# How to eveluate $\int\frac{\sec^4x}{\tan^4x+\tan^2x}dx$?

How to Evaluate $$\int\frac{\sec^4x}{\tan^4x+\tan^2x}dx=?$$

My try:

let $$\tan x=t$$, $$\sec^2x\ dx=dt$$, $$dx=\frac{dt}{t^2+1}$$ $$\int\frac{(t^2+1)^2}{t^4+t^2}\frac{dt}{t^2+1}$$ $$\int\frac{(t^2+1)}{t^4+t^2}dt$$ $$\int\frac{t^2}{t^4+t^2}dt+\int \frac{1}{t^4+t^2}dt$$ $$\int\frac{dt}{t^2+1}+\int \frac{dt}{t^2(t^2+1)}$$ First integration can be evaluated but I don't know how to evaluate second integral. Should I use partial fraction?please help me. thanks

Factor out $$\tan^2x$$ from denominator & use $$\tan^2x+1=\sec^2x$$, $$\int\frac{\sec^4x}{\tan^4x+\tan^2x}dx$$

$$=\int\frac{\sec^4x}{\tan^2x(\tan^2x+1)}dx$$

$$=\int\frac{\sec^4x}{\tan^2x(\sec^2x)}dx$$

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

Here it is an alternative approach for the sake of curiosity: \begin{align*} \frac{\sec^{4}(x)}{\tan^{4}(x) + \tan^{2}(x)} = \frac{1}{\sin^{4}(x) + \sin^{2}(x)\cos^{2}(x)} = \frac{1}{\sin^{2}(x)} = \csc^{2}(x) \end{align*}

• that's nice. can't i evaluate by partial fractions?
– user766881
Jun 19, 2020 at 4:39
• You don't need to. Notice that the derivative of $\cot(x)$ is $-\csc^{2}(x)$. Jun 19, 2020 at 16:25

Your substitution is correct. But in second line, you can factor out $$t^2$$ from the denominator & proceed as follows $$\int \frac{t^2+1}{t^4+t^2}dt=\int\frac{t^2+1}{t^2(t^2+1)}dt=\int \frac{dt}{t^2}=-\frac1t+C$$