How to solve these?

Question

Inverse Trigonometric Functions

They are incomplete and I don't know how to complete them. Who can help me?

1st

$$ \int\frac 1{ x \sqrt{x^{6} - 4}}dx $$ I tried with: $$u = x^3 $$ $$du= 3x^2dx$$ but this is not completed,

2nd

oops, is not $$x^2$$ is only "x" $$ \int \frac 1{ x \sqrt{x-1}}dx $$ $$u = \sqrt x $$ $$du= \frac 1{\sqrt x} dx$$ where is the sqrt(x) on du?

3

$$ \int \frac {e^x}{ \sqrt{4-e^x}}dx $$ I think in this $$u=e^x $$ $$du = e^xdx$$ I have a bad English syntax, but I know read English++ Ok..

Answer 1

For $\int\frac 1{ x \sqrt{x^{6} - 4}}dx$, you didn't go far enough.

Write it as $\int\frac{x^2}{ x^3 \sqrt{x^{6} - 4}}dx $. Make your substitution of $y = x^3$, so $dy = 3 x^2 dx$, and we get $\int \frac{dy}{3 y \sqrt{y^2-4}}$ and this is essentially the same as $\int \frac 1{ x \sqrt{x^2-1}}dx$ with $4$ instead of $1$ (but write $4$ as $2^2$).

Answer 2

Hint

1. $\int\frac {dx}{ x \sqrt{x^{6} - 4}}$, let $x^6=4\sec^2 u$.

2. $\int \frac {dx}{ x \sqrt{x-1}}$, let $x=\sec^2 u$.

3. $\int \frac {e^x }{ \sqrt{4-e^x}}dx$, let $e^x=4\cos^2 u$.

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Keep in mind these trigonometric identities: $\cos^2u + \sin^2u = 1$ and $1 + \tan^2 u = \sec^2 u$.

For $(1)$, observe that in the denominator, you have $\sqrt{x^6-4}$. Now look at the trigonometric identities, you observe that $\tan^2 u = \sec^2 u - 1$. Which is pretty similar to $x^6 - 4$. Number $(2)$ is similar.

For number $(3)$, observe that in the denominator, you have $\sqrt{4-e^x}$. Now look at the trigonometric identities, you observe that $\sin^2u = 1 - \cos^2u$, which is pretty similar to $4-e^x$.

Answer 3

For $\int \frac{dx}{ x \sqrt{x^2-1}}$ let $x = \cosh(y)$, so $\sqrt{x^2-1} = \sinh(y)$ and $dx = \sinh(y) dy$.

The integral becomes $\int \frac{\sinh(y) dy}{\cosh(y)\sinh(y)} =\int \frac{ dy}{\cosh(y)} =\int \frac{2 dy}{e^y + e^{-y}} =\int \frac{2 e^y \,dy}{e^{2y}+1} =\int \frac{2 dz}{z^2+1} $ with $z = e^y$. This last is standard, so just substitute back.

Answer 4

Hint for (3) question

Let $u=4-e^x $ ; $du =- e^xdx$. Substitute and integrate.