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The integral \(\int \frac{\sin ^2 x}{\cos ^4 x} d x\) is
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Verified Answer
The correct answer is:
a polynomial of degree 3 in \(\tan x\)
\(\begin{aligned}
& \text {Let } I_{2,4}=\int \frac{\sin ^2 x}{\cos ^4 x} d x=\frac{\sin x}{3 \cdot \cos ^3 x}-\frac{1}{3} I_{0,2} \\
& =\frac{\sin x}{3 \cos ^3 x}-\frac{1}{3} \tan x+c \\
& =\frac{1}{3} \tan x\left(1+\tan ^2 x\right)-\frac{1}{3} \tan x+c \\
& =\frac{1}{3} \tan ^3 x+c
\end{aligned}\)
ALTERNATE :
\(\begin{aligned}
& \int \frac{\sin ^2 x}{\cos ^4 x} d x=\int \tan ^2 x \cdot \sec ^2 x d x \\
& \text { put } \tan x=t \\
& =\int t^2 \cdot d t=\frac{1}{3} t^3+c=\frac{1}{3} \tan ^3 x+c
\end{aligned}\)
& \text {Let } I_{2,4}=\int \frac{\sin ^2 x}{\cos ^4 x} d x=\frac{\sin x}{3 \cdot \cos ^3 x}-\frac{1}{3} I_{0,2} \\
& =\frac{\sin x}{3 \cos ^3 x}-\frac{1}{3} \tan x+c \\
& =\frac{1}{3} \tan x\left(1+\tan ^2 x\right)-\frac{1}{3} \tan x+c \\
& =\frac{1}{3} \tan ^3 x+c
\end{aligned}\)
ALTERNATE :
\(\begin{aligned}
& \int \frac{\sin ^2 x}{\cos ^4 x} d x=\int \tan ^2 x \cdot \sec ^2 x d x \\
& \text { put } \tan x=t \\
& =\int t^2 \cdot d t=\frac{1}{3} t^3+c=\frac{1}{3} \tan ^3 x+c
\end{aligned}\)
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