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Section 8.3 #4: ∫sin3(3x)dx
Solution: Recall the Pythagorean identity
cos2(θ)+sin2(θ)=1.
We applying this rearranged to sin2(θ)=1−cos2(θ) with θ=3x yields
sin3(3x)=sin(3x)sin2(3x)=sin(3x)(1−cos2(3x)).
Therefore we may compute
∫sin3(3x)dx=∫sin(3x)(1−cos2(3x))dx=∫sin(3x)dx−∫sin(3x)cos2(3x)dxu1=3x,u2=cos(3x)=−13∫sin(u1)du1+13∫u22du2=13cos(u1)−19u32+C=13cos(3x)−19cos3(3x)+C.
Section 8.3 #9: ∫cos2(3x)dx
Solution: Recall the "reduction" identity
cos2(θ)=1+cos(2θ)2.
Apply that identity with θ=3x to see
cos2(3x)=1+cos(6x)2.
Therefore compute
∫cos2(3x)dx=∫1+cos(6x)2dx=12∫1dx+12∫cos(6x)dx=x2+112sin(6x)+C.
Section 8.4 #22: ∫x2√36−x2dx
Solution: Recall that the Pythagorean identity may be rearranged to say
√1−sin2(θ)=cos(θ).
Also recall the "reduction" identity
sin2(θ)=1−cos(2θ)2.
Let x=6sin(θ) so that dx=6cos(θ)dθ. Substituting this variable into the integral yields
∫x236−x2dx=∫36sin2(θ)√36−36sin2(θ)6cos(θ)dθ=36⋅66∫sin2(θ)cos(θ)√1−sin2(θ)dθ=36∫sin2(θ)cos(θ)cos(θ)dθ=36∫sin2(θ)dθ=362∫1dθ−362∫cos(2θ)dθ=18θ−9sin(2θ)+C.
Solving our assignment x=6sin(θ) for θ yields θ=arcsin(x6). Recall the double angle identity
cos(2ψ)=2sin(ψ)cos(ψ).
Applying this shows us that
−9sin(2θ)=−18sin(θ)cos(θ)=−18(x6)cos(θ)=−3cos(θ).
io determine cos(θ) we must draw a triangle:
To find the side labelled "?", we use the Pythagorean theorem:
?2+x2=62.
Solving for ? yields
?=√36−x2.
Therefore cos(θ)=√36−x26. So, finally, we compute
∫x2√36−x2dx=18arcsin(x6)−3xcos(θ)=18arcsin(x6)−x2√36−x2+C.
Section 8.5 #6: ∫29x2−1dx
Solution: The denominator is a difference of squares (i.e. a2−b2=(a−b)(a+b)) and so it factors:
9x2−1=(3x)2−1=(3x−1)(3x+1).
Therefore, we must apply partial fractions to
29x2−1=A3x−1+B3x+1.
Multiply by the common denominator to get
2=A(3x+1)+B(3x−1).
Combining like-terms on the right yields
2=(3A+3B)x+(A−B).
Noting that the "coefficient of x" on the left is zero, we equate coefficients to get the following system of two linear equations in two variables (A and B):
{3A+3B=0(i)A−B=2(ii).
By (ii) we conclude that A=2+B. Plug this into (i) to get
3(2+B)+3B=0,
or equivalently,
6B=−6,
i.e. B=−1, and hence A=2+(−1)=1. This gives us the partial fractions decomposition
29x2−1=13x−1−13x+1.
Therefore we may compute
∫29x2−1dx=∫13x−1dx−∫13x+1dxu=3x−1,v=3x+1=13∫1udu−13∫1vdv=13log(u)−13log(v)+C=13log(3x−1)−13log(3x+1)+C.
Section 8.5 #9: ∫x2+12x+12x3−4xdx
Solution: We may factor this denominator as
x3−4x=x(x2−4)=x(x−2)(x+2),
so the partial fractions decomposition involves three terms (one for each linear factor):
x2+12x+12x3−4x=Ax+Bx−2+Cx+2.
Multiplying by the common denominator x(x−2)(x+2) yields
x2+12x+12=A(x−2)(x+2)+Bx(x+2)+Cx(x−2)=A(x2−4)+B(x2+2x)+C(x2−2x).
Combining like-terms yields
x2+12x+12=(A+B+C)x2+(2B−2C)x+(−4A).
Equating coefficients yields the following system of three linear equations in three variables:
{A+B+C=1(i)2B−2C=12(ii)−4A=12(iii).
From (iii), we may conclude that A=−3. From (ii), we may conclude that B−C=6, and hence B=6+C. Plugging A=−3 and B=6+C into (i), we get
−3+(6+C)+C=1,
from which we may conclude C=−1 and B=6+(−1)=5, yielding the partial fractions decomposition
x2+12x+12x3−4x=−3x+5x−2−1x+2.
Therefore we may compute
∫x2+12x+12x3−4xdx=−∫3xdx+5∫1x−2dx−∫1x+2dx=−3log(x)+5log(x−2)−log(x+2)+C.
Section 8.5 #13: ∫4x2+2x−1x3+x2dx
Solution: Factoring the denominator yields
x3+x2=x2(x+1).
Therefore we set up the partial fraction equation:
4x2+2x−1x3+x2=Ax+Bx2+Cx+1.
This yields a system of three equations in three variables and hence the following partial fractions decomposition:
4x2+2x−1x3+x2=−1x2+1x+1+3x.
Therefore we compute
∫4x2+2x−1x3+x2dx=−∫x−2dx+∫1x+1dx+3∫1xdx=1x+log(x+1)+3log(x)+C.