$\int \tan x\,dx$ equals
A$\log|\cos x|+C$
B$\sec^2 x+C$
C$\log|\sin x|+C$
D$\log|\sec x|+C$
Answer & Solution
Correct answer: D. $\log|\sec x|+C$
1. Write $\tan x=\dfrac{\sin x}{\cos x}$.
2. Put $t=\cos x$, $dt=-\sin x\,dx$.
3. Integral $=-\int \dfrac{dt}{t}=-\log|t|=-\log|\cos x|$.
4. Since $-\log|\cos x|=\log|\sec x|$, the answer is $\log|\sec x|+C$ (option A is the unsimplified negative).
_Source: NCERT Class 12 Mathematics Ch 7 "Integrals", p.300_
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