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The volume element is the absolute Jacobian determinant times the coordinate differentials. In the Cartesian coordinate system, cylindrical coordinates $(r,\theta ,z)$ and spherical coordinates $(\rho ,\vartheta ,\phi )$, respectively,Here $\rho$ is distance from the origin, $\vartheta$ is the polar angle from the positive $z$ axis, and $\phi$ is the azimuthal angle. The spherical coordinates convention fixes which angle enters the sine factor.

For a first-order linear ordinary differential equation $y^{\prime }+p(x)y=q(x)$, an integrating factor is $\exp (\int p(x)dx)$. Here $p(x)=\cot x$, soThe sine is positive on the specified interval, so this integrating factor never vanishes there. Any nonzero constant multiple is equally suitable.

The superposition principle gives the general solution as $y_c+2y_{p,2}+y_{p,1}$:The first boundary condition gives $A=2/(3\sqrt{3})-1/3$. At $x=\pi$, the cosine vanishes and the sine equals $-1$. Comparing with the second boundary condition then givesThe unique solution satisfying both boundary conditions is thereforeUniqueness follows because a homogeneous solution zero at both endpoints would have $A=0$ from the first and $B=0$ from the second.

The Pythagorean trigonometric identity makes the second factor $1+{\cos }^2\theta$, which is strictly positive. It is therefore enough to solve $\sin \theta -\cos \theta \le 0$. The sine addition formula givesThe shifted angle lies between $\pi /4$ and $5\pi /4$. On this interval the sine is nonpositive precisely from $\pi$ onwards. Restoring the original angle and retaining its strict upper bound gives

Apply the product rule to $x\sin (1/x)$ and the chain rule to the sine. Since $d(1/x)/dx=-1/x^2$, the derivative is

Since $A>0$ and $\omega >0$ are fixed, the standard sine limit givesIf $p>-1$, this tends to the assigned value zero. At $p=-1$, the right and left limits are $+A\omega$ and $-A\omega$; if $p<-1$, they diverge with opposite signs. HenceThe cancellation comes from the simple zero of $\sin (\omega x)$ at the origin, not rapid oscillation there.

Apply the chain rule twice: once to the square and once to the argument $x^3$ of the sine. The inner derivative is $y^{\prime }=3x^2\cos (x^3)-2\sin (2x)$. ThereforeThe original PDF has $\sin (x^3)$, rather than the cube of $\sin x$.

Let $a=\pi /3$, $b=2\pi /3$ and take $n$ equal subintervals of width $h=(b-a)/n=\pi /(3n)$. Since sine is continuous, its integral is the limit of the left-endpoint Riemann sumsUsing Euler's formula, the corresponding complex sum is a finite geometric series:Since $(e^{ih}-1)/h\to i$,Taking imaginary parts givesOne can see the same limit entirely within the real sums: the finite trigonometric-sum identity obtained from this geometric sum givesThus the evaluation is explicitly by a Riemann sum, as required.

Both segments start at $(1,1,1)$, so their direction vectors are $\mathbf{a}=(-1,-1,-1)$ and $\mathbf{b}=(1,1,0)$. Their dot product is $-2$, and their norms are $\sqrt{3}$ and $\sqrt{2}$. ThusThe angle between the rays lies in $[0,\pi ]$, so its sine is nonnegative:The negative cosine correctly identifies an obtuse angle. One also obtains the sine from $|\mathbf{a}\times \mathbf{b}|/(|\mathbf{a}||\mathbf{b}|)=\sqrt{2}/\sqrt{6}$, using the cross product.

Write $I=\int e^x\sin xdx$ and $J=\int e^x\cos xdx$. Two applications of integration by parts, first taking the sine and then the cosine as the differentiated factor, giveThus $I+J=e^x\sin x$ and $J-I=e^x\cos x$, in each case up to an arbitrary constant of integration. Keeping that freedom avoids identifying independently chosen antiderivatives too rigidly.

Factor the leading term out of the Maclaurin series for the sine:The binomial theorem gives, through degree four inside the bracket,The coefficient of $x^4$ is $1/20+5/12=7/15$. Multiplying back by $x^6$ yields

The Fourier coefficients follow by multiplying the expansion by $1$, by the appropriate cosine, or by the appropriate sine, then using orthogonality. In the convention with constant term $a_0/2$,The factor $1/L$ comes from the squared norms of the nonconstant sine and cosine functions: the integral of each square over $[-L,L]$ is $L$. The constant function has squared norm $2L$.

Put $s=\sin \theta$. By part (i), $\sin \varphi =\sqrt{2}s$. The acute-angle geometry requires $0<s<1/\sqrt{2}$. Expressing the tangents in terms of their sines in the geometric constraint gives the required implicit equationThe hint suggests $s=1/m$ for an integer $m\ge 2$. Testing $m=2$ givesThus $\boxed{\sin \theta =1/2}$. Each term on the left is strictly increasing in $s$ over the allowed interval, so this solution is unique.

The addition formula for the sine givesEquating the sine and cosine coefficients gives $A\cos \varphi =\sqrt{3}$ and $A\sin \varphi =1$. Squaring and adding yields $A^2=4$, so $A=2$ because $A>0$. Both coefficients are positive, so $\varphi$ is in the first quadrant. Consequently

The sine is nonnegative on $[0,\pi ]$, so the geometric area equals the definite integral, without cancellation:

From the same multiplication, the imaginary part isHere the planar vectors are embedded in three dimensions with zero third coordinate. Its magnitude is the area of their parallelogram; its sign is positive when the oriented angle from ${\mathbf{v}}_2$ to ${\mathbf{v}}_1$ has positive sine. The reversed order matters: $\Im (z_1{z}_2^{\ast })$ is the negative of the usual signed area for the order $({\mathbf{v}}_1,{\mathbf{v}}_2)$.

Put $h=x-\pi /2$. The sine addition identity gives $\sin x=\cos h=1-h^2/2+h^4/24+O(h^6)$. Applying the exponential series givesThus the first three nonzero terms in the Taylor series about $\pi /2$ are

Since $\cos x>0$ throughout the given interval, multiplying by $\cos x$ is reversible. The equation simplifies toThe sine equals $1$ only at $\pi /2+2j\pi$. Here $x+\pi /3\in [\pi /3,5\pi /6)$, so only $x+\pi /3=\pi /2$ is allowed. Hence

Use the real inner productOrthogonality means this integral is zero for any two distinct listed functions. The individual squared norms are positive: the constant function has squared norm $2L$, while every listed sine or cosine has squared norm $L$. Thus they are orthogonal but not normalized to unit length in this inner product.

Set $\ell =\ln 2$. Applying the exponential series and the Maclaurin series of the sine givesMultiply the two power series, retaining every contribution through degree four:Since $0<\ln 2<1$, the coefficient of $x^4$ is nonzero, so these are indeed the first two nonzero terms. The remainder is in Big O notation.