Formula:
Integration of the powers of sine and cosine functions
- Case 1. Has the form ∫sinnax cosax dx. I believe this is the basic and the easiest to evaluate because of the presence of the cosine.
- Case 2. Has the form ∫cosnax sinax dx. In this case instead sine, the one with the exponent is the cosine and it is a lot easier to evaluate because of the presence of sine.
- Case 3. Has the form ∫sinmax cosnax dx where m is odd integer. The first step is to factor sin x, then change the remaining factor into sin2ax = 1 - cos2ax.
- Case 4. Has the form ∫sinmax cosnax dx where n is odd integer. Here we can factor cos x, then change the remaining factor into cos2ax = 1 - sin2ax.
- Case 5. Has the form ∫sinmax cosnax dx where m and n are even integers. In this case we will be using the half-angle identities below.
- Case 6. Has the form ∫sinaxcosbx dx, ∫sinaxsinbx dx, ∫cosaxcosbx dx. To find the integrals for this kind of combination of trigonometric functions, we will be needing the identities below.
Examples 1: ∫sin53x cos3x dx
Examples 2: ∫cos34x sin4x dx
Solution: Let u = cos4x, du = -4sin4x dx
= ![answer for the integral of cos^3(4x)sin(4x) dx solution for the integral of cos^3(4x)sin(4x) dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_unjN2TAVfnisPq0HD3WKYNZOsz8Lu-M3MrpvKtnvY-VhH__xQysWWHndIOLKUH6LPHnm5PC_YyqWXVhf0MtFbFHWcnGdP9rjJkAyb1fg2WIyKDtglNSKpDIh0fCuuZgTw8ozV2BDDACRmanI4BUOxx6zCEAJtwrnrOaw4dN9hUCByxDT7EJv4Wp8PhHI-zZuyvoMN4GZhnJX4=s0-d)
Examples 3:
∫sin33x dx = ∫sin23x ∙ sin3x dx
= ∫(1 - cos23x) sin3x dx
= ∫sin3x dx - ∫cos23x sin3x dx
=![answer for the integral of sin^3(3x) dx solution for the integral of sin^3(3x) dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tA3h2263kxRTQeVd6k8B5kOSDdR6rH9lg1LtP6GwVLJuoZFY-QQ-_VtvikIQ2T3PN-NPaKwB8v_fiJqNYIiCtO64TkabsGWu2w75aBUPRmax2xsIpY-HX9ZSNHmaYqkQK-P42HWhC-X-duKPPZ2x5mmNAsVzyUS84t3h6-bA5Ck1g_9gdlaXgQP1K7uSmGt_MkUFgKY5P151sWh4XyYcQPdFv1XZGKhuej_rbDeSMeyR4XMvkE1Btrert4h3Q=s0-d)
Examples 4: = ∫(1 - cos23x) sin3x dx
= ∫sin3x dx - ∫cos23x sin3x dx
=
∫sin52x cos22x dx = ∫sin42x ∙ sin2x cos22x dx
= ∫(sin22x)2 cos22x sin2x dx
= ∫(1 - cos22x)2 cos22x sin2x dx
= ∫(1 - 2cos22x + cos42x) cos22x sin2x dx
= ∫cos22x sin2x dx - 2∫cos42x sin2x dx + ∫cos62x sin2x dx
Let u = cos2x, du = -2sin2x dx
=![answer for the integral of sin^5(2x)cos^2(2x) dx solution for the integral of sin^5(2x)cos^2(2x) dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_txevkmEnJ79VETajOT6mtpxVi3J1BnpgCFk3mq7kJ976KBgWlX6OxN1nETRTmdYgiklZQWdxEKX3Dqp9PqYJE_Pep2pSJIQTD80r1BSNqQrq-t2PpKH_UEh1ctWLqxua2hxMFDwIdoLu1dugkCoQi-uxPzh8Lnf3tLN4bF2DYcIN7jkiSQxlhE93vPC5Wn-4VUhqc1Z97pYGb2SavMmOLzkZew4rjOSXbY-AOn1ga8s6huK5M0ChSSd277G4DWne7MfLlnVluNCInTT6txhn7OxO-GDqsmtu4vCK044jAJY-YvAM9lofk0WwiLT6nAyKCMDGMzL2oFostD58BH_L_kCjIih7Ja01qcpsbptY8RZg=s0-d)
= ∫(sin22x)2 cos22x sin2x dx
= ∫(1 - cos22x)2 cos22x sin2x dx
= ∫(1 - 2cos22x + cos42x) cos22x sin2x dx
= ∫cos22x sin2x dx - 2∫cos42x sin2x dx + ∫cos62x sin2x dx
