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Solve the integral

4-xx-2dx

Short Answer

Expert verified

sin-1(x-3)2+14sin(2sin-1(x-3))+C

Step by step solution

01

Step 1. Given 

4-xx-2dx

02

Step 2. Simplified the integral 

Theintegralissimplifiedbelow.4-x)(x-2)dx=-x2+6x-8dx=-(x-3)2+1dx=-u2+1dx[u=x-3,du=dx]

03

Step 3.  Using the identity 

Tosolvetheintegral,letx=sin(v).Now,derivationoftheaboveequationissolvedbelow.u=sin(v)du=cos(v)dvNow,usetheidentity-sin2(u)+1=cos2(u)So,(-u2+1)=(-sin2(v)+1)=cos2(v)=cos(y)

04

Step 4. Calculation 

Theintegralissolvedbelow.=u²+1dx=cos²(v)dv=12cos(2v)+12dv=12dv-12cos(2v)dv=v2+1212cos(w)dw[w=2v,dw=2dv]=v2+14sin(w)+C=v2+14sin(2v)+C=sin-1(u)2+14sin(2sin-1(u))+C=sin-1(x-3)2+14sin(2sin-1(x-3))+C

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