# Trigonometric substitution integrals

CellCoree
what would be the appropriate trigonometic substition for the following:

1.) $$\int(7x^2-7)^{3/2}) dx$$

2.) $$\int \frac{x^2}{\sqrt{(7x^2 + 8)}}$$

ok i have a book with a list of trigonometric functions in front of me, but i don't get it. i don't see anything that i can sub in for.

ok i kinda get the first one...

$$7(x^2-1)^{3/2}$$

$$sin^2(\theta) + cos^2(\theta) = 1/$$ so...
divided by $$cos^2(\theta)$$ and get $$tan^2(\theta)$$

$$7(x^2 - tan^2(\theta))^{3/2}$$

$$7(x^3 - tan^3(\theta))$$

$$7x^3 - 7tan^3(\theta)$$
$$7x^3=7tan^3(\theta)$$

divide by 7... (lol, don't think i can do this)

$$x^3 = tan^3(\theta)$$

cube root...

$$x= tan(\theta)$$ is that right?

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CellCoree
lol, thanks. i just made a dumb mistake

CellCoree
$$\int \frac{x^2}{\sqrt{(7x^2 + 8)}}$$

can i do $$\int \frac{x^2}{\sqrt{(7x^2 + 7 + 1)}}$$
?

$$sin^2(\theta) + cos^2(\theta) = 1$$ so...

well i know it's $$sec^2(\theta)$$ from solving the previous problem

$$\int \frac{x^2}{\sqrt{(7x^2 + 7 + sec^2(\theta))}}$$

$$\int \frac{x^2}{\sqrt{7(x^2 + 1 + sec^2(\theta))}}$$

agian, there's another 1! so i use $$sec^2(\theta)$$
(just working with the bottom part now)

$$\sqrt{7(x^2 + sec^2(\theta) + sec^2(\theta)$$

if i squared 7, it would be a mess, can someone help?

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Parth Dave
You might want to review trigonometric substitution in your textbook.

Homework Helper
Check what are you doing again, it seems a bit... mixed up...

Here's a hint

$$\sqrt{7}x = \sqrt{8}tan\theta$$

so you should have

$$\sqrt{7x^2 + 8} = \sqrt{8}sec\theta$$

CellCoree
i get it now! thanks.

the trig. sub. should be:

$$\frac{\sqrt(8)tan(t)}{\sqrt(7)}$$