انتگرال توابع مثلثاتی

در ادامه فهرستی از انتگرال تابع‌هایمثلثاتی نوشته شده‌است. برای آگاهی از انتگرال تابع‌های نمایی و مثلثاتی فهرست انتگرال تابع‌های نمایی را نگاه کنید، همچنین برای داشتن یک فهرست کامل صفحهٔ فهرست انتگرال‌ها را نگاه کنید.

اگر تابع sin ⁡ ( x ) {\displaystyle \sin(x)} را شکل کلی تابع مثلثاتی در نظر بگیریم و cos ⁡ ( x ) {\displaystyle \cos(x)} را به عنوان مشتق آن، آنگاه:

∫ a cos ⁡ n x d x = a n sin ⁡ n x + c {\displaystyle \int a\cos nx\;dx={\frac {a}{n}}\sin nx+c}

در تمامی رابطه‌ها فرض می‌شود که a ناصفر است و C ثابت انتگرال‌گیری است.

انتگرال‌هایی که تنها تابع سینوس دارند[ویرایش]

∫ sin ⁡ a x d x = − 1 a cos ⁡ a x + C {\displaystyle \int \sin ax\;dx=-{\frac {1}{a}}\cos ax+C\,\!}

∫ sin 2 ⁡ a x d x = x 2 − 1 4 a sin ⁡ 2 a x + C = x 2 − 1 2 a sin ⁡ a x cos ⁡ a x + C {\displaystyle \int \sin ^{2}{ax}\;dx={\frac {x}{2}}-{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}-{\frac {1}{2a}}\sin ax\cos ax+C\!}

∫ x sin 2 ⁡ a x d x = x 2 4 − x 4 a sin ⁡ 2 a x − 1 8 a 2 cos ⁡ 2 a x + C {\displaystyle \int x\sin ^{2}{ax}\;dx={\frac {x^{2}}{4}}-{\frac {x}{4a}}\sin 2ax-{\frac {1}{8a^{2}}}\cos 2ax+C\!}

∫ x 2 sin 2 ⁡ a x d x = x 3 6 − ( x 2 4 a − 1 8 a 3 ) sin ⁡ 2 a x − x 4 a 2 cos ⁡ 2 a x + C {\displaystyle \int x^{2}\sin ^{2}{ax}\;dx={\frac {x^{3}}{6}}-\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax-{\frac {x}{4a^{2}}}\cos 2ax+C\!}

∫ sin ⁡ b 1 x sin ⁡ b 2 x d x = sin ⁡ ( ( b 1 − b 2 ) x ) 2 ( b 1 − b 2 ) − sin ⁡ ( ( b 1 + b 2 ) x ) 2 ( b 1 + b 2 ) + C (for  | b 1 | ≠ | b 2 | ) {\displaystyle \int \sin b_{1}x\sin b_{2}x\;dx={\frac {\sin((b_{1}-b_{2})x)}{2(b_{1}-b_{2})}}-{\frac {\sin((b_{1}+b_{2})x)}{2(b_{1}+b_{2})}}+C\qquad {\mbox{(for }}|b_{1}|\neq |b_{2}|{\mbox{)}}\,\!}

∫ sin n ⁡ a x d x = − sin n − 1 ⁡ a x cos ⁡ a x n a + n − 1 n ∫ sin n − 2 ⁡ a x d x (for  n > 0 ) {\displaystyle \int \sin ^{n}{ax}\;dx=-{\frac {\sin ^{n-1}ax\cos ax}{na}}+{\frac {n-1}{n}}\int \sin ^{n-2}ax\;dx\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!}

∫ d x sin ⁡ a x = 1 a ln ⁡ | tan ⁡ a x 2 | + C {\displaystyle \int {\frac {dx}{\sin ax}}={\frac {1}{a}}\ln \left|\tan {\frac {ax}{2}}\right|+C}

∫ d x sin n ⁡ a x = cos ⁡ a x a ( 1 − n ) sin n − 1 ⁡ a x + n − 2 n − 1 ∫ d x sin n − 2 ⁡ a x (for  n > 1 ) {\displaystyle \int {\frac {dx}{\sin ^{n}ax}}={\frac {\cos ax}{a(1-n)\sin ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\sin ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}\,\!}

