Question

Let $q\in N$ and $p\in (-\frac{\pi }{2},\frac{\pi }{2})$. Then, the value of $\underset{0}{\overset{p+q\pi }{\int }}|\cos x|dx$ is

• A. $2q+\sin p$
• B. $2q+1-\sin p$
• C. $2q+\cos p$
• D. $2q-1+\cos p$

Answer 1

The correct option is A $2q+\sin p$

Let $I=\underset{0}{\overset{p+q\pi }{\int }}|\cos x|dx$
$\Rightarrow I=\underset{0}{\overset{p}{\int }}|\cos x|dx+\underset{p}{\overset{p+q\pi }{\int }}|\cos x|dx$
Using the property:
$\underset{a}{\overset{a+nT}{\int }}f\left(x\right)dx=n\underset{0}{\overset{T}{\int }}f\left(x\right)dx\left[T\text{is the period of}f\right(x\left)\right]\Rightarrow I=\underset{0}{\overset{p}{\int }}|\cos x|dx+q\underset{0}{\overset{\pi }{\int }}|\cos x|dx\Rightarrow I=[\sin x{]}_0^p+2q\underset{0}{\overset{\frac{\pi }{2}}{\int }}\left(\cos x\right)dx\Rightarrow I=2q+\sin p$