Grade 12Integral Calculus

Respected sir / madam

I need value of ∫sin^4x / sin^4x+cos^4x

Tejaswani

10 Years agoGrade 12

1 Answer

jagdish singh singh

10 Years ago

$\hspace{-0.6 cm}$Let $\bf{I = \frac{1}{2}\int\frac{2\sin^4 x}{\sin^4 x+\cos^4 x}dx = \frac{1}{2}\int\frac{\sin^4 x+\cos^4 x+\sin^4x-\cos^4 x}{\sin^4 x+\cos^4 x}dx}$\\\\\\So $\bf{I = \frac{1}{2}\int 1dx-\frac{1}{2}\int\frac{(\cos^2 x-\sin^2 x)}{\sin^4 x+\cos^4 x}dx = \frac{1}{2}x-\frac{1}{2}\int\frac{2\cos 2x}{2-(\sin 2x)^2}dx}$\\\\\\Now Put $\bf{\sin 2x=t\;,}$ Then $\bf{2\cos 2x dx =dt}$$

$\hspace{-0.6 cm}$So We get $\bf{I = \frac{1}{2}x-\frac{1}{2}\int\frac{1}{2-t^2}dt = \frac{x}{2}-\frac{1}{4\sqrt{2}}\ln\left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\mathcal{C}}$\\\\\\So we get $\bf{I = \frac{x}{2}-\frac{1}{4\sqrt{2}}\ln\left|\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right|+\mathcal{C}}$$