Question

${\int }_{-\pi /2}^{\pi /2}\frac{dx}{{\theta }^{sinx}+1}$ is equal to

• A. $-\frac{\pi }{2}$
• B. $\frac{\pi }{2}$
• C. 0
• D. 1

Answer 1

Answer: (B)

$\frac{\pi }{2}$

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{{\theta }^{\sin x}+1}⟶1$

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

$\therefore I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{{\theta }^{(\frac{n}{2}-\frac{n}{2}-\sin x)}+1}$

$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{{\theta }^{-\sin x}+1}⟶2$

Add equation 1 and 2

$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{{\theta }^{\sin x}+1}+{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{dx}{{\theta }^{-\sin x}+1}$

$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}[\frac{1}{{\theta }^{\sin x}+1}+\frac{1}{{\theta }^{-\sin x}+1}]dx$

$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left[\frac{{\theta }^{-\sin x}+1+{\theta }^{\sin x}+1}{1+{\theta }^{\sin x}+{\theta }^{-\sin x}+1}\right]dx$

$2I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}1\cdot dx$

$2I={\left|x\right|}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}=\frac{\pi }{2}+\frac{\pi }{2}$

$2I=\pi$

$I=\frac{\pi }{2}$