Question

Find the integral of $\frac{\sin x-\cos x}{1+\sin x\cos x}$with limits zero to $\frac{\mathrm{\pi }}{2}$.

Answer 1

Find the integral of the given function.

Given: ${\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$

Let, $I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$

We know that ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

By using this we can write,

$$\begin{gathered} I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin \left(\frac{\mathrm{\pi }}{2}-x\right)-\cos \left(\frac{\mathrm{\pi }}{2}-x\right)}{1+\sin \left(\frac{\mathrm{\pi }}{2}-x\right)\cos \left(\frac{\mathrm{\pi }}{2}-x\right)}dx \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\cos x-\sin x}{1+\cos x\sin x}dx\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}\mathbf{}\mathbf{a}\mathbf{n}\mathbf{d}\mathbf{}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{(}\frac{\mathbf{\pi }}{\mathbf{2}}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{=}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{]} \\ \Rightarrow \mathrm{I}=-{\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\mathrm{sinx}-\mathrm{cosx}}{1+\mathrm{cosx}\mathrm{sinx}}\mathrm{dx} \\ \Rightarrow \mathrm{I}=-\mathrm{I}\mathbf{}\mathbf{}\mathbf{[}\mathbf{\because }{\mathbf{\int }}_{\mathbf{0}}^{\frac{\mathbf{\pi }}{\mathbf{2}}}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}}{\mathbf{1}\mathbf{+}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{x}\mathbf{}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{x}}\mathbf{}\mathbf{d}\mathbf{x}\mathbf{=}\mathbf{I}\mathbf{]} \\ \Rightarrow \mathrm{I}+\mathrm{I}=0 \\ \Rightarrow 2\mathrm{I}=0 \\ \Rightarrow \mathrm{I}=0 \end{gathered}$$

$\therefore {\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx=0$

Hence, the integral of $\frac{\sin x-\cos x}{1+\sin x\cos x}$with limits zero to $\frac{\mathrm{\pi }}{2}$ is $0$