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There's

$$ \int \frac{\sin^2x}{\cos^4x} \,dx$$

Is there the way to solve it bypass general substitute like $\,t = \tan\dfrac x2$?

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  • $\begingroup$ have you tried manipulating the expression into sec then tan, then using u = tan x? $\endgroup$
    – Cato
    Feb 21, 2017 at 10:38

2 Answers 2

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$$ \int \frac{\sin^2x \operatorname{dx}}{\cos^4x}= \int \frac{\tan^2x \operatorname{dx}}{\cos^2x}=\int \tan^2 x\sec^2x dx=\frac{\tan^3x}{3}+C $$

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By parts:

$$\begin{cases}u=\sin x,&u'=\cos x\\{}\\v'=\frac{\sin x}{\cos^4x},&v=-\frac1{3\cos^3}\end{cases}\implies\int\frac{\sin^2x}{\cos^4x}dx=-\frac13\frac{\sin x}{\cos^2x}+\frac13\int\frac{dx}{\cos^2x}$$

and now remember that $\;(\tan x)'=\frac1{\cos^2x}\;$

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