∫ x sin ⁡ a x d x = sin ⁡ a x a 2 − x cos ⁡ a x a + C {\displaystyle \int x\sin ax\;dx={\frac {\sin ax}{a^{2}}}-{\frac {x\cos ax}{a}}+C\,\!}

∫ x n sin ⁡ a x d x = − x n a cos ⁡ a x + n a ∫ x n − 1 cos ⁡ a x d x = ∑ k = 0 2 k ≤ n ( − 1 ) k + 1 x n − 2 k a 1 + 2 k n ! ( n − 2 k ) ! cos ⁡ a x + ∑ k = 0 2 k + 1 ≤ n ( − 1 ) k x n − 1 − 2 k a 2 + 2 k n ! ( n − 2 k − 1 ) ! sin ⁡ a x (for  n > 0 ) {\displaystyle \int x^{n}\sin ax\;dx=-{\frac {x^{n}}{a}}\cos ax+{\frac {n}{a}}\int x^{n-1}\cos ax\;dx=\sum _{k=0}^{2k\leq n}(-1)^{k+1}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\cos ax+\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-1-2k}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\sin ax\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!}

∫ − a 2 a 2 x 2 sin 2 ⁡ n π x a d x = a 3 ( n 2 π 2 − 6 ) 24 n 2 π 2 (for  n = 2 , 4 , 6... ) {\displaystyle \int _{\frac {-a}{2}}^{\frac {a}{2}}x^{2}\sin ^{2}{\frac {n\pi x}{a}}\;dx={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(for }}n=2,4,6...{\mbox{)}}\,\!}

∫ sin ⁡ a x x d x = ∑ n = 0 ∞ ( − 1 ) n ( a x ) 2 n + 1 ( 2 n + 1 ) ⋅ ( 2 n + 1 ) ! + C {\displaystyle \int {\frac {\sin ax}{x}}dx=\sum _{n=0}^{\infty }(-1)^{n}{\frac {(ax)^{2n+1}}{(2n+1)\cdot (2n+1)!}}+C\,\!}

∫ sin ⁡ a x x n d x = − sin ⁡ a x ( n − 1 ) x n − 1 + a n − 1 ∫ cos ⁡ a x x n − 1 d x {\displaystyle \int {\frac {\sin ax}{x^{n}}}dx=-{\frac {\sin ax}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}\int {\frac {\cos ax}{x^{n-1}}}dx\,\!}

∫ d x 1 ± sin ⁡ a x = 1 a tan ⁡ ( a x 2 ∓ π 4 ) + C {\displaystyle \int {\frac {dx}{1\pm \sin ax}}={\frac {1}{a}}\tan \left({\frac {ax}{2}}\mp {\frac {\pi }{4}}\right)+C}

∫ x d x 1 + sin ⁡ a x = x a tan ⁡ ( a x 2 − π 4 ) + 2 a 2 ln ⁡ | cos ⁡ ( a x 2 − π 4 ) | + C {\displaystyle \int {\frac {x\;dx}{1+\sin ax}}={\frac {x}{a}}\tan \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)+{\frac {2}{a^{2}}}\ln \left|\cos \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)\right|+C}

∫ x d x 1 − sin ⁡ a x = x a cot ⁡ ( π 4 − a x 2 ) + 2 a 2 ln ⁡ | sin ⁡ ( π 4 − a x 2 ) | + C {\displaystyle \int {\frac {x\;dx}{1-\sin ax}}={\frac {x}{a}}\cot \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)+{\frac {2}{a^{2}}}\ln \left|\sin \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)\right|+C}

∫ sin ⁡ a x d x 1 ± sin ⁡ a x = ± x + 1 a tan ⁡ ( π 4 ∓ a x 2 ) + C {\displaystyle \int {\frac {\sin ax\;dx}{1\pm \sin ax}}=\pm x+{\frac {1}{a}}\tan \left({\frac {\pi }{4}}\mp {\frac {ax}{2}}\right)+C}

