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Integral Trigonometri – Fungsi Beserta Contoh Soal dan Jawaban

Daftar integral dari fungsi trigonometri

Daftar integral trigonometri (antiderivatif: integral tak tentu) dari fungsi trigonometri. Untuk antiderivatif yang melibatkan baik fungsi eksponensial dan trigonometri, lihat Daftar integral dari fungsi eksponensial. Untuk daftar lengkap fungsi-fungsi antiderivatif, lihat Tabel integral. Untuk antiderivatif khusus yang melibatkan fungsi trigonometri.

Umumnya, jika fungsi {\displaystyle \sin(x)} adalah suatu fungsi trigonometri, dan {\displaystyle \cos(x)} adalah turunannya,

{\displaystyle \int a\cos nx\;\mathrm {d} x={\frac {a}{n}}\sin nx+C}

Dalam semua rumus, konstanta a diasumsikan bukan nol, dan C melambangkan konstanta integrasi.

 

Integral trigonometri – Integrand melibatkan hanya sinus

{\displaystyle \int [fusion_builder_container hundred_percent=
{\displaystyle \int \sin ^{2}{ax}\;\mathrm {d} x={\frac {x}{2}}-{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}-{\frac {1}{2a}}\sin ax\cos ax+C\!}
{\displaystyle \int \sin ^{3}{ax}\;\mathrm {d} x={\frac {\cos 3ax}{12a}}-{\frac {3\cos ax}{4a}}+C\!}
{\displaystyle \int x\sin ^{2}{ax}\;\mathrm {d} x={\frac {x^{2}}{4}}-{\frac {x}{4a}}\sin 2ax-{\frac {1}{8a^{2}}}\cos 2ax+C\!}
{\displaystyle \int x^{2}\sin ^{2}{ax}\;\mathrm {d} x={\frac {x^{3}}{6}}-\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax-{\frac {x}{4a^{2}}}\cos 2ax+C\!}
{\displaystyle \int \sin [fusion_builder_container hundred_percent=
{\displaystyle \int \sin ^{n}{ax}\;\mathrm {d} x=-{\frac {\sin ^{n-1}ax\cos ax}{na}}+{\frac {n-1}{n}}\int \sin ^{n-2}ax\;\mathrm {d} x\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{\sin ax}}={\frac {1}{a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\mathrm {d} x}{\sin ^{n}ax}}={\frac {\cos ax}{a(1-n)\sin ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {\mathrm {d} x}{\sin ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}\,\!}
{\displaystyle \int x\sin ax\;\mathrm {d} x={\frac {\sin ax}{a^{2}}}-{\frac {x\cos ax}{a}}+C\,\!}
{\displaystyle \int x^{n}\sin ax\;\mathrm {d} x=-{\frac {x^{n}}{a}}\cos ax+{\frac {n}{a}}\int x^{n-1}\cos ax\;\mathrm {d} x=\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{)}}\,\!}
{\displaystyle \int {\frac {\sin ax}{x}}\mathrm {d} x=\sum _{n=0}^{\infty }(-1)^{n}{\frac {(ax)^{2n+1}}{(2n+1)\cdot (2n+1)!}}+C\,\!}
{\displaystyle \int {\frac {\sin ax}{x^{n}}}\mathrm {d} x=-{\frac {\sin ax}{(n-1)x^{n-1}}}+{\frac {a}{n-1}}\int {\frac {\cos ax}{x^{n-1}}}\mathrm {d} x\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{1\pm \sin ax}}={\frac {1}{a}}\tan \left({\frac {ax}{2}}\mp {\frac {\pi }{4}}\right)+C}
{\displaystyle \int {\frac {x\;\mathrm {d} x}{1+\sin ax}}={\frac {x}{a}}\tan \left({\frac {ax}{2}}-{\frac {\pi }{4}}\right)+{\frac {2}{a^{2}}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {x\;\mathrm {d} x}{1-\sin ax}}={\frac {x}{a}}\cot \left({\frac {\pi }{4}}-{\frac {ax}{2}}\right)+{\frac {2}{a^{2}}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{1\pm \sin ax}}=\pm x+{\frac {1}{a}}\tan \left({\frac {\pi }{4}}\mp {\frac {ax}{2}}\right)+C}

