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Question Number 336 by Vishal Bhardwaj last updated on 25/Jan/15

$$\int \frac{1}{{sin}^{4}x+{cos}^{4}x+{sin}^{2}x{cos}^{2}x}dx$$

Commented by 123456 last updated on 22/Dec/14

$$f\left(x\right)=\frac{1}{{\sin }^{4}x+{\cos }^{4}x+{\sin }^{2}x{\cos }^{2}x}$$$$=\frac{1}{{\sin }^{4}x+{\cos }^{4}x+{2\sin }^{2}x{\cos }^{2}x-{\sin }^{2}x{\cos }^{2}x}$$$$=\frac{1}{{({\sin }^{2}x+{\cos }^{2}x)}^2-{\sin }^{2}x{\cos }^{2}x}$$$$=\frac{1}{1-{\sin }^{2}x{\cos }^{2}x}$$

Commented by 123456 last updated on 23/Dec/14

$$\frac{{\tan }^{-1}\left[\frac{\sqrt{3}}{2}\tan \left(2x\right)\right]}{\sqrt{3}}$$

Answered by prakash jain last updated on 27/Dec/14

$$\frac{1}{1-{\sin }^{2}x{\cos }^{2}x}=\frac{4}{4-{\sin }^{2}2x}$$$$\tan 2x=t$$$${\sec }^{2}2x\cdot 2\cdot dx=dt$$$$dx=\frac{dt}{2(1+t^2)}$$$${\sin }^{2}2x=\frac{t^2}{1+t^2}$$$$\mathrm{Given}\mathrm{integral}I$$$$I=\int \frac{\frac{4dt}{2(1+t^2)}}{4-\frac{t^2}{(1+t^2)}}$$$$=\int \frac{4dt}{2(4+4t^2-t^2)}$$$$=2\int \frac{dt}{3t^2+4}$$$$=2\int \frac{dt}{t^2+\frac{4}{3}}=\frac{2}{3}\int \frac{dt}{t^2+{\left(\frac{2}{\sqrt{3}}\right)}^2}$$$$=\frac{2}{3}\cdot \frac{\sqrt{3}}{2}{\tan }^{-1}\frac{\sqrt{3}t}{2}$$$$=\frac{1}{\sqrt{3}}{\tan }^{-1}\frac{\sqrt{3}\tan 2x}{2}$$

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