انتگرال‌هایی که تنها تابع کسینوس دارند[ویرایش]

∫ cos ⁡ a x d x = 1 a sin ⁡ a x + C {\displaystyle \int \cos ax\;dx={\frac {1}{a}}\sin ax+C\,\!} ∫ cos 2 ⁡ a x d x = x 2 + 1 4 a sin ⁡ 2 a x + C = x 2 + 1 2 a sin ⁡ a x cos ⁡ a x + C {\displaystyle \int \cos ^{2}{ax}\;dx={\frac {x}{2}}+{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}+{\frac {1}{2a}}\sin ax\cos ax+C\!} ∫ cos n ⁡ a x d x = cos n − 1 ⁡ a x sin ⁡ a x n a + n − 1 n ∫ cos n − 2 ⁡ a x d x (for  n > 0 ) {\displaystyle \int \cos ^{n}ax\;dx={\frac {\cos ^{n-1}ax\sin ax}{na}}+{\frac {n-1}{n}}\int \cos ^{n-2}ax\;dx\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!} ∫ x cos ⁡ a x d x = cos ⁡ a x a 2 + x sin ⁡ a x a + C {\displaystyle \int x\cos ax\;dx={\frac {\cos ax}{a^{2}}}+{\frac {x\sin ax}{a}}+C\,\!} ∫ x 2 cos 2 ⁡ a x d x = x 3 6 + ( x 2 4 a − 1 8 a 3 ) sin ⁡ 2 a x + x 4 a 2 cos ⁡ 2 a x + C {\displaystyle \int x^{2}\cos ^{2}{ax}\;dx={\frac {x^{3}}{6}}+\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax+{\frac {x}{4a^{2}}}\cos 2ax+C\!} ∫ x n cos ⁡ a x d x = x n sin ⁡ a x a − n a ∫ x n − 1 sin ⁡ a x d x = ∑ k = 0 2 k + 1 ≤ n ( − 1 ) k x n − 2 k − 1 a 2 + 2 k n ! ( n − 2 k − 1 ) ! cos ⁡ a x + ∑ k = 0 2 k ≤ n ( − 1 ) k x n − 2 k a 1 + 2 k n ! ( n − 2 k ) ! sin ⁡ a x {\displaystyle \int x^{n}\cos ax\;dx={\frac {x^{n}\sin ax}{a}}-{\frac {n}{a}}\int x^{n-1}\sin ax\;dx\,=\sum _{k=0}^{2k+1\leq n}(-1)^{k}{\frac {x^{n-2k-1}}{a^{2+2k}}}{\frac {n!}{(n-2k-1)!}}\cos ax+\sum _{k=0}^{2k\leq n}(-1)^{k}{\frac {x^{n-2k}}{a^{1+2k}}}{\frac {n!}{(n-2k)!}}\sin ax\!} ∫ cos ⁡ a x x d x = ln ⁡ | a x | + ∑ k = 1 ∞ ( − 1 ) k ( a x ) 2 k 2 k ⋅ ( 2 k ) ! + C {\displaystyle \int {\frac {\cos ax}{x}}dx=\ln |ax|+\sum _{k=1}^{\infty }(-1)^{k}{\frac {(ax)^{2k}}{2k\cdot (2k)!}}+C\,\!} ∫ cos ⁡ a x x n d x = − cos ⁡ a x ( n − 1 ) x n − 1 − a n − 1 ∫ sin ⁡ a x x n − 1 d x (for  n ≠ 1 ) {\displaystyle \int {\frac {\cos ax}{x^{n}}}dx=-{\frac {\cos ax}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\int {\frac {\sin ax}{x^{n-1}}}dx\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ d x cos ⁡ a x = 1 a ln ⁡ | tan ⁡ ( a x 2 + π 4 ) | + C {\displaystyle \int {\frac {dx}{\cos ax}}={\frac {1}{a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C} ∫ d x cos n ⁡ a x = sin ⁡ a x a ( n − 1 ) cos n − 1 ⁡ a x + n − 2 n − 1 ∫ d x cos n − 2 ⁡ a x (for  n > 1 ) {\displaystyle \int {\frac {dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}\,\!} ∫ d x 1 + cos ⁡ a x = 1 a tan ⁡ a x 2 + C {\displaystyle \int {\frac {dx}{1+\cos ax}}={\frac {1}{a}}\tan {\frac {ax}{2}}+C\,\!} ∫ d x 1 − cos ⁡ a x = − 1 a cot ⁡ a x 2 + C {\displaystyle \int {\frac {dx}{1-\cos ax}}=-{\frac {1}{a}}\cot {\frac {ax}{2}}+C\,\!} ∫ x d x 1 + cos ⁡ a x = x a tan ⁡ a x 2 + 2 a 2 ln ⁡ | cos ⁡ a x 2 | + C {\displaystyle \int {\frac {x\;dx}{1+\cos ax}}={\frac {x}{a}}\tan {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\cos {\frac {ax}{2}}\right|+C} ∫ x d x 1 − cos ⁡ a x = − x a cot ⁡ a x 2 + 2 a 2 ln ⁡ | sin ⁡ a x 2 | + C {\displaystyle \int {\frac {x\;dx}{1-\cos ax}}=-{\frac {x}{a}}\cot {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left|\sin {\frac {ax}{2}}\right|+C} ∫ cos ⁡ a x d x 1 + cos ⁡ a x = x − 1 a tan ⁡ a x 2 + C {\displaystyle \int {\frac {\cos ax\;dx}{1+\cos ax}}=x-{\frac {1}{a}}\tan {\frac {ax}{2}}+C\,\!} ∫ cos ⁡ a x d x 1 − cos ⁡ a x = − x − 1 a cot ⁡ a x 2 + C {\displaystyle \int {\frac {\cos ax\;dx}{1-\cos ax}}=-x-{\frac {1}{a}}\cot {\frac {ax}{2}}+C\,\!} ∫ cos ⁡ a 1 x cos ⁡ a 2 x d x = sin ⁡ ( a 1 − a 2 ) x 2 ( a 1 − a 2 ) + sin ⁡ ( a 1 + a 2 ) x 2 ( a 1 + a 2 ) + C (for  | a 1 | ≠ | a 2 | ) {\displaystyle \int \cos a_{1}x\cos a_{2}x\;dx={\frac {\sin(a_{1}-a_{2})x}{2(a_{1}-a_{2})}}+{\frac {\sin(a_{1}+a_{2})x}{2(a_{1}+a_{2})}}+C\qquad {\mbox{(for }}|a_{1}|\neq |a_{2}|{\mbox{)}}\,\!}