 

Integral trigonometri – Integrand melibatkan hanya kosinus

{\displaystyle \int \cos ax\;\mathrm {d} x={\frac {1}{a}}\sin ax+C\,\!}
{\displaystyle \int \cos ^{2}{ax}\;\mathrm {d} x={\frac {x}{2}}+{\frac {1}{4a}}\sin 2ax+C={\frac {x}{2}}+{\frac {1}{2a}}\sin ax\cos ax+C\!}
{\displaystyle \int \cos ^{n}ax\;\mathrm {d} x={\frac {\cos ^{n-1}ax\sin ax}{na}}+{\frac {n-1}{n}}\int \cos ^{n-2}ax\;\mathrm {d} x\qquad {\mbox{(for }}n>0{\mbox{)}}\,\!}
{\displaystyle \int x\cos ax\;\mathrm {d} x={\frac {\cos ax}{a^{2}}}+{\frac {x\sin ax}{a}}+C\,\!}
{\displaystyle \int x^{2}\cos ^{2}{ax}\;\mathrm {d} x={\frac {x^{3}}{6}}+\left({\frac {x^{2}}{4a}}-{\frac {1}{8a^{3}}}\right)\sin 2ax+{\frac {x}{4a^{2}}}\cos 2ax+C\!}
{\displaystyle \int x^{n}\cos ax\;\mathrm {d} x={\frac {x^{n}\sin ax}{a}}-{\frac {n}{a}}\int x^{n-1}\sin ax\;\mathrm {d} x\,=\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\!}
{\displaystyle \int {\frac {\cos ax}{x}}\mathrm {d} x=\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|ax|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\cos ax}{x^{n}}}\mathrm {d} x=-{\frac {\cos ax}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\int {\frac {\sin ax}{x^{n-1}}}\mathrm {d} x\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{\cos ax}}={\frac {1}{a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\mathrm {d} x}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}+{\frac {n-2}{n-1}}\int {\frac {\mathrm {d} x}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n>1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{1+\cos ax}}={\frac {1}{a}}\tan {\frac {ax}{2}}+C\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{1-\cos ax}}=-{\frac {1}{a}}\cot {\frac {ax}{2}}+C}
{\displaystyle \int {\frac {x\;\mathrm {d} x}{1+\cos ax}}={\frac {x}{a}}\tan {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {x\;\mathrm {d} x}{1-\cos ax}}=-{\frac {x}{a}}\cot {\frac {ax}{2}}+{\frac {2}{a^{2}}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{1+\cos ax}}=x-{\frac {1}{a}}\tan {\frac {ax}{2}}+C\,\!}
{\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{1-\cos ax}}=-x-{\frac {1}{a}}\cot {\frac {ax}{2}}+C\,\!}
{\displaystyle \int \cos [/fusion_text][fusion_builder_container hundred_percent=

 

Integral trigonometri – Integrand melibatkan hanya tangen

{\displaystyle \int \tan ax\;\mathrm {d} x=-{\frac {1}{a}}\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|\cos ax|[fusion_builder_container hundred_percent=
{\displaystyle \int \tan ^{2}{x}\,\mathrm {d} x=\tan {x}-x+C}
{\displaystyle \int \tan ^{n}ax\;\mathrm {d} x={\frac {1}{a(n-1)}}\tan ^{n-1}ax-\int \tan ^{n-2}ax\;\mathrm {d} x\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{q\tan ax+p}}={\frac {1}{p^{2}+q^{2}}}(px+{\frac {q}{a}}\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\mathrm {d} x}{\tan ax+1}}={\frac {x}{2}}+{\frac {1}{2a}}\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\mathrm {d} x}{\tan ax-1}}=-{\frac {x}{2}}+{\frac {1}{2a}}\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\tan ax\;\mathrm {d} x}{\tan ax+1}}={\frac {x}{2}}-{\frac {1}{2a}}\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|\sin ax+\cos ax|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\tan ax\;\mathrm {d} x}{\tan ax-1}}={\frac {x}{2}}+{\frac {1}{2a}}\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|\sin ax-\cos ax|[fusion_builder_container hundred_percent=

