i have 2 calculus problem, please help me solve?

Question

* note: big f (F) is the antidervitive of the function, little f (f) is the normal function, df is the dervitive of the function f , and i'll use @ as the degree symbol

1) F (xsin^2 (x) dx)
2) dr/d@ = sin^4 (@pie)

10 points for the first person to solve both

Answer 1

Qn:1

x Sin^2 x = x( 1-Cos 2x)/2

= x/2-xCos2x/2

Integral of x/2= x^2/4

Integral of xCos2x /2 = x/2Int(Cos2x)-Inte( 1/2*Inte(Cos2x))

(by using Integration by parts)

= (x/2)*(Sin2x)/2 - Inte(Sin2x/4) = (x Sin2x)/4 + (Cos2x)/4

Final Answer = 1/4{ x^2-xSin2x-Cos2x} +C

where C is the constant of Integration.

Qn:2

dr/dt= Sin^4(Pi*t)

Note: ( You have used @ inplace of "theta' & I have used t)
Let A= Pi*t
Sin^3 A= 1/4 {3 SinA-Sin 3A}
Therefore Sin^4 A= 1/4{ 3 Sin^2 A- SinASin3A}

= 1/4{ 3( 1-Cos 2A)/2 - 1/2( Cos 2A - Cos 4A)}
=1/8{ 3- 3Cos2A-Cos2A+Cos4A}
=1/8{ 3-4Cos 2A+Cos4A}

Re substituting A= Pi*t & integrating

Integral = 1/8 { 3t +4 Sin(2(Pi*t))/2Pi)- Sin(4Pi*t)/(4Pi)}

r = 1/32Pi{ 12Pi + 8 Sin(2Pi*t) -Sin(4Pi*t)} +C

Answer 2

Sorry, but I do not understand the degree symbol (@).
dr/d@ makes no sense. Where is df used?

Answer 3

∫ xsin²(x) dx =

-(cos²x )/4 - ( xsin(x)cos(x) )/2 - x²/4 + C

Clarify the second problem.