Consider the given integral
I = $\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}$ 2 sin x cos x $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ (sin x) dx
Let t = sinx
⇒ dt = cos x dx
When x = $\frac{\mathit{\pi }}{2}$ , t = 1
When x = 0 , t = 0
Now, ∫ 2 sin x cos x $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ (sin x) dx
= ∫ 2t $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ t dt
= $\left[\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{t}\right]$ ∫ 2t dt - ∫ $\left[\frac{\mathit{d}}{\mathit{d}\mathit{t}}\left(\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{t}\right)\int 2\mathit{t}\mathit{d}\mathit{t}\right]$ dt
$\left[\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{t}\right]$ $\left[2.\frac{{\mathit{t}}^4}{2}\right]$ - ∫ $\left(\frac{1}{1+{\mathit{t}}^2}\times 2.\frac{{\mathit{t}}^2}{2}\right)$ dt
= ${\mathit{t}}^2\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ t - ∫ $\frac{{\mathit{t}}^2}{1+{\mathit{t}}^2}$ dt
= ${\mathit{t}}^2\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ t - ∫ $\left[1-\frac{1}{1+{\mathit{t}}^2}\right]$ dt
= ${\mathit{t}}^2\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ t - t + $$tan^{-1} t
∴ I = $\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}$ 2 sin x cos x $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ (sin x) dx
= ${\left[{\mathit{t}}^2\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{t}-\mathit{t}+\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{t}\right]}_0^1$
= $\left[1^2\mathit{t}\mathit{a}{\mathit{n}}^{-1}1-1+\mathit{t}\mathit{a}{\mathit{n}}^{-1}1\right]$ - $\left[0^2\mathit{t}\mathit{a}{\mathit{n}}^{-1}0-0+\mathit{t}\mathit{a}{\mathit{n}}^{-1}0\right]$
= $\left[1\times \frac{\mathit{\pi }}{4}-1+\frac{\mathit{\pi }}{4}\right]$ - 0
= $\frac{\mathit{\pi }}{4}-1+\frac{\mathit{\pi }}{4}$
= $\frac{\mathit{\pi }}{2}$ - 1
OR
I = $\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}$ $\frac{\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathit{x}}{\mathit{s}\mathit{i}{\mathit{n}}^4\mathit{x}+\mathit{c}\mathit{o}{\mathit{s}}^4\mathit{x}}$ dx ... (1)
Using the property $\underset{0}{\overset{\mathit{a}}{\int }}$ f (x) dx = $\underset{0}{\overset{\mathit{a}}{\int }}$ f (a - x) dx
I = dx
⇒ I = $\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}\frac{\left(\frac{\mathit{\pi }}{2}-\mathit{x}\right)\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathit{x}}{\mathit{c}\mathit{o}{\mathit{s}}^4\mathit{x}+\mathit{s}\mathit{i}{\mathit{n}}^4\mathit{x}}$ dx ... (2)
Adding (1) and (2)
2I = $\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}\frac{\left(\frac{\mathit{\pi }}{2}.\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\right)}{\mathit{s}\mathit{i}{\mathit{n}}^4\mathit{x}+\mathit{c}\mathit{o}{\mathit{s}}^4\mathit{x}}$ dx
⇒ I = $\frac{\mathit{\pi }}{4}\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}|\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathit{x}}{\mathit{s}\mathit{i}{\mathit{n}}^4\mathit{x}+\mathit{c}\mathit{o}{\mathit{s}}^4\mathit{x}}|$ dx
= $\frac{\mathit{\pi }}{4}\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}|\frac{\frac{\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathit{x}}{\mathit{c}\mathit{o}{\mathit{s}}^4\mathit{x}}}{\frac{\mathit{s}\mathit{i}{\mathit{n}}^4\mathit{x}}{\mathit{c}\mathit{o}{\mathit{s}}^4\mathit{x}}+1}|$ dx
$\frac{\mathit{\pi }}{4}\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}\frac{\mathit{t}\mathit{a}\mathit{n}\mathit{x}\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{x}}{\mathit{t}\mathit{a}{\mathit{n}}^4\mathit{x}+1}$ dx
Put $\mathit{t}\mathit{a}{\mathit{n}}^2$ x = z
∴ 2 tan x $\mathit{s}\mathit{e}{\mathit{c}}^2$ x dx = dz
⇒ tan x $\mathit{s}\mathit{e}{\mathit{c}}^2$ x dx = $\frac{\mathit{d}\mathit{z}}{2}$
When x = 0 , z = 0 and when x = $\frac{\mathit{\pi }}{2}$ , z = ∞
∴ I = $\frac{\mathit{\pi }}{4}\underset{0}{\overset{\mathit{\infty }}{\int }}\frac{\frac{\mathit{d}\mathit{z}}{2}}{{\mathit{z}}^2+1}$
⇒ I = $\frac{\mathit{\pi }}{8}\underset{0}{\overset{\mathit{\infty }}{\int }}\frac{\mathit{d}\mathit{z}}{1+{\mathit{z}}^2}$
= $\frac{\mathit{\pi }}{8}|\mathit{t}\mathit{a}{\mathit{n}}^{-1}\left(\mathit{z}\right){|}_0^{\mathit{\infty }}$
= $\frac{\mathit{\pi }}{8}\mathit{t}\mathit{a}{\mathit{n}}^{-1}\mathit{\infty }-\mathit{t}\mathit{a}{\mathit{n}}^{-1}0$
= $\frac{\mathit{\pi }}{8}\left(\frac{\mathit{\pi }}{2}-0\right)$
= $\frac{{\mathit{\pi }}^2}{16}$