 

Integral trigonometri – Integrand melibatkan hanya sekan

{\displaystyle \int \sec {ax}\,\mathrm {d} x={\frac {1}{a}}\ln {\left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int \sec ^{2}{x}\,\mathrm {d} x=\tan {x}+C}
{\displaystyle \int \sec ^{3}x\,dx={\frac {1}{2}}\sec x\tan x+{\frac {1}{2}}\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int \sec ^{n}{ax}\,\mathrm {d} x={\frac {\sec ^{n-2}{ax}\tan {ax}}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \sec ^{n-2}{ax}\,\mathrm {d} x\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{\sec {x}+1}}=x-\tan {\frac {x}{2}}+C}

 

Integral trigonometri – Integrands melibatkan hanya kosekan

{\displaystyle \int csc(ax)\mathrm {d} x=-{\frac {1}{a}}\ln {\left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int \csc ^{2}{x}\,\mathrm {d} x=-\cot {x}+C}
{\displaystyle \int \csc ^{n}{ax}\,\mathrm {d} x=-{\frac {\csc ^{n-1}\left(ax\right)\cos \left(ax\right)}{a(n-1)}}\,+\,{\frac {n-2}{n-1}}\int \csc ^{n-2}{ax}\,\mathrm {d} x\qquad {\mbox{ (for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{\csc {x}+1}}=x-{\frac {2\sin {\frac {x}{2}}}{\cos {\frac {x}{2}}+\sin {\frac {x}{2}}}}+C}
{\displaystyle \int {\frac {\mathrm {d} x}{\csc {x}-1}}={\frac {2\sin {\frac {x}{2}}}{\cos {\frac {x}{2}}-\sin {\frac {x}{2}}}}-x+C}

 

Integral trigonometri – Integrand melibatkan hanya kotangen

{\displaystyle \int \cot ax\;\mathrm {d} x={\frac {1}{a}}\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|\sin ax|[fusion_builder_container hundred_percent=
{\displaystyle \int \cot ^{n}ax\;\mathrm {d} x=-{\frac {1}{a(n-1)}}\cot ^{n-1}ax-\int \cot ^{n-2}ax\;\mathrm {d} x\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{1+\cot ax}}=\int {\frac {\tan ax\;\mathrm {d} x}{\tan ax+1}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{1-\cot ax}}=\int {\frac {\tan ax\;\mathrm {d} x}{\tan ax-1}}\,\!}

 

