Question

The value of definite integral ${\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{{\cos }^{2}x+{\sin }^{4}x}dx$, is

• A. $\frac{1}{2\sqrt{3}}\log \left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)-\frac{\pi }{4}$
• B. $-\frac{1}{3\sqrt{3}}\log \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)+\frac{\pi }{4}$
• C. $-\frac{1}{2\sqrt{3}}\log \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)+\frac{\pi }{4}$
• D. $\frac{1}{3\sqrt{3}}\log \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)+\frac{\pi }{4}$

Answer 1

The correct option is C $-\frac{1}{2\sqrt{3}}\log \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)+\frac{\pi }{4}$

$I={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{{\cos }^{2}x+si{n}^{4}x}dx$
$={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{1-{\sin }^{2}x(1-{\sin }^{2}x)}dx={\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{1-{\sin }^{2}x.{\cos }^{2}x}dx$
$=4{\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{4-4{\sin }^{2}x.{\cos }^{2}x}dx$
$I=4{\int }_0^{\frac{\pi }{4}}\frac{\sin x+\cos x}{4-{\{1-{(\sin x-\cos x)}^2\}}^2}dx$
Put $\sin x-\cos x=t⟹(\cos x+\sin x)dx=dt$
$I=4{\int }_{-1}^0\frac{dt}{4-{(1-t^2)}^2}=4{\int }_{-1}^0\frac{dt}{(2-1+t^2)(2+1-t^2)}$
$=4{\int }_{-1}^0\frac{dt}{(1+t^2)(3-t^2)}=4{\int }_{-1}^0\{\frac{1}{4(3-t^2)}+\frac{1}{4(1+t^2)}\}dt$ (using partial fraction)
$={\int }_{-1}^0\frac{dt}{3-t^2}+{\int }_{-1}^0\frac{dt}{1+t^2}$
$={\left[\frac{1}{2\sqrt{3}}\log \left(\frac{\sqrt{3}+t}{\sqrt{3}-t}\right)\right]}_{-1}^0+{\left[{\tan }^{-1}t\right]}_{-1}^0$
$=\frac{1}{2\sqrt{3}}[0-\log \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)]+[0-{\tan }^{-1}(-1)]$

$I=-\frac{1}{2\sqrt{3}}\log \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)+\frac{\pi }{4}$