Ответы. Интегрирование тригонометрических функций
Подготовка к экзамену «Элементы высшей математики» · Вопрос №15 · 50 задач
Уровень А (№1–20)
- \(-\dfrac{\cos 3x}{3}+C\)
- \(\dfrac{\sin 5x}{5}+C\)
- \(-2\cos\dfrac{x}{2}+C\)
- \(\dfrac{\sin(2x+1)}{2}+C\)
- \(\dfrac{x}{2}+\dfrac{\sin 2x}{4}+C\)
- \(\dfrac{x}{2}-\dfrac{\sin 4x}{8}+C\)
- \(\dfrac{x}{2}+\dfrac{\sin x}{2}+C\)
- \(-\cos x+\dfrac{\cos^3 x}{3}+C\)
- \(\sin x-\dfrac{\sin^3 x}{3}+C\)
- \(\dfrac{\sin^2 x}{2}+C\)
- \(\dfrac{\sin^3 x}{3}+C\)
- \(-\dfrac{\cos^4 x}{4}+C\)
- \(-\ln|\cos x|+C\)
- \(\ln|\sin x|+C\)
- \(\operatorname{tg} x-x+C\)
- \(-\dfrac{\cos 2x}{4}-\dfrac{\cos 8x}{16}+C\)
- \(\dfrac{\sin 2x}{4}-\dfrac{\sin 6x}{12}+C\)
- \(\dfrac{\sin 4x}{8}+\dfrac{\sin 8x}{16}+C\)
- \(\dfrac{\cos 2x}{4}-\dfrac{\cos 4x}{8}+C\)
- \(x-\dfrac{\cos 2x}{2}+C\)
Уровень Б (№21–40)
- \(\dfrac{3x}{8}-\dfrac{\sin 2x}{4}+\dfrac{\sin 4x}{32}+C\)
- \(\dfrac{3x}{8}+\dfrac{\sin 2x}{4}+\dfrac{\sin 4x}{32}+C\)
- \(\dfrac{x}{8}-\dfrac{\sin 4x}{32}+C\)
- \(\dfrac{\cos^7 x}{7}-\dfrac{\cos^5 x}{5}+C\)
- \(\sin x-\dfrac{2\sin^3 x}{3}+\dfrac{\sin^5 x}{5}+C\)
- \(-\cos x+\dfrac{2\cos^3 x}{3}-\dfrac{\cos^5 x}{5}+C\)
- \(\dfrac{\sin^3 x}{3}-\dfrac{\sin^5 x}{5}+C\)
- \(\dfrac{\operatorname{tg}^2 x}{2}+\ln|\cos x|+C\)
- \(-\dfrac{\operatorname{ctg}^2 x}{2}-\ln|\sin x|+C\)
- \(\dfrac{\operatorname{tg}^3 x}{3}-\operatorname{tg} x+x+C\)
- \(\ln\left|\operatorname{tg}\dfrac{x}{2}\right|+C\)
- \(\ln\left|\operatorname{tg}\left(\dfrac{x}{2}+\dfrac{\pi}{4}\right)\right|+C\)
- \(\dfrac{\sin 3x}{6}+\dfrac{\sin 5x}{10}+C\)
- \(\dfrac{\sin 2x}{4}-\dfrac{\sin 8x}{16}+C\)
- \(\operatorname{tg}\dfrac{x}{2}+C\)
- \(-\operatorname{ctg}\dfrac{x}{2}+C\)
- \(\operatorname{tg} x-\dfrac{1}{\cos x}+C\)
- \(\operatorname{tg} x-\operatorname{ctg} x+C\)
- \(\dfrac{\operatorname{tg} x}{2}+C\)
- \(\dfrac{\sin^5 x}{5}-\dfrac{\sin^7 x}{7}+C\)
Уровень В (№41–50)
- Замена \(t=\operatorname{tg}\dfrac{x}{2}\): \(\dfrac{2}{\sqrt{5}}\,\operatorname{arctg}\dfrac{\operatorname{tg}\frac{x}{2}}{\sqrt{5}}+C\)
- Замена \(t=\operatorname{tg}\dfrac{x}{2}\): \(\dfrac{2}{3}\,\operatorname{arctg}\dfrac{5\operatorname{tg}\frac{x}{2}-4}{3}+C\)
- \(\dfrac{1}{\sqrt{2}}\,\ln\left|\operatorname{tg}\left(\dfrac{x}{2}+\dfrac{\pi}{8}\right)\right|+C\)
- Замена \(t=\operatorname{tg} x\): \(\dfrac{1}{\sqrt{2}}\,\operatorname{arctg}\bigl(\sqrt{2}\,\operatorname{tg} x\bigr)+C\)
- Замена \(t=\operatorname{tg} x\): \(\dfrac{1}{6}\,\operatorname{arctg}\dfrac{2\operatorname{tg} x}{3}+C\)
- Замена \(t=\operatorname{tg} x\): \(\dfrac{1}{ab}\,\operatorname{arctg}\dfrac{a\operatorname{tg} x}{b}+C\)
- \(\dfrac{5x}{16}-\dfrac{15\sin 2x}{64}+\dfrac{3\sin 4x}{64}-\dfrac{\sin 6x}{192}+C\)
- Замена \(t=\operatorname{tg}\dfrac{x}{2}\): \(\ln\left|1+\operatorname{tg}\dfrac{x}{2}\right|+C\)
- \(\dfrac{x}{16}-\dfrac{\sin 4x}{64}+\dfrac{\sin^3 2x}{48}+C\)
- \(x-\operatorname{tg} x+\dfrac{1}{\cos x}+C\)