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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.

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    Split terms and put as I put in my answer. – 2017-01-05

4 Answers 4

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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.

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    I knew I was missing something, didn't think of using that trigonometric identity. – 2017-01-05
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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

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

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    How'd you spot $\int{2\sec^2{x}\tan{x} = \tan^2{x}}$? – 2017-01-05
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    I guess this is how the question was intended to be solved, as it was originally formatted as ".. $\int{x^2 2\sec^2{x}\tan{x}}$" – 2017-01-05
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    $\frac{d\tan^2(x)}{dx}=2\tan(x)\sec^2(x)$ – 2017-01-05
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    I mean, looking at the question, how would you have spotted that? I get it's correct. Is it something you've remembered, brute force? – 2017-01-05
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    Tobi, I remember that the derivative of the tangent function is the square of the secant function. So, $2\tan(x) \sec^2(x)=2\tan(x)\frac{d\tan(x)}{dx}=\frac{d\tan^2(x)}{dx}$. – 2017-01-05
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    Isn't $\int{\tan{x}} \equiv \ln{\sec{x}}$? – 2017-01-05
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    $\int \tan(x)\,dx=-\log(\cos(x))+C$ since $\frac{d\log(\cos(x))}{dx}=\frac{1}{\cos(x)}\frac{d\cos(x)}{dx}=-\tan(x)$. And $- \log(\cos(x))=\log(\sec(x))$. So, yes we can write it either way. – 2017-01-05
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    Shouldn't ...$\tan{x}+x^2−2\log$... be ...$\tan{x}+3x^2−2\log$...? $-2(-x^2+\int{-x})$ – 2017-01-05
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    Tobi, I'm sorry. I don't understand your question. Have a close look at the steps. $-x^2+\frac12x^2=-\frac12 x^2$ – 2017-01-05
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    Isn't $(-x^2+\int{-x}) =(-x^2-\int{x}) = -\frac32 x^2$ Never mind, there, it's minus the integral – 2017-01-05
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    I surrender to this question. – 2017-01-05
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    It is $-x^2-\int (-x)\,dx=-x^2-(-x^2/2)=-x^2+x^2/2=-x^2/2$. – 2017-01-05
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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

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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.