Question

What is the value of ${\int }_{-\pi /4}^{\pi /4}(\sin x-\tan x)dx$

• A. $-\frac{1}{\sqrt{2}}+\ln \left(\frac{1}{\sqrt{2}}\right)$
• B. $\frac{1}{\sqrt{2}}$
• C. $0$
• D. $\sqrt{2}$

Answer 1

Answer: (C)

$0$

We need to find ${\int }_{-\pi /4}^{\pi /4}(sinx-tanx)dx$

Let $f\left(x\right)=sinx-tanx$

Then $f(-x)=sin(-x)-tan(-x)=-sinx+tanx=-(sinx-tanx)=-f\left(x\right)$

Since $f(-x)=-f\left(x\right),f\left(x\right)$ is an odd function.

We know that ${\int }_{-a}^{a}f\left(x\right)dx=0$ if $f\left(x\right)$ is an odd function.

Therefore ${\int }_{-\pi /4}^{\pi /4}(sinx-tanx)dx=0$