انتگرال‌هایی که تنها تابع تانژانت دارند[ویرایش]

انتگرال‌هایی که تنها تابع سکانت دارند[ویرایش]

∫ sec ⁡ a x d x = 1 a ln ⁡ | sec ⁡ a x + tan ⁡ a x | + C {\displaystyle \int \sec {ax}\,dx={\frac {1}{a}}\ln {\left|\sec {ax}+\tan {ax}\right|}+C} ∫ sec 2 ⁡ x d x = tan ⁡ x + C {\displaystyle \int \sec ^{2}{x}\,dx=\tan {x}+C} ∫ sec n ⁡ a x d x = sec n − 1 ⁡ a x tan ⁡ a x a ( n − 1 ) + n − 2 n − 1 ∫ sec n − 2 ⁡ a x d x  (for  n ≠ 1 ) {\displaystyle \int \sec ^{n}{ax}\,dx={\frac {\sec ^{n-1}{ax}\tan {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{ax}\,dx\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!} ∫ sec n ⁡ x d x = sec n − 2 ⁡ x tan ⁡ x n − 1 + n − 2 n − 1 ∫ sec n − 2 ⁡ x d x {\displaystyle \int \sec ^{n}{x}\,dx={\frac {\sec ^{n-2}{x}\tan {x}}{n-1}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{x}\,dx} [۱] ∫ d x sec ⁡ x + 1 = x − tan ⁡ x 2 + C {\displaystyle \int {\frac {dx}{\sec {x}+1}}=x-\tan {\frac {x}{2}}+C} ∫ d x sec ⁡ x − 1 = − x − cot ⁡ x 2 + C {\displaystyle \int {\frac {dx}{\sec {x}-1}}=-x-\cot {\frac {x}{2}}+C}

