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SOLUTION KEY: Level 1 (Integrals)
Teacher/Staff Use Only Class: 12 Subject: Mathematics
Topic 1: Fundamental Formulas & Anti-derivatives
1.
Answer: $\frac{x^4}{4} - \frac{2x^3}{3} + 5x + C$.
2.
Answer: $\int (x^{1/2} + x^{-1/2}) dx = \frac{2}{3}x^{3/2} + 2x^{1/2} + C$.
3.
Answer: $\int \left(x + 5 - \frac{4}{x^2}\right) dx = \frac{x^2}{2} + 5x + \frac{4}{x} + C$.
4.
Answer: $\frac{2^x}{\ln 2} + e^x + C$.
5.
Answer: $-2\cos x - 3\sin x + C$.
6.
Answer: $\int (\sec^2 x + \sec x \tan x) dx = \tan x + \sec x + C$.
7.
Answer: $\int (\sec^2 x - \sec x \tan x) dx = \tan x - \sec x + C$.
8.
Answer: $\frac{x^{a+1}}{a+1} + \frac{a^x}{\ln a} + C$.
9.
Answer: $F(x) = \int (4x^3 - 6) dx = x^4 - 6x + C$. Given $F(0)=3 \implies C=3$. So, $F(x) = x^4 - 6x + 3$.
10.
Answer: $2\tan^{-1}x + C$.
Topic 2: Integration by Substitution
11.
Answer: Let $u = x^2, du = 2x dx$. $\int e^u du = e^{x^2} + C$.
12.
Answer: Let $u = \log x, du = \frac{1}{x} dx$. $\int u^2 du = \frac{u^3}{3} + C = \frac{(\log x)^3}{3} + C$.
13.
Answer: Let $u = \tan^{-1}x, du = \frac{1}{1+x^2} dx$. $\int \sin u du = -\cos u + C = -\cos(\tan^{-1}x) + C$.
14.
Answer: Let $u = e^x + e^{-x}, du = (e^x - e^{-x}) dx$. $\int \frac{du}{u} = \ln|e^x + e^{-x}| + C$.
15.
Answer: Let $u = \tan x, du = \sec^2 x dx$. $\int u^3 du = \frac{u^4}{4} + C = \frac{\tan^4 x}{4} + C$.
16.
Answer: Let $u = \log x, du = \frac{1}{x} dx$. $\int \frac{du}{u} = \ln|u| + C = \ln|\log x| + C$.
17.
Answer: Let $u = 1+\sin x, du = \cos x dx$. $\int u^{-1/2} du = 2u^{1/2} + C = 2\sqrt{1+\sin x} + C$.
18.
Answer: Let $u = x^3, du = 3x^2 dx$. $\int e^u du = e^{x^3} + C$.
19.
Answer: Let $u = x^2+3x+5, du = (2x+3) dx$. $\int \frac{du}{u} = \ln|x^2+3x+5| + C$.
20.
Answer: Let $u = \sin x, du = \cos x dx$. $\int u du = \frac{\sin^2 x}{2} + C$ (or $-\frac{\cos^2 x}{2} + C$).
Topic 3: Integration using Trigonometric Identities
21.
Answer: $\int \frac{1-\cos 2x}{2} dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$.
22.
Answer: $\int \frac{1+\cos 2x}{2} dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$.
23.
Answer: $\frac{1}{2}\int (\sin 7x - \sin x)dx = -\frac{\cos 7x}{14} + \frac{\cos x}{2} + C$.
24.
Answer: $\int (1-\sin^2 x)\cos x dx$. Let $u=\sin x \implies \int (1-u^2)du = \sin x - \frac{\sin^3 x}{3} + C$.
25.
Answer: $\int \sqrt{2\cos^2 x} dx = \sqrt{2} \int \cos x dx = \sqrt{2}\sin x + C$.
26.
Answer: $\int \frac{2\sin^2 x}{2\cos^2 x} dx = \int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x + C$.
27.
Answer: $\int (\sec^2 x - 1) dx = \tan x - x + C$.
28.
Answer: $\int (\csc^2 x - 1) dx = -\cot x - x + C$.
29.
Answer: $\frac{1}{2}\int (\cos x - \cos 5x)dx = \frac{\sin x}{2} - \frac{\sin 5x}{10} + C$.
30.
Answer: Let $x-a = t \implies dx = dt$. $\int \frac{\sin(t+a)}{\sin t} dt = \int (\cos a + \sin a \cot t) dt = t\cos a + \sin a \ln|\sin t| = (x-a)\cos a + \sin a \ln|\sin(x-a)| + C$.
Topic 4: Special Integrals & Partial Fractions
31.
Answer: $\frac{1}{2(4)}\ln\left|\frac{x-4}{x+4}\right| + C = \frac{1}{8}\ln\left|\frac{x-4}{x+4}\right| + C$.