Integral trigonometri – Integrand melibatkan sinus dan kosinus

{\displaystyle \int {\frac {\mathrm {d} x}{\cos ax\pm \sin ax}}={\frac {1}{a{\sqrt {2}}}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\mathrm {d} x}{(\cos ax\pm \sin ax)^{2}}}={\frac {1}{2a}}\tan \left(ax\mp {\frac {\pi }{4}}\right)+C}
{\displaystyle \int {\frac {\mathrm {d} x}{(\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 {\mathrm {d} x}{(\cos x+\sin x)^{n-2}}}\right)}
{\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\cos ax+\sin ax}}={\frac {x}{2}}+{\frac {1}{2a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\cos ax-\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ax+\sin ax}}={\frac {x}{2}}-{\frac {1}{2a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|\sin ax+\cos ax\right|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ax-\sin ax}}=-{\frac {x}{2}}-{\frac {1}{2a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|\sin ax-\cos ax\right|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\sin ax(1+\cos ax)}}=-{\frac {1}{4a}}\tan ^{2}{\frac {ax}{2}}+{\frac {1}{2a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|\tan {\frac {ax}{2}}\right|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\sin ax(1-\cos ax)}}=-{\frac {1}{4a}}\cot ^{2}{\frac {ax}{2}}-{\frac {1}{2a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|\tan {\frac {ax}{2}}\right|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ax(1+\sin ax)}}={\frac {1}{4a}}\cot ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)+{\frac {1}{2a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ax(1-\sin ax)}}={\frac {1}{4a}}\tan ^{2}\left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)-{\frac {1}{2a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|\tan \left({\frac {ax}{2}}+{\frac {\pi }{4}}\right)\right|[fusion_builder_container hundred_percent=
{\displaystyle \int \sin ax\cos ax\;\mathrm {d} x=-{\frac {1}{2a}}\cos ^{2}ax+C\,\!}
{\displaystyle \int \sin a_{1}x\cos a_{2}x\;\mathrm {d} x=-{\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 }}[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|a_{1}|\neq |a_{2}|[fusion_builder_container hundred_percent=
{\displaystyle \int \sin ^{n}ax\cos ax\;\mathrm {d} x={\frac {1}{a(n+1)}}\sin ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
{\displaystyle \int \sin ax\cos ^{n}ax\;\mathrm {d} x=-{\frac {1}{a(n+1)}}\cos ^{n+1}ax+C\qquad {\mbox{(for }}n\neq -1{\mbox{)}}\,\!}
{\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;\mathrm {d} x=-{\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\;\mathrm {d} x\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}
juga: {\displaystyle \int \sin ^{n}ax\cos ^{m}ax\;\mathrm {d} x={\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\;\mathrm {d} x\qquad {\mbox{(for }}m,n>0{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{\sin ax\cos ax}}={\frac {1}{a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\mathrm {d} x}{\sin ax\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+\int {\frac {\mathrm {d} x}{\sin ax\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\mathrm {d} x}{\sin ^{n}ax\cos ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+\int {\frac {\mathrm {d} x}{\sin ^{n-2}ax\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\sin ax\;\mathrm {d} x}{\cos ^{n}ax}}={\frac {1}{a(n-1)\cos ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\sin ^{2}ax\;\mathrm {d} x}{\cos ax}}=-{\frac {1}{a}}\sin ax+{\frac {1}{a}}\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\sin ^{2}ax\;\mathrm {d} x}{\cos ^{n}ax}}={\frac {\sin ax}{a(n-1)\cos ^{n-1}ax}}-{\frac {1}{n-1}}\int {\frac {\mathrm {d} x}{\cos ^{n-2}ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\sin ^{n}ax\;\mathrm {d} x}{\cos ax}}=-{\frac {\sin ^{n-1}ax}{a(n-1)}}+\int {\frac {\sin ^{n-2}ax\;\mathrm {d} x}{\cos ax}}\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\sin ^{n}ax\;\mathrm {d} x}{\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\;\mathrm {d} x}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
juga: {\displaystyle \int {\frac {\sin ^{n}ax\;\mathrm {d} x}{\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\;\mathrm {d} x}{\cos ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}
juga: {\displaystyle \int {\frac {\sin ^{n}ax\;\mathrm {d} x}{\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\;\mathrm {d} x}{\cos ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\cos ax\;\mathrm {d} x}{\sin ^{n}ax}}=-{\frac {1}{a(n-1)\sin ^{n-1}ax}}+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}
{\displaystyle \int {\frac {\cos ^{2}ax\;\mathrm {d} x}{\sin ax}}={\frac {1}{a}}\left(\cos ax+\ln \left[/fusion_text][/fusion_builder_column][/fusion_builder_row][/fusion_builder_container]|\tan {\frac {ax}{2}}\right|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\cos ^{2}ax\;\mathrm {d} x}{\sin ^{n}ax}}=-{\frac {1}{n-1}}\left({\frac {\cos ax}{a\sin ^{n-1}ax)}}+\int {\frac {\mathrm {d} x}{\sin ^{n-2}ax}}\right)\qquad {\mbox{(for }}n\neq 1{\mbox{)}}}
{\displaystyle \int {\frac {\cos ^{n}ax\;\mathrm {d} x}{\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\;\mathrm {d} x}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}
juga: {\displaystyle \int {\frac {\cos ^{n}ax\;\mathrm {d} x}{\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\;\mathrm {d} x}{\sin ^{m}ax}}\qquad {\mbox{(for }}m\neq n{\mbox{)}}\,\!}
juga: {\displaystyle \int {\frac {\cos ^{n}ax\;\mathrm {d} x}{\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\;\mathrm {d} x}{\sin ^{m-2}ax}}\qquad {\mbox{(for }}m\neq 1{\mbox{)}}\,\!}

