JEE Advance - Mathematics (1994 - No. 4)
Show that $$\int\limits_0^{n\pi + v} {\left| {\sin x} \right|dx = 2n + 1 - \cos \,v} $$ where $$n$$ is a positive integer and $$\,0 \le v < \pi .$$
The integral of |sin(x)| from 0 to nπ + v is 2n - cos(v).
The integral of |sin(x)| from 0 to nπ + v is n + 1 - cos(v).
The integral of |sin(x)| from 0 to nπ + v is 2n + 1 - cos(v).
The integral of |sin(x)| from 0 to nπ + v is 2n + cos(v).
The integral of |sin(x)| from 0 to nπ + v is n + cos(v).
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