Integral how to choose the good substitution? I have a problem with this integral. I don't know what kind of substitution to use.
$$\int t\sqrt{1+9t^4}dt$$
 A: $$
\Big( t\,dt\Big)
$$
What you see above suggests at least considering $u=t^2$ so that $du=2t\,dt$, and then what you see above is $\frac 1 2 \, du$.
Since you have $9t^4 = (3t^2)^2$, using $u=3t^2$ works; then $du = 6t\,dt$ and $\frac 1 6 \, du = t\,dt$.
And then you have $9t^4 = (3t^2)^2 = u^2$, so $\displaystyle\int \sqrt{1+9t^4}\ t\, dt = \frac 1 6 \int\sqrt{1+u^2}\  du$.
A: $$\int t\sqrt{1+9t^4}\space\text{d}t=$$

Substitute $u=t^2$ and $\text{d}u=2t\space\text{d}t$:

$$\frac{1}{2}\int\sqrt{1+9u^2}\space\text{d}u=$$

Substitute $u=\frac{\tan(s)}{3}$ and $\text{d}u=\frac{\sec^2(s)}{3}\space\text{d}s$.
Then $\sqrt{1+9u^2}=\sqrt{1+\tan^2(s)}=\sec(s)$ and $s=\arctan\left(3u\right)$:

$$\frac{1}{6}\int\sec^3(s)\space\text{d}s=$$
$$\frac{\tan(s)\sec(s)}{12}+\frac{1}{12}\int\sec(s)\space\text{d}s=$$
$$\frac{\tan(s)\sec(s)}{12}+\frac{\ln\left|\tan(s)+\sec(s)\right|}{12}+\text{C}=$$
$$\frac{\tan(s)\sec(s)+\ln\left|\tan(s)+\sec(s)\right|}{12}+\text{C}=$$
$$\frac{\tan\left(\arctan\left(3u\right)\right)\sec\left(\arctan\left(3u\right)\right)+\ln\left|\tan\left(\arctan\left(3u\right)\right)+\sec\left(\arctan\left(3u\right)\right)\right|}{12}+\text{C}=$$
$$\frac{3u\sqrt{1+9u^2}+\sinh^{-1}(3u)}{12}+\text{C}=$$
$$\frac{3t^2\sqrt{1+9t^4}+\sinh^{-1}(3t^2)}{12}+\text{C}$$
A: $$\int t \sqrt{1-9t^4} \, dt $$
$$=\frac{1}{6}\int 6t \sqrt{1-9t^4} \, dt $$
$$=\frac{1}{6}\int \sqrt{1-9t^4} \, d(3t^2) $$
$$=\frac{1}{6}\int \sqrt{1-(3t^2)^2} \, d(3t^2) $$
Now apply standard formula.
A: $3t^2 = \sin x$, $t\,dt=\frac{1}{6} \cos x \,dx$. So, we have,
$$\int \cos x (1/6) \cos x \, dx =\int \frac{1}{12}(1+\cos2x) \, x 
= \frac {1} {12} (x + \sin x \cos x)+c=\frac{1}{12}\left(\arcsin(3t^2)+3t^2\sqrt{1-9t^4}\right)+c.$$