Integral trigonometri – Integrand melibatkan baik sinus dan tangen

{\displaystyle \int \sin ax\tan ax\;\mathrm {d} x={\frac {1}{a}}(\ln[/fusion_text][/fusion_builder_column] [/fusion_builder_row][/fusion_builder_container]|[fusion_builder_container hundred_percent=
{\displaystyle \int {\frac {\tan ^{n}ax\;\mathrm {d} x}{\sin ^{2}ax}}={\frac {1}{a(n-1)}}\tan ^{n-1}(ax)+C\qquad {\mbox{(for }}n\neq 1{\mbox{)}}\,\!}

Integral trigonometri – Integrand melibatkan baik kosinus dan tangen

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

Integral trigonometri – Integrand melibatkan baik sinus dan kotangen

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

Integral trigonometri – Integrand melibatkan baik kosinus dan kotangen

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

Integral trigonometri – Integrand melibatkan baik sekan dan tangen

{\displaystyle \int \sec x\tan x\;\mathrm {d} x=\sec x+C}

Integral dengan limit simetris

{\displaystyle \int _{-c}^{c}\sin {x}\;\mathrm {d} x=0\!}
{\displaystyle \int _{-c}^{c}\cos {x}\;\mathrm {d} x=2\int _{0}^{c}\cos {x}\;\mathrm {d} x=2\int _{-c}^{0}\cos {x}\;\mathrm {d} x=2\sin {c}\!}
{\displaystyle \int _{-c}^{c}\tan {x}\;\mathrm {d} x=0\!}
{\displaystyle \int _{-{\frac {a}{2}}}^{\frac {a}{2}}x^{2}\cos ^{2}{\frac {n\pi x}{a}}\;\mathrm {d} x={\frac {a^{3}(n^{2}\pi ^{2}-6)}{24n^{2}\pi ^{2}}}\qquad {\mbox{(for }}n=1,3,5...{\mbox{)}}\,\!}
{\displaystyle \int _{\frac {-a}{2}}^{\frac {a}{2}}x^{2}\sin ^{2}{\frac {n\pi x}{a}}\;\mathrm {d} x={\frac {a^{3}(n^{2}\pi ^{2}-6(-1)^{n})}{24n^{2}\pi ^{2}}}={\frac {a^{3}}{24}}(1-6{\frac {(-1)^{n}}{n^{2}\pi ^{2}}})\qquad {\mbox{(for }}n=1,2,3,...{\mbox{)}}\,\!}

Integral satu lingkaran penuh

{\displaystyle \int _{0}^{2\pi }\sin ^{2m+1}{x}\cos ^{2n+1}{x}\;\mathrm {d} x=0\!\qquad \{n,m\}\in \mathbb {Z} }

 


 

Contoh Soal dan Jawaban Integral trigonometri

1. Soal: Tentukan hasil dari ∫sin4 x dx =…

Jawaban:

∫sin4 x dx
=∫ (sin2 x)2 dx
= ∫ (1/2 – 1/2 cos 2x)2 dx
= ∫ (1/4 – 1/2 cos 2x + 1/4 cos2 2x) dx
= ∫ (1/4 – 1/2 cos 2x + 1/4 (1/2 + 1/2 cos 4x)) dx
= ∫ (1/4 – 1/2 cos 2x + 1/8 + 1/8 cos 4x) dx
= ∫ (3/8 – 1/2 cos 2x + 1/8 cos 4x) dx
= 3/8 x – 1/4 sin 2x + 1/32 sin 4x + c

 

 

2. Soal: ∫ (x2 – 4x) cos (x3 – 6x2 + 7) dx =…

Jawaban:

misal y = x3 – 6x2 + 7
maka dy/dx = 3x2 – 12x
sehingga dx = dy/(3x2 – 12x)
atau  dx = 1/3 dy/(x2 – 4x)