انتگرال‌هایی که تنها تابع کسکانت دارند[ویرایش]

انتگرال‌هایی که تنها تابع کتانژانت دارند[ویرایش]

انتگرال‌هایی که سینوس و کسینوس دارند[ویرایش]

∫ d x cos ⁡ a x ± sin ⁡ a x = 1 a 2 ln ⁡ | tan ⁡ ( a x 2 ± π 8 ) | + C {\displaystyle \int {\frac {dx}{\cos ax\pm \sin ax}}={\frac {1}{a{\sqrt {2}}}}\ln \left|\tan \left({\frac {ax}{2}}\pm {\frac {\pi }{8}}\right)\right|+C} ∫ d x ( cos ⁡ a x ± sin ⁡ a x ) 2 = 1 2 a tan ⁡ ( a x ∓ π 4 ) + C {\displaystyle \int {\frac {dx}{(\cos ax\pm \sin ax)^{2}}}={\frac {1}{2a}}\tan \left(ax\mp {\frac {\pi }{4}}\right)+C} ∫ d x ( cos ⁡ x + sin ⁡ x ) n = 1 n − 1 ( sin ⁡ x − cos ⁡ x ( cos ⁡ x + sin ⁡ x ) n − 1 − 2 ( n − 2 ) ∫ d x ( cos ⁡ x + sin ⁡ x ) n − 2 ) {\displaystyle \int {\frac {dx}{(\cos x+\sin x)^{n}}}={\frac {1}{n-1}}\left({\frac {\sin x-\cos x}{(\cos x+\sin x)^{n-1}}}-2(n-2)\int {\frac {dx}{(\cos x+\sin x)^{n-2}}}\right)} ∫ cos ⁡ a x d x cos ⁡ a x + sin ⁡ a x = x 2 + 1 2 a ln ⁡ | sin ⁡ a x + cos ⁡ a x | + C {\displaystyle \int {\frac {\cos ax\;dx}{\cos ax+\sin ax}}={\frac {x}{2}}+{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C} ∫ cos ⁡ a x d x cos ⁡ a x − sin ⁡ a x = x 2 − 1 2 a ln ⁡ | sin ⁡ a x − cos ⁡ a x | + C {\displaystyle \int {\frac {\cos ax\;dx}{\cos ax-\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C} ∫ sin ⁡ a x d x cos ⁡ a x + sin ⁡ a x = x 2 − 1 2 a ln ⁡ | sin ⁡ a x + cos ⁡ a x | + C {\displaystyle \int {\frac {\sin ax\;dx}{\cos ax+\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax+\cos ax\right|+C} ∫ sin ⁡ a x d x cos ⁡ a x − sin ⁡ a x = − x 2 − 1 2 a ln ⁡ | sin ⁡ a x − cos ⁡ a x | + C {\displaystyle \int {\frac {\sin ax\;dx}{\cos ax-\sin ax}}=-{\frac {x}{2}}-{\frac {1}{2a}}\ln \left|\sin ax-\cos ax\right|+C} ∫ cos ⁡ a x d x sin ⁡ a x ( 1 + cos ⁡ a x ) = − 1 4 a tan 2 ⁡ a x 2 + 1 2 a ln ⁡ | tan ⁡ a x 2 | + C {\displaystyle \int {\frac {\cos ax\;dx}{\sin ax(1+\cos ax)}}=-{\frac {1}{4a}}\tan ^{2}{\frac {ax}{2}}+{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C} ∫ cos ⁡ a x d x sin ⁡ a x ( 1 − cos ⁡ a x ) = − 1 4 a cot 2 ⁡ a x 2 − 1 2 a ln ⁡ | tan ⁡ a x 2 | + C {\displaystyle \int {\frac {\cos ax\;dx}{\sin ax(1-\cos ax)}}=-{\frac {1}{4a}}\cot ^{2}{\frac {ax}{2}}-{\frac {1}{2a}}\ln \left|\tan {\frac {ax}{2}}\right|+C} ∫ sin ⁡ a x d x cos ⁡ a x ( 1 + sin ⁡ a x ) = 1 4 a cot 2 ⁡ ( a x 2 + π 4 ) + 1 2 a ln ⁡ | tan ⁡ ( a x 2 + π 4 ) | + C {\displaystyle \int {\frac {\sin ax\;dx}{\cos ax(1+\sin ax)}}={\frac {1}{4a}}\cot ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C} ∫ sin ⁡ a x d x cos ⁡ a x ( 1 − sin ⁡ a x ) = 1 4 a tan 2 ⁡ ( a x 2 + π 4 ) − 1 2 a ln ⁡ | tan ⁡ ( a x 2 + π 4 ) | + C {\displaystyle \int {\frac {\sin ax\;dx}{\cos ax(1-\sin ax)}}={\frac {1}{4a}}\tan ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2a}}\ln \left|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|+C} ∫ sin ⁡ a x cos ⁡ a x d x = − 1 2 a cos 2 ⁡ a x + C {\displaystyle \int \sin ax\cos ax\;dx=-{\frac {1}{2a}}\cos ^{2}ax+C\,\!