?(dx/?36-x^2) My solution: x=6sin? dx= 6cos?d? Therefore, =?(6cos?d?/6cos?) =?d? What to do next???
S sareen New member Joined Oct 9, 2009 Messages 24 Oct 28, 2009 #1 ?(dx/?36-x^2) My solution: x=6sin? dx= 6cos?d? Therefore, =?(6cos?d?/6cos?) =?d? What to do next???
D Deleted member 4993 Guest Oct 28, 2009 #2 sareen said: ?(dx/?36-x^2) My solution: x=6sin? ? = sin[sup:jz7wrb9v]-1[/sup:jz7wrb9v](x/6) dx= 6cos?d? Therefore, =?(6cos?d?/6cos?) =?d? = ? + C = sin[sup:jz7wrb9v]-1[/sup:jz7wrb9v](x/6) + C What to do next??? Click to expand...
sareen said: ?(dx/?36-x^2) My solution: x=6sin? ? = sin[sup:jz7wrb9v]-1[/sup:jz7wrb9v](x/6) dx= 6cos?d? Therefore, =?(6cos?d?/6cos?) =?d? = ? + C = sin[sup:jz7wrb9v]-1[/sup:jz7wrb9v](x/6) + C What to do next??? Click to expand...
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Oct 28, 2009 #3 ∫dx(36−x2). Let x = 6sin(θ), then dx = 6cos(θ)dθ.\displaystyle \int \frac{dx}{\sqrt(36-x^{2})}. \ Let \ x \ = \ 6sin(\theta), \ then \ dx \ = \ 6cos(\theta)d\theta.∫(36−x2)dx. Let x = 6sin(θ), then dx = 6cos(θ)dθ. Ergo, 6∫cos(θ)dθ(36−36sin2(θ))\displaystyle Ergo, \ 6\int \frac{cos(\theta)d\theta}{\sqrt(36-36sin^{2}(\theta))}Ergo, 6∫(36−36sin2(θ))cos(θ)dθ = ∫cos(θ)dθcos2(θ) = ∫dθ = θ+C\displaystyle = \ \int\frac{cos(\theta)d\theta}{\sqrt cos^{2}(\theta)} \ = \ \int d\theta \ = \ \theta+C= ∫cos2(θ)cos(θ)dθ = ∫dθ = θ+C Now, x = 6sin(θ), sin(θ) = x6, hence θ = arcsin(x/6)\displaystyle Now, \ x \ = \ 6sin(\theta), \ sin(\theta) \ = \ \frac{x}{6}, \ hence \ \theta \ = \ arcsin(x/6)Now, x = 6sin(θ), sin(θ) = 6x, hence θ = arcsin(x/6) Therefore, ∫dx(36−x2) = arcsin(x/6)+C.\displaystyle Therefore, \ \int \frac{dx}{\sqrt(36-x^{2})} \ = \ arcsin(x/6)+C.Therefore, ∫(36−x2)dx = arcsin(x/6)+C.
∫dx(36−x2). Let x = 6sin(θ), then dx = 6cos(θ)dθ.\displaystyle \int \frac{dx}{\sqrt(36-x^{2})}. \ Let \ x \ = \ 6sin(\theta), \ then \ dx \ = \ 6cos(\theta)d\theta.∫(36−x2)dx. Let x = 6sin(θ), then dx = 6cos(θ)dθ. Ergo, 6∫cos(θ)dθ(36−36sin2(θ))\displaystyle Ergo, \ 6\int \frac{cos(\theta)d\theta}{\sqrt(36-36sin^{2}(\theta))}Ergo, 6∫(36−36sin2(θ))cos(θ)dθ = ∫cos(θ)dθcos2(θ) = ∫dθ = θ+C\displaystyle = \ \int\frac{cos(\theta)d\theta}{\sqrt cos^{2}(\theta)} \ = \ \int d\theta \ = \ \theta+C= ∫cos2(θ)cos(θ)dθ = ∫dθ = θ+C Now, x = 6sin(θ), sin(θ) = x6, hence θ = arcsin(x/6)\displaystyle Now, \ x \ = \ 6sin(\theta), \ sin(\theta) \ = \ \frac{x}{6}, \ hence \ \theta \ = \ arcsin(x/6)Now, x = 6sin(θ), sin(θ) = 6x, hence θ = arcsin(x/6) Therefore, ∫dx(36−x2) = arcsin(x/6)+C.\displaystyle Therefore, \ \int \frac{dx}{\sqrt(36-x^{2})} \ = \ arcsin(x/6)+C.Therefore, ∫(36−x2)dx = arcsin(x/6)+C.
