Question

Integrals Giving Inverse Trigonometric Functions

1. ∫(xdx)/√(3-2x-x²)

2. ∫(dy)/(16+y²) from -4 to 4

Answer 1

1)

$\int{\frac{x}{\sqrt{3-2x-x^2}}dx}$

focus on the denominator

$3-2x-x^2$

$\left(-1\right)\left(x^2+2x-3\right)$

$\left(-1\right)\left(x^2+2x+1-1-3\right)$

$\left(-1\right)\left(\left(x+1\right)^2-4\right)$

$4-\left(x+1\right)^2$

$\int{\frac{x}{\sqrt{4-\left(x+1\right)^2}}dx}$

$\int{\frac{x+1-1}{\sqrt{4-\left(x+1\right)^2}}dx}$

$\int{\frac{x+1}{\sqrt{4-\left(x+1\right)^2}}dx}-\ \int{\frac{1}{\sqrt{4-\left(x+1\right)^2}}dx\ }$

$let\ u=\sqrt{4-\left(x+1\right)^2}$

$du=\ -\frac{2\left(x+1\right)}{\sqrt{4-\left(x+1\right)^2}}dx=>dx=\ -\frac{\sqrt{4-\left(x+1\right)^2}}{2\left(x+1\right)}du$

$\int{\frac{x+1}{\sqrt{4-\left(x+1\right)^2}}dx}=\ -\frac{1}{2}\int d u=-\frac{u}{2}+c=\ -\frac{\sqrt{4-\left(x+1\right)^2}}{2}+c$

$let\ v=x+1=>dv=dx$

$\int{\frac{1}{\sqrt{4-\left(x+1\right)^2}}dx\ }=\ \int{\frac{1}{\sqrt{2^2-v^2}}dv=\sin^{-1}\left(\frac{v}{2}\right)}+d=\sin^{-1}{\left(\frac{x+1}{2}\right)}+d$

$-\frac{\sqrt{4-\left(x+1\right)^2}}{2}-\ \sin^{-1}{\left(\frac{x+1}{2}\right)}+K$

2)

$\int_{-4}^{4}\frac{dy}{16+y^2}$

$\int_{-4}^{4}\frac{dy}{4^2+y^2}$

$\frac{1}{4}\tan^{-1}{\left(\frac{y}{4}\right)}\ \ \left(-4,\ 4\right)$

$\frac{1}{4}\left(\tan^{-1}{\left(1\right)-\tan^{-1}{\left(-1\right)}}\right)$

$\frac{1}{4}\left(\frac{\pi}{4}--\frac{\pi}{4}\right)=\frac{\pi}{8}$