Jadi

∫ (x2 – 4x) cos (x3 – 6x2 + 7) dx
= ∫ (x2 – 4x) cos y 1/3 dy/(x2 – 4x)
= 1/3 ∫ cos y dy
= 1/3 sin y + c
= 1/3 sin (x3 – 6x2 + 7) + c

 

 

3. Soal:  \int sin^2\;3x\; dx = …..

Jawaban:

Gunakan rumus trigonometri
{\color{Red} sin^2\;x=\frac 12-\frac 12\;cos\;2x}   sehingga
 \begin{align*}sin^2\;3x&=&\frac 12-\frac 12\;cos\;2(3x)\\&=&\frac 12-\frac 12\;cos\;6x \end{align*}
Maka :
\begin{align*}\int sin^2\;3x\;dx&=&\int \left (\frac 12-\frac 12\;cos\;6x \right )dx\\&=&\frac 12x-\frac 12.\frac 16\;sin\;6x+C\\&=&\frac 12x-\frac{1}{12}\;sin\;6x+C \end{align*}

 

 

4. Soal: ∫ cos4 x dx

Jawaban:

∫ cos4 x dx
=∫ (cos2 x)2 dx
= ∫ (1/2 + 1/2 cos 2x)2 dx
= ∫ (1/4 + 1/2 cos 2x + 1/4 cos2 2x) dx
= ∫ (1/4 + 1/2 cos 2x + 1/4 (1/2 + 1/2 cos 4x)) dx
= ∫ (1/4 + 1/2 cos 2x + 1/8 + 1/8 cos 4x) dx
= ∫ (3/8 + 1/2 cos 2x + 1/8 cos 4x) dx
= 3/8 x + 1/4 sin 2x + 1/32 sin 4x + c

 

 

5. Soal: ∫ (tan 2x − sec 2x)2 dx =…

Jawaban:
⇒ ∫ (tan22x + sec22x − 2 sec 2x tan 2x) dx
⇒ ∫ (sec22x − 1 + sec22x − 2 sec 2x tan 2x) dx
⇒ ∫ (2sec22x − 2 sec 2x tan 2x − 1) dx
2/2tan 2x − 2/2sec 2x − x + C
= tan 2x − sec 2x − x + C

 

 

6. Soal: ∫ (tan24x + 3) dx =…

Jawaban:
⇒ ∫ (sec24x − 1 + 3) dx
⇒ ∫ (sec24x + 2) dx
= ¼tan 4x + 2x + C

 

7. Soal: \displaystyle \int \sqrt{1+\sin 2x} \, \mathrm{d}x = ...

Jawaban:

Karena   \cos^2x+\sin^2x = 1    dan   \sin 2x = 2 \sin x \cos x   sehingga

\displaystyle \begin{aligned} \int \sqrt{1 + \sin 2x} &= \int \sqrt{\cos^2 x + \sin 2x + \sin^2 x} \, \mathrm{d}x \\ & = \int \sqrt{\cos^2x + 2\sin x \cos x+ \sin^2x} \, \mathrm{d}x \\ & = \int \sqrt{(\cos x + \sin x)^2} \, \mathrm{d}x \\ & = \int (\cos x + \sin x) \, \mathrm{d}x \\ & = \sin x - \cos x + C \end{aligned}

Jawaban : \sin x - \cos x + C

 

 

8. Soal: \displaystyle \int \left( \sin x + \sin^3 x + \sin^5 + \dots \right) \, \mathrm{d}x = ...

Jawaban:

Bentuk dalam integral merupakan Deret Geometri tak hingga dengan suku pertama a = \sin xdan rasio r = \sin^2 x, sehingga bentuk integral tersebut dapat ditulis

\displaystyle \begin{aligned} \int \left( \sin x + \sin^3 x + \sin^5 + \dots \right) \, \mathrm{d}x &= \int \dfrac{\sin x}{1-\sin^2 x} \, \mathrm{d}x \end{aligned}

Dengan memisalkan u = \cos x \rightarrow -\mathrm{d}u = \sin x \, \mathrm{d}x dan mengganti 1 - \sin^2 x = \cos^2 x = u^2 maka

