Question Please provide step by step solution! Thanks! (1 point) For each of the following integrals find an appropriate trigonometric substitution of the form x = f(t) to simplify the integral. a. ((3x? – 73312 dx x = os viento x = – 16x + 14 dx X = a./ V-25 - 3 - Trade X =

(a){:[I=int(3x^(2)-7)^(3//2)dx],[" Put "x=sqrt((7)/(3))sec(t)=sqrt((7)/(3))sec(t)],[dx=sqrt((7)/(3))*sec(t)tan(t)dt],[I=int(7sec^(2)(t)-7)^(3//2)(sec(t)*tan(t))dt],[I=int(7)^(3//2)(sec^(2)(t)-1)^(3//2)sec(t)tan(t)dt],[I=7sqrt7int(tan^(2)(t))^(3//2)*sec(t)tan(t)dt],[(:'1+tan^(2)(t)=sec^(2)(t))],[I=7sqrt7inttan^(3)(t)sec(t)tan(t)dt],[I=7sqrt7inttan^(4)(t)*sec(t)dt]:}(b){:[I=int(x^(2))/(sqrt(8x^(2)+8))dx],[" Put "x=tan(t)],[dx=sec^(2)(t)dt]:}{:[I=int(tan^(2)(t))/(sqrt8sqrt(tan^(2)(t)t+))sec^(2)(t)dt],[I=(1)/(2sqrt2)int(tan^(2)(t)sec^(2)(t))/(sqrt(sec^(2)(t)))dt],[I=(1)/(2sqrt2)int(tan^(2)(t)sec^(2)(t))/(sec(t))dt],[I=(1)/(2sqrt2)inttan^(2)(t)*sec(t)dt],[" (c) "I=int xsqrt(4x^(2)-16 x+14)dx],[I=int xsqrt(4(x^(2)-4x+4-4)+14)dx],[I=int xsqrt(4(x-2)^(2)-16+14)dx],[I=int xsqrt(4(x-2)^(2)-2)dx],[I=sqrt4int xsqrt((x-2)^(2)-(1)/(2))dx],[I=2int xsqrt((x-2)^(2)-(1)/(2))dx]:}lut x-2=(1)/(sqrt2)sec(phi){:[dx-0=(1)/(sqrt2)sec(t)tan(t)dt],[dx=(1)/(sqrt2)sec(t)tan(t)dt]:}New,{:[I=(2)/(sqrt2)int(2+(1)/(sqrt2)sec(t))sqrt((1)/(2)sec^(2)( ... See the full answer