} ∫ sin ⁡ a 1 x cos ⁡ a 2 x d x = − cos ⁡ ( ( a 1 − a 2 ) x ) 2 ( a 1 − a 2 ) − cos ⁡ ( ( a 1 + a 2 ) x ) 2 ( a 1 + a 2 ) + C (for  | a 1 | ≠ | a 2 | ) {\displaystyle \int \sin a_{1}x\cos a_{2}x\;dx=-{\frac {\cos((a_{1}-a_{2})x)}{2(a_{1}-a_{2})}}-{\frac {\cos((a_{1}+a_{2})x)}{2(a_{1}+a_{2})}}+C\qquad {\mbox{(for }}|a_{1}|\neq |a_{2}|{\mbox{)}}\,\!} ∫ sin n ⁡ a x cos ⁡ a x d x = 1 a ( n + 1 ) sin n + 1 ⁡ a x + C (for  n ≠ − 1 ) {\displaystyle \int \sin ^{n}ax\cos ax\;dx={\frac {1}{a(n+1)}}\sin ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!} ∫ sin ⁡ a x cos n ⁡ a x d x = − 1 a ( n + 1 ) cos n + 1 ⁡ a x + C (for  n ≠ − 1 ) {\displaystyle \int \sin ax\cos ^{n}ax\;dx=-{\frac {1}{a(n+1)}}\cos ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!} ∫ sin n ⁡ a x cos m ⁡ a x d x = − sin n − 1 ⁡ a x cos m + 1 ⁡ a x a ( n + m ) + n − 1 n + m ∫ sin n − 2 ⁡ a x cos m ⁡ a x d x (for  m , n > 0 ) {\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;dx=-{\frac {\sin ^{n-1}ax\cos ^{m+1}ax}{a(n+m)}}+{\frac {n-1}{n+m}}\int \sin ^{n-2}ax\cos ^{m}ax\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!} ∫ sin n ⁡ a x cos m ⁡ a x d x = sin n + 1 ⁡ a x cos m − 1 ⁡ a x a ( n + m ) + m − 1 n + m ∫ sin n ⁡ a x cos m − 2 ⁡ a x d x (for  m , n > 0 ) {\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;dx={\frac {\sin ^{n+1}ax\cos ^{m-1}ax}{a(n+m)}}+{\frac {m-1}{n+m}}\int \sin ^{n}ax\cos ^{m-2}ax\;dx\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!} ∫ d x sin ⁡ a x cos ⁡ a x = 1 a ln ⁡ | tan ⁡ a x | + C {\displaystyle \int {\frac {dx}{\sin ax\cos ax}}={\frac {1}{a}}\ln \left|\tan ax\right|+C} ∫ d x sin ⁡ a x cos n ⁡ a x = 1 a ( n − 1 ) cos n − 1 ⁡ a x + ∫ d x sin ⁡ a x cos n − 2 ⁡ a x (for  n ≠ 1 ) {\displaystyle \int {\frac {dx}{\sin ax\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+\int {\frac {dx}{\sin ax\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ d x sin n ⁡ a x cos ⁡ a x = − 1 a ( n − 1 ) sin n − 1 ⁡ a x + ∫ d x sin n − 2 ⁡ a x cos ⁡ a x (for  n ≠ 1 ) {\displaystyle \int {\frac {dx}{\sin ^{n}ax\cos ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+\int {\frac {dx}{\sin ^{n-2}ax\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ sin ⁡ a x d x cos n ⁡ a x = 1 a ( n − 1 ) cos n − 1 ⁡ a x + C (for  n ≠ 1 ) {\displaystyle \int {\frac {\sin ax\;dx}{\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ sin 2 ⁡ a x d x cos ⁡ a x = − 1 a sin ⁡ a x + 1 a ln ⁡ | tan ⁡ ( π 4 + a x 2 ) | + C {\displaystyle \int {\frac {\sin ^{2}ax\;dx}{\cos ax}}=-{\frac {1}{a}}\sin ax+{\frac {1}{a}}\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {ax}{2}}\right)\right|+C} ∫ sin 2 ⁡ a x d x cos n ⁡ a x = sin ⁡ a x a ( n − 1 ) cos n − 1 ⁡ a x − 1 n − 1 ∫ d x cos n − 2 ⁡ a x (for  n ≠ 1 ) {\displaystyle \int {\frac {\sin ^{2}ax\;dx}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}-{\frac {1}{n-1}}\int {\frac {dx}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ sin n ⁡ a x d x cos ⁡ a x = − sin n − 1 ⁡ a x a ( n − 1 ) + ∫ sin n − 2 ⁡ a x d x cos ⁡ a x (for  n ≠ 1 ) {\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ax}}=-{\frac {\sin ^{n-1}ax}{a(n-1)}}+\int {\frac {\sin ^{n-2}ax\;dx}{\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ sin n ⁡ a x d x cos m ⁡ a x = sin n + 1 ⁡ a x a ( m − 1 ) cos m − 1 ⁡ a x − n − m + 2 m − 1 ∫ sin n ⁡ a x d x cos m − 2 ⁡ a x (for  m ≠ 1 ) {\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ^{m}ax}}={\frac {\sin ^{n+1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-m+2}{m-1}}\int {\frac {\sin ^{n}ax\;dx}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!} ∫ sin n ⁡ a x d x cos m ⁡ a x = − sin n − 1 ⁡ a x a ( n − m ) cos m − 1 ⁡ a x + n − 1 n − m ∫ sin n − 2 ⁡ a x d x cos m ⁡ a x (for  m ≠ n ) {\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ^{m}ax}}=-{\frac {\sin ^{n-1}ax}{a(n-m)\cos ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\sin ^{n-2}ax\;dx}{\cos ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!} ∫ sin n ⁡ a x d x cos m ⁡ a x = sin n − 1 ⁡ a x a ( m − 1 ) cos m − 1 ⁡ a x − n − 1 m − 1 ∫ sin n − 2 ⁡ a x d x cos m − 2 ⁡ a x (for  m ≠ 1 ) {\displaystyle \int {\frac {\sin ^{n}ax\;dx}{\cos ^{m}ax}}={\frac {\sin ^{n-1}ax}{a(m-1)\cos ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\sin ^{n-2}ax\;dx}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!} ∫ cos ⁡ a x d x sin n ⁡ a x = − 1 a ( n − 1 ) sin n − 1 ⁡ a x + C (for  n ≠ 1 ) {\displaystyle \int {\frac {\cos ax\;dx}{\sin ^{n}ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!} ∫ cos 2 ⁡ a x d x sin ⁡ a x = 1 a ( cos ⁡ a x + ln ⁡ | tan ⁡ a x 2 | ) + C {\displaystyle \int {\frac {\cos ^{2}ax\;dx}{\sin ax}}={\frac {1}{a}}\left(\cos ax+\ln \left|\tan {\frac {ax}{2}}\right|\right)+C} ∫ cos 2 ⁡ a x d x sin n ⁡ a x = − 1 n − 1 ( cos ⁡ a x a sin n − 1 ⁡ a x ) + ∫ d x sin n − 2 ⁡ a x ) (for  n ≠ 1 ) {\displaystyle \int {\frac {\cos ^{2}ax\;dx}{\sin ^{n}ax}}=-{\frac {1}{n-1}}\left({\frac {\cos ax}{a\sin ^{n-1}ax)}}+\int {\frac {dx}{\sin ^{n-2}ax}}\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}} ∫ cos n ⁡ a x d x sin m ⁡ a x = − cos n + 1 ⁡ a x a ( m − 1 ) sin m − 1 ⁡ a x − n − m − 2 m − 1 ∫ cos n ⁡ a x d x sin m − 2 ⁡ a x (for  m ≠ 1 ) {\displaystyle \int {\frac {\cos ^{n}ax\;dx}{\sin ^{m}ax}}=-{\frac {\cos ^{n+1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-m-2}{m-1}}\int {\frac {\cos ^{n}ax\;dx}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!} ∫ cos n ⁡ a x d x sin m ⁡ a x = cos n − 1 ⁡ a x a ( n − m ) sin m − 1 ⁡ a x + n − 1 n − m ∫ cos n − 2 ⁡ a x d x sin m ⁡ a x (for  m ≠ n ) {\displaystyle \int {\frac {\cos ^{n}ax\;dx}{\sin ^{m}ax}}={\frac {\cos ^{n-1}ax}{a(n-m)\sin ^{m-1}ax}}+{\frac {n-1}{n-m}}\int {\frac {\cos ^{n-2}ax\;dx}{\sin ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!} ∫ cos n ⁡ a x d x sin m ⁡ a x = − cos n − 1 ⁡ a x a ( m − 1 ) sin m − 1 ⁡ a x − n − 1 m − 1 ∫ cos n − 2 ⁡ a x d x sin m − 2 ⁡ a x (for  m ≠ 1 ) {\displaystyle \int {\frac {\cos ^{n}ax\;dx}{\sin ^{m}ax}}=-{\frac {\cos ^{n-1}ax}{a(m-1)\sin ^{m-1}ax}}-{\frac {n-1}{m-1}}\int {\frac {\cos ^{n-2}ax\;dx}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}

