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\(\int x\left(\tan ^2 x\right) d x=\)
Options:
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2045 Upvotes
Verified Answer
The correct answer is:
\(x \tan (x)-\log _e(\sec x)-\frac{x^2}{2}+c\)
\(\begin{aligned}
I & =\int x\left(\tan ^2 x\right) d x=x \int\left(\sec ^2 x-1\right) d x -\int\left(1 \int\left(\sec ^2 x-1\right) d x\right) d x \\
& =x(\tan x-x)-\int(\tan x-x) d x \\
& =x \tan x-x^2-\log _e(\sec x)+\frac{x^2}{2}+C \\
& =x \tan x-\log _e(\sec x)-\frac{x^2}{2}+C
\end{aligned}\)
Hence, option (a) is correct.
I & =\int x\left(\tan ^2 x\right) d x=x \int\left(\sec ^2 x-1\right) d x -\int\left(1 \int\left(\sec ^2 x-1\right) d x\right) d x \\
& =x(\tan x-x)-\int(\tan x-x) d x \\
& =x \tan x-x^2-\log _e(\sec x)+\frac{x^2}{2}+C \\
& =x \tan x-\log _e(\sec x)-\frac{x^2}{2}+C
\end{aligned}\)
Hence, option (a) is correct.
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