Find an appropriate trigonometric substitution of the form x=f(t) to simplify the integral

Question

Can someone help me find x=f(t) with these questions?

Number 1: $$\int x\sqrt{8x^2+32x+29} \, dx $$

and

Number 2: $$\int {\frac{x}{\sqrt{-7x^2+28x-24}}} \, dx $$

I thought it was $\ x= {\frac{\sqrt{3}}{\sqrt{8}}}\tan(\theta)+2 $ for #1 and $\ x= {\frac{{2}}{\sqrt{7}}}\tan(\theta)-2 $ for #2.

Thank you so much!!!

Answer 1

for your first integral substitute $$\sqrt{8x^2+32x+29}=x\sqrt{8}+t$$ for your second integral Substitute $$\sqrt{-7x^2+28x-24}=\left(x-2-\sqrt{\frac{24}{7}}\right)t$$ it is the so-called Euulerian substitution

Answer 2

If you really want to use a trigonometric substitution, note that $$ 8x^2+32x+29=\frac{(4x+8)^2-6}{2} $$ so you can set $$ 4x+8=\sqrt{6}\sec\theta $$ but the integral will be quite messy.

For the second integral, note that $$ -7x^2+28x-24=4-7(x^2-4x+4) $$ so you can set $\sqrt{7}(x-2)=2\sin\theta$. Not the best way, in my opinion.