Question: I am trying to solve the following problem using strong induction, the problem is the following:

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> For any positive integer $n$, let $T\_n$ be the number $1$ if $n<4$ and the number $T\_{n − 1} +T\_{n − 2} +T\_{n − 3}$ if $n\ge 4$.
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> 
> We have $T\_1 =1, T\_2 =1, T\_3 =1, T\_4 =T\_3 +T\_2+T\_1=1+1+1=3$, $T\_5 =T\_4 +T\_3 +T\_2 =3+1+1=5$, etc.
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> 
> Prove that: $\forall n \in \Bbb{Z}^+$, $T\_n <2^n$ $\;$
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> 
> 

Sadly, I don't even know where to start this question!

Answer: 
Another way:

$$I=\int\frac{dx}{\cos^4x+\sin^4x}=\int\frac{(1+\tan^2x)\sec^2x dx}{1+\tan^4x}$$

Setting $\displaystyle\tan x=u,$

$$I=\frac{(1+u^2)du}{1+u^4}=\int\frac{1+\dfrac1{u^2}}{u^2+\dfrac1{u^2}}du$$

$$=\int\frac{1+\dfrac1{u^2}}{\left(u-\dfrac1u\right)^2+2}du$$

Set $\displaystyle u-\dfrac1u=v$
Let me try something mixing the work of chndn and lab bhattacharjee. First let's rewrite the denominator as chndn has tried.

$$\int\dfrac1{\cos^4x+\sin^4x}dx=\int\dfrac1{(\cos^2x+\sin^2x)^2-2\sin^2x\cos^2x}dx=$$
$$\int\dfrac1{1-\frac12\sin^22x}dx$$

Now let's do a move similar to lab bhattacharjee's.

$$\int\dfrac{\csc^22xdx}{\csc^22x-\frac12}=\int\dfrac{\csc^22xdx}{\cot^22x+\frac12}$$

Now let $u=\cot 2x$.