# solving integrals

### solving integrals

Hi!
I can't solve such integrals.

f$$\frac{dx}{\sqrt{2-x^{2}}}$$

f$$\frac{dx}{2x^{2}-6}$$

Guest

### Re: solving integrals

Those are fairly standard integrals. You should be able to find them in any Calculus text.

The first comes from the fact that $$sin^2(t)+ cos^2(t)= 1$$ so that $$cos^2(t)= 1- sin^2(t)$$ and $$cos(t)= \sqrt{1- sin^2(t)}$$.

Compare that to $$\sqrt{2- x^2}= \sqrt{2}\sqrt{1- \left(\frac{x}{\sqrt{2}}\right)^2}$$.
Use the substitution $$\frac{x}{\sqrt{2}}= sin(t)$$. Then $$\frac{1}{\sqrt{2}}dx= cos(t)dt$$ so $$dx= \sqrt{2}cos(t)dt$$ and $$\sqrt{1- \left(\frac{x}{\sqrt{2}}\right)^2}= \sqrt{1- sin^2(t)}= cos(t)$$. The integral becomes $$2\int cos(t) dt$$.

For the second one use $$a^2- b^2= (a- b)(a+ b)$$ with $$a= \sqrt{2}x$$ and $$b= 6$$ to write the integral as $$\int \frac{dx}{(\sqrt{2}x- \sqrt{6})(\sqrt{2}x+ \sqrt{6})}$$.

Now use "partial fraction": Find A and B so that $$\frac{1}{(\sqrt{2}x- \sqrt{6})(\sqrt{2}x+ \sqrt{6})}=$$$$\frac{A}{\sqrt{2}x- \sqrt{6}}+ \frac{B}{\sqrt{2}x+ \sqrt{6}}$$.

HallsofIvy

Posts: 341
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 119

### Re: solving integrals

From the formula: $$\int \frac{dx}{\sqrt{a^2-x^2}}=Arcsin \frac{x}{a}+C$$
So we have, $$\int \frac{dx}{\sqrt{2-x^2}}=Arcsin \frac{x}{\sqrt{2}}+C$$

For $$\int \frac{dx}{2x^{2}-6}$$, you can factor out 2, so you have,

$$\frac{1}{2} \int \frac{dx}{x^{2}-3} = \frac{1}{2} \int \frac{dx}{(x - \sqrt{3})(x+\sqrt{3})}$$

Then use partial fractions:

$$= \frac{1}{2} \int \frac{dx}{(2 \sqrt{3})(x+\sqrt{3})} - \frac{1}{2} \int \frac{dx}{(2 \sqrt{3})(x - \sqrt{3})}$$

$$=\frac{1}{4 \sqrt{3}} ( \ln \mid {x+ \sqrt{3}} \mid - \ln \mid {x - \sqrt{3}} \mid ) +C$$

$$=\frac{1}{4 \sqrt{3}} \ln \mid \frac { {x+ \sqrt{3} } } { {x - \sqrt{3} } } \mid +C$$

depending if its ok that there is a radical in the denominator..
Guest