انتگرال‌هایی که سینوس و تانژانت دارند[ویرایش]

انتگرال‌هایی که کسینوس و تانژانت دارند[ویرایش]

∫ tan n ⁡ a x d x cos 2 ⁡ a x = 1 a ( n + 1 ) tan n + 1 ⁡ a x + C (for  n ≠ − 1 ) {\displaystyle \int {\frac {\tan ^{n}ax\;dx}{\cos ^{2}ax}}={\frac {1}{a(n+1)}}\tan ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}

انتگرال‌هایی که سینوس و کتانژانت دارند[ویرایش]

∫ cot n ⁡ a x d x sin 2 ⁡ a x = 1 a ( n + 1 ) cot n + 1 ⁡ a x + C (for  n ≠ − 1 ) {\displaystyle \int {\frac {\cot ^{n}ax\;dx}{\sin ^{2}ax}}={\frac {1}{a(n+1)}}\cot ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}

انتگرال‌هایی که کسینوس و کتانژانت دارند[ویرایش]

∫ cot n ⁡ a x d x cos 2 ⁡ a x = 1 a ( 1 − n ) tan 1 − n ⁡ a x + C (for  n ≠ 1 ) {\displaystyle \int {\frac {\cot ^{n}ax\;dx}{\cos ^{2}ax}}={\frac {1}{a(1-n)}}\tan ^{1-n}ax+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}

انتگرال‌های با بازه‌های متقارن[ویرایش]

منابع[ویرایش]

1. ↑ Stewart, James. Calculus: Early Transcendentals, 6th Edition. Thomson: 2008