32.
Answer: $\sin^{-1}\left(\frac{x}{5}\right) + C$.
33.
Answer: $\int \frac{dx}{(x+1)^2+4} = \frac{1}{2}\tan^{-1}\left(\frac{x+1}{2}\right) + C$.
34.
Answer: $\ln|x+\sqrt{x^2-9}| + C$.
35.
Answer: $\int \left(\frac{1}{x} - \frac{1}{x+1}\right) dx = \ln|x| - \ln|x+1| + C = \ln\left|\frac{x}{x+1}\right| + C$.
36.
Answer: $\frac{2x}{(x+1)(x+2)} = \frac{-2}{x+1} + \frac{4}{x+2}$. Integral is $4\ln|x+2| - 2\ln|x+1| + C$.
37.
Answer: $\frac{1}{(x-3)(x-2)} = \frac{1}{x-3} - \frac{1}{x-2}$. Integral is $\ln\left|\frac{x-3}{x-2}\right| + C$.
38.
Answer: $\frac{x}{(x-1)(x-2)} = \frac{-1}{x-1} + \frac{2}{x-2}$. Integral is $2\ln|x-2| - \ln|x-1| + C$.
39.
Answer: $\int \frac{dx}{4 + (3x)^2}$. Let $u=3x, du=3dx$. $\frac{1}{3} \cdot \frac{1}{2}\tan^{-1}\left(\frac{3x}{2}\right) = \frac{1}{6}\tan^{-1}\left(\frac{3x}{2}\right) + C$.
40.
Answer: $\frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\left(\frac{x}{4}\right) + C = \frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) + C$.
Topic 5: Integration by Parts
41.
Answer: $x(-\cos x) - \int 1(-\cos x) dx = -x\cos x + \sin x + C$.
42.
Answer: $x(e^x) - \int 1(e^x) dx = x e^x - e^x + C$.
43.
Answer: Take $v=1$. $\log x \cdot x - \int \frac{1}{x} \cdot x dx = x\log x - x + C$.
44.
Answer: $x^2 e^x - \int 2x e^x dx = x^2 e^x - 2(x e^x - e^x) + C = e^x(x^2 - 2x + 2) + C$.
45.
Answer: $x\tan x - \int 1\tan x dx = x\tan x - \ln|\sec x| + C$.
46.
Answer: Using $\int e^x(f(x)+f'(x))dx$. Here $f(x) = \sin x$. Result: $e^x\sin x + C$.
47.
Answer: Using $\int e^x(f(x)+f'(x))dx$. Here $f(x) = \frac{1}{x}$. Result: $\frac{e^x}{x} + C$.
48.
Answer: $\log x \cdot \frac{x^2}{2} - \int \frac{1}{x} \cdot \frac{x^2}{2} dx = \frac{x^2}{2}\log x - \frac{x^2}{4} + C$.
49.
Answer: Here $f(x) = \tan x$. $f'(x) = \sec^2 x$. Result: $e^x\tan x + C$.
50.
Answer: Take $v=1$. $\sin^{-1} x \cdot x - \int \frac{x}{\sqrt{1-x^2}} dx = x\sin^{-1} x + \sqrt{1-x^2} + C$.
Topic 6: Definite Integrals & Properties
51.
Answer: $[x^3/3]_1^2 = 8/3 - 1/3 = 7/3$.
52.
Answer: $[-\cos x]_0^{\pi/2} = -\cos(\pi/2) - (-\cos 0) = 0 + 1 = 1$.
53.
Answer: $[\tan^{-1} x]_0^1 = \tan^{-1}(1) - \tan^{-1}(0) = \pi/4$.
54.
Answer: $[\ln|\sec x|]_0^{\pi/4} = \ln(\sqrt{2}) - \ln(1) = \frac{1}{2}\ln 2$.
55.
Answer: $x^3 \cos x$ is an odd function. Thus, integral over $[-1, 1]$ is $0$.
56.
Answer: $I = \int_0^{\pi/2} \frac{\cos x}{\cos x + \sin x} dx$. $2I = \int_0^{\pi/2} 1 dx = \pi/2 \implies I = \pi/4$.
57.
Answer: $2 \int_0^a x^4 dx = 2 [\frac{x^5}{5}]_0^a = \frac{2a^5}{5}$.
58.
Answer: Using $\int_0^{2a} f(x)dx$. $f(\pi-x) = \cos^3(\pi-x) = -\cos^3 x = -f(x)$. Thus, integral is $0$.
59.
Answer: $[\ln x]_1^3 = \ln 3 - \ln 1 = \ln 3$.
60.
Answer: $\int_0^1 (1-x) dx + \int_1^2 (x-1) dx = [x-x^2/2]_0^1 + [x^2/2-x]_1^2 = 1/2 + 1/2 = 1$.