\tan{(u)} - x + C \\ &= -\frac{1}{3}\tan{(1 - 3x)} - x + C \end{align*}

6. Re: Integrating Trigonometry

Originally Posted by nav92
I tried doing 2 questions here which I fail to get the answer.

1) ∫ $tan^2 (1-3x) dx$

f(x) = tan(1-3x)
f'(x) = -3 sec^2 (1-3x)

By using the integration technique, that is ,
∫f(x)^n * f'(x) = f(x)^(n+1) /(n+1) + c
The problem is that you don't have $\int f^n f' dx$ because you don't have the f'.
You can't just write it as
$\frac{1}{-3sec^2(1- 3x)} \int tan^2(1- 3x)(-3sec^2(1- 3x))dx$
because you cannot pass a function of x through the integral like that.

I'll get,
(1/ (-3sec^2 (1-3x)) * ( tan(1-3x)^3 / 3).
I know something is wrong somewhere, because the answer is $-1/3 tan(1-3x) -x + c$

2) The same goes to the second question, ∫ $sin^2 (2x) dx$.