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Respected sir / madam
I need value ofsin^4x / sin^4x+cos^4x

Tejaswani , 10 Years ago
Grade 12
anser 1 Answers
jagdish singh singh
\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 $\bf{I = \frac{x}{2}-\frac{1}{4\sqrt{2}}\ln\left|\frac{\sqrt{2}+\sin 2x}{\sqrt{2}-\sin 2x}\right|+\mathcal{C}}$
Last Activity: 10 Years ago
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