Integral of $2x^2 \sec^2{x} \tan{x}$

Question

I've been trying for a while to find $\int{( 2x^2 \sec^2{x} \tan{x} )} dx$, using integration by parts.

I always end up getting a more complicated integral in the second part of the equation.

For example: $$ \int{( 2x^2 \sec^2{x} \tan{x} )} dx = \\ 2x^2 \tan^2x - \int{\tan{x} \cdot \frac{d}{dx}(2x^2 \tan{x})} \\ \frac{d}{dx}(2x^2 \tan{x})=4x\tan{x} + 2x\sec^2{x} \rightarrow \\ 2x^2 \tan^2x - \int{4x \tan^2{x}+2x\tan{x}\sec^2{x}} $$

I've tried integrating with different value for $u$ and $v$, such as:

$$ 1:( 2x^2 \sec^2{x} \tan{x} ), \\ \tan{x} : 2x^2 \sec^2{x}, \\ 2x^2: \sec^2{x} \tan{x}, \\ \sin{x}: 2x^2 \sec^3{x} $$ etc, however, haven't succeeded.

Answer 1

You can proceed from where you ended up by integrating $4x\tan^2 x$ using the trigonometric identity $\sec^2x = 1 + \tan^2x$ to express it as $4x(\sec^2x - 1)$ first.

Always remember that $\tan^2 x$ is easy to integrate after applying that identity.

Answer 2

Use integration by parts with $u=x^2$ and $v=\tan^2(x)$. Then, we have

$$\int 2x^2 \tan(x)\sec^2(x)\,dx=x^2\tan^2(x)-2\int x\tan^2(x)\,dx$$

Continue with a subsequent integration by parts with $u=x$ and $v=\tan(x)-x$ to obtain

$$\begin{align} \int 2x^2 \tan(x)\sec^2(x)\,dx&=x^2\tan^2(x)-2\left(x\tan(x)-x^2-\int(\tan(x)-x)\,dx\right)\\\\ &=x^2\tan^2(x)-2x\tan(x)+x^2-2\log(\cos(x))+C\\\\ &=x^2\sec^2(x)-2(x\tan(x)+\log(\cos(x)))+C \end{align}$$

Answer 3

Hint consider whole $\sec^2 (x)\tan (x)=v $ as its integral then apply by parts to get $x^2\tan^2 (x)-(\int 4x\tan^2 (x)) $ again consider $\tan^2 (x) $ as v and apply by parts . Note use $\tan^2 (x)=\sec^2 (x)-1$ and proceed with the integration join both these parts to get the integral

Answer 4

We have,

$\int \sec ^{2}x\tan {x}dx$

Put v = $\sec{x}$

Then dv = 2$\sec{x}\tan{x}$

So $\int (v) dv=\frac{1}{2}v^{2}=\frac{1}{2}\sec ^{2}x.$

Now use this result in question,

2$\int{(x^2 \sec^2{x} \tan{x} )} dx$

Integrating by parts, integrate $\sec^2x \tan{x}$ and differentiate $x^2$

We have,

2$\left[\frac{1}{2}x^{2}\sec ^{2}x-\int x\sec ^{2}x\,dx \right]$

Now again integration by parts on $ x\sec ^{2}x\,dx $ by integrating $\sec ^{2}x$ and differentiate x.

You got an answer.