Definite and Indefinite Integrals Questions and Answers

Answer: A

The key gives \(2I = \pi\int_{-1}^{1}\frac{x^{4}}{1 - x^{4}}\,dx\).

Use \(\int_{-a}^{a} F(x)\,dx = \int_{-a}^{a} F(-x)\,dx\). The factor \(\frac{x^{4}}{1 - x^{4}}\) is even, and \(\cos^{-1}(-y) = \pi - \cos^{-1}y\), so

\[ \begin{aligned} I &= \int_{-1}^{1}\frac{x^{4}}{1 - x^{4}}\left[\pi - \cos^{-1}\tfrac{2x}{1 + x^{2}}\right]dx \\ &= \pi\int_{-1}^{1}\frac{x^{4}}{1 - x^{4}}\,dx - I \end{aligned} \]

Adding \(I\) to both sides gives \(2I = \pi\int_{-1}^{1}\frac{x^{4}}{1 - x^{4}}\,dx\). (The factor \(\frac{x^{4}}{1 - x^{4}}\) grows without bound at \(x = \pm 1\); the question tests the symmetry property, and the relation is the one that property gives.)

\(2\pi\) and \(-\pi\) misapply the identity, and \(1 + x^{4}\) changes the integrand.

Answer: C

The integral equals \(\frac{\pi}{2}\).

For any \(u \gt 0\), \(\tan^{-1}\frac{1}{u} + \tan^{-1}u = \frac{\pi}{2}\). Here \(u = 1 + x + x^{2} + x^{3}\), which is at least 1 on \([0, 1]\), so the integrand is the constant \(\frac{\pi}{2}\):

\[ \int_{0}^{1}\frac{\pi}{2}\,dx = \frac{\pi}{2} \]

\(\frac{\pi}{4}\) takes one of the two terms only, and \(2\pi\), \(4\pi\) and 1 do not follow from the identity.

Answer: C

The value is \(\frac{1}{30}\).

Use \(\int_{0}^{1} f(x)\,dx = \int_{0}^{1} f(1 - x)\,dx\) to move the power onto a single \(x\):

\[ \begin{aligned} \int_{0}^{1} x(1 - x)^{4}dx &= \int_{0}^{1}(1 - x)x^{4}\,dx \\ &= \tfrac{1}{5} - \tfrac{1}{6} = \tfrac{1}{30} \end{aligned} \]

This agrees with the beta-function value \(\frac{1!\,4!}{6!} = \frac{24}{720}\). \(\frac{1}{20}\) and \(\frac{1}{60}\) come from slips in the subtraction, and \(\frac{1}{15}\) doubles the result.

Answer: B

The integral is \(y + 2\log|y| - 6\log|1 + y| + C\).

Divide first, then use partial fractions:

\[ \begin{aligned} \frac{y^{2} - 3y + 2}{y^{2} + y} &= 1 + \frac{2 - 4y}{y(y + 1)} \\ \frac{2 - 4y}{y(y + 1)} &= \frac{2}{y} - \frac{6}{y + 1} \end{aligned} \]

(At \(y = 0\) the numerator gives 2, and at \(y = -1\) it gives \(\frac{6}{-1} = -6\).) Integrating term by term gives \(y + 2\log|y| - 6\log|1 + y| + C\). The other options have a wrong coefficient or sign on one of the logarithms.

Answer: D

The integral is \(\sin\theta - \dfrac{1}{2}\theta - \dfrac{\sin 2\theta}{4} + C\).

Simplify \(\tan\frac{\theta}{2}\sin\theta\) using \(\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\):

\[ \begin{aligned} \tan\tfrac{\theta}{2}\sin\theta &= 2\sin^{2}\tfrac{\theta}{2} = 1 - \cos\theta \\ (1 - \cos\theta)\cos\theta &= \cos\theta - \tfrac{1}{2} - \tfrac{1}{2}\cos 2\theta \end{aligned} \]

Integrating gives \(\sin\theta - \frac{\theta}{2} - \frac{\sin 2\theta}{4} + C\). The other options differ only in a single coefficient, the kind of slip made in integrating \(\cos 2\theta\) or in the half-angle step.

Answer: B

The integral is \(\log\left|\tan^{-1}\left(\dfrac{1 + x}{1 - x}\right)\right| + C\).

For \(x \lt 1\), \(\tan^{-1}\frac{1 + x}{1 - x} = \frac{\pi}{4} + \tan^{-1}x\), so its derivative is \(\frac{1}{1 + x^{2}}\). With \(t = \tan^{-1}\frac{1 + x}{1 - x}\):

\[ \int \frac{dt}{t} = \log|t| + C \]

No constant factor appears, since the numerator \(\frac{dx}{1 + x^{2}}\) is exactly \(dt\); that rules out the \(\frac{1}{2}\) and \(\frac{1}{4}\) options. \(\tan^{-1}\frac{1 - x}{1 + x}\) has derivative \(-\frac{1}{1 + x^{2}}\), the wrong sign.

Answer: C

The integral is \(e^{x}(e^{x} + \sin x + 2) + C\).

Use \(\int e^{x}[g(x) + g'(x)]\,dx = e^{x}g(x) + C\). With \(g(x) = e^{x} + \sin x + 2\):

\[ \begin{aligned} g'(x) &= e^{x} + \cos x \\ g + g' &= 2e^{x} + \sin x + \cos x + 2 \end{aligned} \]

That is exactly the bracket in the integrand, so the integral is \(e^{x}(e^{x} + \sin x + 2) + C\). Differentiating each other option fails to reproduce the integrand; for example \(e^{x}(e^{x} + \cos x + 2)\) gives \(-\sin x\) where \(+\cos x\) is needed.

Answer: E

The integral is \(t - \log(1 + e^{t}) + C\).

Write the integrand as 1 minus a fraction whose numerator is the derivative of its denominator:

\[ \begin{aligned} \frac{1}{1 + e^{t}} &= 1 - \frac{e^{t}}{1 + e^{t}} \\ \int \frac{dt}{1 + e^{t}} &= t - \log(1 + e^{t}) + C \end{aligned} \]

Check by differentiating: \(1 - \frac{e^{t}}{1 + e^{t}} = \frac{1}{1 + e^{t}}\). The option with \(+\log\) has the wrong sign, and \(\log(1 - e^{t})\) is undefined for \(t \gt 0\).

Answer: D

The integral is \(\dfrac{\pi}{12}\).

Replace \(x\) by \(\frac{\pi}{6} + \frac{\pi}{3} - x = \frac{\pi}{2} - x\), which swaps \(\sin x\) and \(\cos x\) and leaves the limits unchanged. Adding the two forms of the integral \(I\):

\[ \begin{aligned} 2I &= \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \\ &= \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \end{aligned} \]

So \(I = \frac{\pi}{12}\). \(\frac{\pi}{6}\) is \(2I\), the usual slip of forgetting to halve.

Answer: E

The integral is \(\dfrac{1}{16\sqrt{2}}\).

Substitute \(t = \tan^{-1}(x^{16})\), so \(dt = \frac{16x^{15}}{1 + x^{32}}dx\), with \(t\) running from 0 to \(\frac{\pi}{4}\):

\[ \begin{aligned} I &= \frac{1}{16}\int_{0}^{\pi/4} \cos t\,dt \\ &= \frac{1}{16}\sin\frac{\pi}{4} = \frac{1}{16\sqrt{2}} \end{aligned} \]

The factor \(\frac{1}{16}\) comes from the 16 in \(dt\); dropping it or using another power gives the other options.