Question #8d4f0

Question

How do you find the integral $\int \frac{1}{x^{2} \cdot \sqrt{x^{2} - 9}} d x$ $int1/(x^2*sqrt(x^2-9))dx$ ?
How do you find the integral $\int \frac{x^{3}}{\sqrt{x^{2} + 9}} d x$ $intx^3/(sqrt(x^2+9))dx$ ?
How do you find the integral $\int x^{3} \cdot \sqrt{9 - x^{2}} d x$ $intx^3*sqrt(9-x^2)dx$ ?
How do you find the integral $\int \frac{x^{3}}{\sqrt{16 - x^{2}}} d x$ $intx^3/(sqrt(16-x^2))dx$ ?
How do you find the integral $\int \frac{\sqrt{x^{2} - 1}}{x} d x$ $intsqrt(x^2-1)/xdx$ ?
How do you find the integral $\int \frac{\sqrt{x^{2} - 9}}{x^{3}} d x$ $intsqrt(x^2-9)/x^3dx$ ?
How do you find the integral $\int \frac{x}{\sqrt{x^{2} + x + 1}} d x$ $intx/(sqrt(x^2+x+1))dx$ ?
How do you find the integral $\int \frac{d t}{\sqrt{t^{2} - 6 t + 13}}$ $intdt/(sqrt(t^2-6t+13))$ ?
How do you find the integral $\int x \cdot \sqrt{1 - x^{4}} d x$ $intx*sqrt(1-x^4)dx$ ?
How do you prove the integral formula $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + C$ $intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C$ ?

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Answer 1 · 2016-04-17T19:15:53

Please see below.

Form a
$\sqrt{a - b x^{2}}$ $sqrt(a-bx^2)$ $" "$

Recall $1 - {sin}^{2} θ = {cos}^{2} θ$ $1-sin^2theta = cos^2theta$

So turn $\sqrt{a (1 - \frac{b}{a} x^{2})}$ $sqrt(a(1-b/ax^2))$ into $\sqrt{a (1 - {sin}^{2} θ)}$ $sqrt(a(1-sin^2theta)$

using $sin θ = \sqrt{\frac{b}{a}} x$ $sintheta = sqrt(b/a)x$

Form b
$\sqrt{a x^{2} - b}$ $sqrt(ax^2-b)$ $" "$

Recall ${sec}^{2} θ - 1 = {tan}^{2} θ$ $sec^2theta - 1 = tan^2theta$

So turn $\sqrt{a (\frac{b}{a} x^{2} - 1)}$ $sqrt(a(b/ax^2-1))$ into $\sqrt{a ({sec}^{2} θ - 1)}$ $sqrt(a(sec^2theta - 1)$

using $sec θ = \sqrt{\frac{b}{a}} x$ $sectheta = sqrt(b/a)x$

Form c
$\sqrt{a + b x^{2}}$ $sqrt(a+bx^2)$ $" "$

Recall $1 + {tan}^{2} θ = {sec}^{2} θ$ $1+tan^2theta = sec^2theta$

So turn $\sqrt{a (1 + \frac{b}{a} x)}$ $sqrt(a(1+b/ax))$ into $\sqrt{a (1 + {tan}^{2} θ)}$ $sqrt(a(1+tan^2theta)$

Don't forget to find $d x$ $dx$ !