question is $$\int\frac{1}{\sin^6(x) + \cos^6(x)}\,dx$$
My method :
$$\sin^6\left(x\right)+\cos^6\left(x\right)$$
$$=\left(\sin^2\left(x\right)+\cos^2\left(x\right)\right)\left(\sin^4\left(x\right)-\cos^2\left(x\right)\sin^2\left(x\right)+\cos^4\left(x\right)\right)$$
$$=1-3\cos^2\left(x\right)\sin^2\left(x\right)$$
so integration is
$$\int\dfrac{1}{1-3\cos^2\left(x\right)\sin^2\left(x\right)}\,dx$$
divide numerator and denominator by $$cos^2x$$
$$\int\dfrac{\sec^2\left(x\right)}{\sec^2\left(x\right)-3\tan^2\left(x\right)} \, dx$$
Now put $$t=\tan x$$
So $$dt = \sec^2x\,dx$$
So
$$\int\dfrac{t}{1-2t^2} \, dt$$
$$=\frac{1}{2\sqrt{2}}\ln\left(\left|\frac{\sqrt{2}\tan x+1}{\sqrt{2}\tan x-1}\right|\right)$$
But answer is $$\arctan\left(\tan\left(x\right)-\cot\left(x\right)\right)$$
So can there be two answers of different forms of same integration?
Thanks!
Best Answer
You made a mistake : \begin{aligned}\int{\frac{\mathrm{d}x}{1-3\cos^{2}{x}\sin^{2}{x}}}&=\int{\frac{\sec^{2}{x}}{\sec^{2}{x}-3\color{red}{\sin^{2}{x}}}\,\mathrm{d}x}\\ &=\int{\frac{\mathrm{d}y}{1+y^{2}-3\left(1-\frac{1}{1+y^{2}}\right)}}\\ &=\int{\frac{y^{2}+1}{y^{4}-y^{2}+1}\,\mathrm{d}y}\\ &=\int{\frac{1+\frac{1}{y^{2}}}{\left(y-\frac{1}{y}\right)^{2}+1}\,\mathrm{d}y}\\ &=\int{\frac{\mathrm{d}u}{u^{2}+1}}\\ &=\arctan{u}+C\\ &=\arctan{\left(y-\frac{1}{y}\right)}+C\\ \int{\frac{\mathrm{d}x}{1-3\cos^{2}{x}\sin^{2}{x}}}&=\arctan{\left(\tan{x}-\cot{x}\right)}+C\end{aligned}