Integrals: questions with answers

Only sums and constant multiples pass through the integral sign; a product does not. So the product is multiplied out first, and the resulting sum of powers is integrated term by term: $\frac{x^3}{3} + \frac{3x^2}{2} + 2x + C$. Integrating each factor and multiplying gives a function whose derivative (by the product rule) has extra terms that are not in the integrand.

$v$ is an antiderivative of $\tan^{-1}x$, not its derivative. Differentiate $x\tan^{-1}x-\frac12\log(1+x^2)$ to recover $\tan^{-1}x$. Choosing the polynomial as $u$ is valid.

$\cos^2(2\pi - x) = \cos^2 x$, so $I = 2\int_0^{\pi}\cos^2 x\,dx = 2\cdot\frac{\pi}{2}$, so $I = \pi$. When $f(2a - x) = f(x)$, $P_6$ gives $2\int_0^a f$; the doubling is part of the property. Oddness says nothing on $[0, 2a]$, and this integrand is never negative anyway.

With $t$ = the linear argument, $dt$ is the coefficient of $x$ times $dx$, so $dx$ is $dt$ DIVIDED by that coefficient, sign included: $\int f(ax + b)\,dx = \frac{1}{a}F(ax + b) + C$. That gives $\frac{2}{9}(3x + 1)^{3/2} + C$. Ignoring the coefficient, or multiplying by it, produces a function whose derivative is the integrand times a wrong constant.

With $t$ = the base of the power, $dt$ is the other factor times a constant, so the integrand is $\frac{1}{c}t^n\,dt$ and the integral is $\frac{1}{c}\cdot\frac{t^{n+1}}{n+1}$: $\frac{\sin^4 2x}{8} + C$. The constant from $dt$ must be divided away, and the factor that became $dt$ must not be integrated separately as if it were a spectator.

The $e^x[f + f']$ result rests on $\frac{d}{dx}e^x = e^x$. For $e^{-x}$ the derivative is $-e^{-x}$, so $\frac{d}{dx}\left(-e^{-x}f\right) = e^{-x}f - e^{-x}f' = e^{-x}[f - f']$: the bracket must be $f - f'$ and the answer is $-e^{-x}f$. Here $f = \cos x$, $f' = -\sin x$, $f - f' = \cos x + \sin x$, so $-e^{-x}\cos x + C$. Copying $e^x f$ with $e^{-x}$ gets the sign wrong.

The three moduli vanish at $x = 2$, $x = 3$ and $x = 5$, all in the interval, so the interval is cut at EVERY one of them and the pieces are added by $P_2$. Equivalently, integrate each modulus on its own: over $[a, b]$, $\int|x - c|\,dx = \frac{(c - a)^2 + (b - c)^2}{2}$, two triangles. Adding the three gives $32$. Cutting only at the middle zero gives $|x - 2|$ and $|x - 5|$ the wrong sign on part of the interval.

Line 1 is correct. Line 2 is the first wrong line: the new limits have been computed from the $dt$ relation instead of from the substitution itself. The limits are the values of $t$ at the $x$-limits, and the value is $e - 1$.

A reciprocal does not split over a sum. The sum of roots is cleared by its conjugate: multiplying numerator and denominator by the difference of the roots turns the denominator into the difference of the radicands, a constant, and the numerator into two separate roots, each a power-rule integral. That gives $\frac{2}{9}\left[(x + 4)^{3/2} - (x + 1)^{3/2}\right] + C$.

$P_3$: $\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx$. The substitution $t = a + b - x$ sends $x = a$ to $t = b$ and $x = b$ to $t = a$, and $dx = -dt$; the two minus signs cancel. With $a - x$ or $b - x$ the limits do not come back to $[a, b]$.

Line 1 is the correct identity. Line 2 is the first wrong line: $\theta = 3x + 5$, so $2\theta = 6x + 10$ -- the whole argument is doubled, not just the outer square. Lines 3 and 4 integrate the wrong integrand correctly. The integral is $\frac{x}{2} - \frac{\sin(6x + 10)}{ 12 } + C$.

$A$ multiplies the WHOLE derivative, sign included; with a negative leading coefficient the derivative has a negative $x$-coefficient, so $A$ is negative for a positive numerator slope. Comparing $x$-coefficients and then constants gives $A = -\frac{5}{2},\ B = -6$.

Lines 1 and 2 are correct. Line 3 is the first wrong line: integrating $\operatorname{cosec}^2\dfrac{x}{2}$ gives twice the tangent or cotangent of $\frac{x}{2}$, because differentiating $\tan\frac{x}{2}$ produces a $\frac{1}{2}$ by the chain rule. The factor $2$ cancels the $\frac{1}{2}$ of Line 2, so the integral is $-\cot\dfrac{x}{2} + C$.

With a negative leading coefficient the minus is factored out FIRST: $c - 2px - x^2 = c - (x^2 + 2px) = (c + p^2) - (x + p)^2$, the shape $k^2 - t^2$. Under a root that is the $\sin^{-1}$ form, $\sin^{-1}\frac{t}{k}$ with $t = x + p$ put back, shift and all: $\sin^{-1}\frac{x + 1}{3} + C$.

The numerator is the derivative of the trigonometric function inside, so with $t = \sin x$ (or $t = \tan x$) the integral becomes one of the six forms in $t$. Then the root and the sign choose the form exactly as for $x$: a root with $t^2 + a^2$ is a log, a root with $a^2 - t^2$ is $\sin^{-1}$, no root with $t^2 + a^2$ is $\tan^{-1}$. Resubstituting gives $\log\left|\tan x + \sqrt{\tan^2 x + 4}\right| + C$.