\displaystyle \begin{aligned} \int \dfrac{\sin x}{1-\sin^2 x} \, \mathrm{d}x &= \int \dfrac{1}{u^2} \, (-\mathrm{d}u) \\ &= u^{-1} + C \\ &= \sec x + C \\ \end{aligned}

9. Soal: \displaystyle \int \sqrt{1-\cos x} \, \mathrm{d}x = ...

Jawaban:

Dari persamaan trigonometri \cos x = 1 - 2\sin^2\frac{1}{2}x

\displaystyle \begin{aligned} \int \sqrt{1-\cos x} \, \mathrm{d}x &= \int \sqrt{1-(1-2\sin ^2 \frac{1}{2}x)} \, \mathrm{d}x \\ &= \int \sqrt{2\sin^2\frac{1}{2}x} \, \mathrm{d}x \\ &= \sqrt{2} \int \sin \frac{1}{2}x \, \mathrm{d}x \\ &= -2\sqrt{2} \cos \frac{1}{2}x + C \end{aligned}

Jawaban : -2\sqrt{2} \cos \frac{1}{2}x + C

 

 

10. Soal: 

∫ (x + 3) cos (2x − π)dx =…..
|____| |__________|
u                  dv

Jawaban:

Langkah pertama yaitu tentukan terlebih dulu mana u dan mana dv
Misalkan

(x + 3) adalah u, dan sisanya, cos (2x − π)dx sebagai dv,
u = (x + 3) …(Persamaan 1)
dv = cos (2x − π) dx … (Persamaan 2)

Langkah pertama selesai, kita tengok lagi rumus dasar integral parsial:

∫ u dv = uv − ∫v du

Terlihat di situ kita perlu u, perlu v dan perlu du. u nya sudah ada, tinggal mencari du dan v nya.

Dari persamaan 1, untuk menentukan du, caranya turunkan u nya,
u = (x + 3)
du/dx = 1
du = dx

Dari persamaan 2, untuk menentukan v,
dv = cos (2x − π)dx
atau
dv/dx = cos (2x − π)

dv/dx artinya turunan dari v adalah cos (2x − π), untuk mendapatkan v, berarti kita harus integralkan cos (2x − π) jika lupa, tengok lagi cara integral fungsi trigonometri,

v = ∫ cos (2x − π) dx = 1/2 sin (2x − π) + C

Kita rangkum lagi :
u = (x + 3)
v = 1/2 sin (2x − π)
du = dx

masukkan nilai-nilai yang sudah dicari tadi sesuai rumus integral parsial:
16 ∫ (x + 3) cos (2x − π)dx
Simpan dulu 16 nya, terakhir nanti hasilnya baru di kali 16
= uv − ∫v du
= (x + 3) 1/2 sin (2x − π) − ∫ 1/2 sin (2x − π) du
1/2 (x + 3) sin (2x − π) − ∫ 1/2 sin (2x − π) dx
1/2 (x + 3) sin (2x − π) − 1/2 {− 1/2 cos (2x − π) }
1/2 (x + 3) sin (2x − π) + 1/4 cos (2x − π)

kalikan 16, tambahkan + C nya

= 16 { 1/2 (x + 3) sin (2x − π) + 1/4 cos (2x − π) } + C
= 8 (x + 3) sin (2x − π) + 4 cos (2x − π) + C

 

 

11. Soal:   \[ \int cos \left( 2x + 5 \right) \; dx = ...\]

Pembahasan:
Misalkan:

  \[ u = 2x + 5 \]

  \[ du = 2 dx \rightarrow dx = \frac{du}{2} \]

Sehingga,

  \[ \int cos \left( 2x + 5 \right) \; dx = \int cos \; u \; \frac{du}{2} \]

  \[ = \frac{1}{2} \int cos \; u \; du \]

  \[ = \frac{1}{2} sin \; u + C \]

  \[ = \frac{1}{2} sin \; \left(2x + 5 \right) + C \]

 

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By | 2018-07-12T20:22:42+00:00 Juli 10th, 2018|Matematika|0 Comments

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