Question

The value of ${\int }_{-\pi }^{\pi }{\sin }^{3}x{\cos }^{2}xdx$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is D $0$

applying, ${\int }_a^b+\left(x\right)dx={\int }_a^{b}f(a+b-x)dx...$

${\int }_{-\pi }^{\pi }si{n}^{3}xco{s}^{2}dx={\int }_{-\pi }^{\pi }si{n}^{3}x(1-si{n}^{2}x)dx$

$\Rightarrow {\int }_{-\pi }^{\pi }(si{n}^{3}x-si{n}^{5}x)dx=I....\left(1\right)$

$\Rightarrow I={\int }_{-\pi }^{\pi }\left(si{n}^3\right(\pi -\pi -x)-si{n}^{5}x(\pi -\pi -x\left)\right)dx$

$I={\int }_{-\pi }^{\pi }(-si{n}^{3}x+si{n}^{5}x)dx....\left(2\right)$

$\left(1\right)+\left(2\right)\Rightarrow 2\pi =0$

$I=0$