Let u = cos2x, du = -2sin2x dx
=
Examples 5:
∫cos34x dx = ∫cos24x ∙ cos4x dx
= ∫(1 - sin24x) cos4x dx
= ∫cos4x dx - ∫sin24x cos4x dx
=![answer for the integral of cos^3(4x) dx solution for the integral of cos^3(4x) dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sb_bk-GElmgCOk-C9RG2wMvl-i1ed9MArmiAJy16d_PkUWGIKwDIhXDVk3BZ2miWJewkGRBNmZaXyY-exVJuDZpddwvqYZYJUnNCo3f_7ljyV3x_n-opkwDlUk-wRS0Kn4CtAU7uyT6qOQlgpSkt4lAk-JAFu3odC-AenanSL4w4BevsA-6ywgx_VfxNmHj3e-EKmRW76DtTNz4UhVJppWK1raMyuGRSkb3gIs-DPC0dhv9TJkvR_zZgCi5XA=s0-d)
= ∫(1 - sin24x) cos4x dx
= ∫cos4x dx - ∫sin24x cos4x dx
=
Examples 6:
∫sin2x cos3x dx = ∫sin2x cos2x ∙ cosx dx
= ∫sin2x (1 - sin2x) cosx dx
= ∫(sin2x - sin4x) cosx dx
Let u = sinx, du = cosx dx
=![answer for the integral of sin^2xcos^3x dx solution for the integral of sin^2xcos^3x dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vKnDRKNPO6ZwVfzvEAxQ83R1cL115IoYSZgPvuj4FbSYAhOnuZ8P90AZ_vEz_0YC5Cmvr2DW8X-v_oN7ocOdV6Pabzr1lO-EQnseg6MJOyWw_KUl6DrFSf9FSmPxapz0r7eH0103g9e-O0uQelkctTg39zYeiJ5ldqd8C5WM27QdP2g99RsD4DyO50WPFnWtstrRJJ7BF0bswL9hGpQE3ytGBvnp0TrqKe0ToZOlFgMa2KaqVNfK8yQVF-JUh-6pEudwMWGgbI73--wutJydhZ=s0-d)
= ∫sin2x (1 - sin2x) cosx dx
= ∫(sin2x - sin4x) cosx dx
Let u = sinx, du = cosx dx
=
Examples 7:
∫sin2x dx = ½∫(1 - cos2x)dx
= ½∫dx - ½∫cos2x dx
=![answer for the integral of sin^2x dx solution for the integral of sin^2x dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u_XbR25yVslyCv_ZjvU4JJJcmkHKiMlBYLbJnsbpZL0M9o_fY_FiGHh3P_3ftMcFM5_suFNzSPQvU5EUZjRR27vJ-O4y945umpHxZLCwe-rddO1xWGpV-ytNYScsmIz0ypXpHeDKLkKDJdSMvGqm_qnKbMUZb8dlgajCwY6PcB87GgC3zwfGtysRVsKCI2f3CGF1EMM-Wqm4-47A=s0-d)
= ½∫dx - ½∫cos2x dx
=
Examples 8:
∫sin2x cos2x dx = ∫½(1 - cos2x)½(1 + cos2x) dx
= ¼∫(1 - cos22x) dx
= ¼∫[1 - ½(1 + cos4x)] dx
= ⅛∫(1 - cos4x) dx
=![answer for the integral of sin^2xcos^x dx solution for the integral of sin^2xcos^x dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uz5Hkj9uuR-joNAqUJZAe-Rp1J7V9yKCwabbQ5UxJDlwXP-ZXSHCBp-Wac1XxAWpK_f1DB-taYKeBJzhr9MDYZjT6a-l7RPNEWPm3esTg-Gseqm6uEF0nqpI_QkcyzpFjqgB4w54_uv4lLqJMKx_5J5yGK4JD5q7VNZ8P5qcJI-5CX-CF1B4bT2q5EvTxphIJyQAxKzTR17JE3GrwuYQ=s0-d)
= ¼∫(1 - cos22x) dx
= ¼∫[1 - ½(1 + cos4x)] dx
= ⅛∫(1 - cos4x) dx
=
Examples 9:
∫sin6xcos3x dx = ½∫(sin9x + sin3x) dx
= ½∫sin9x dx + ½∫sin3x dx
=![answer for the integral of sin(6x)cos(3x) dx solution for the integral of sin(6x)cos(3x) dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sHakhgjcTMsLlR8WG4c9-TY-QN_pfzYS-U41Cg3HyIw9kUJoCzjpDTo6UEZNpPqtUVNhrAzJc3z0MqCbdkUKMVYYJ5iIviHosCRiBpwhK2cwuJcAzF-TZfROLcRdO2Ia7SGI0zrnlhPvSI-YpmCUbW8GTYoDAZZRrJJrv12zfHrLzMJDGZRqoTmVGqfqn6qtT69S-CaGsduivDRjbLaDVxbnB4CFvnPQ=s0-d)
= ½∫sin9x dx + ½∫sin3x dx
=
Examples 10:
∫sin5xsin2x dx = ½∫(cos3x - cos7x) dx
= ½∫cos3x dx - ½∫cos7x dx
=![](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ullX3appW6y_RtNJ8VeGoSk3SGWQtRGdWQkpeJnf-ZwsjpoEaHaxey1Bk7aKV17sGmSUpBk8OxGgUlJOY6aVLyfKga6xKw8yGYyrIoYkiZ8nr6Cp8uqT-tXzJrIccp2LzxIwfxGv2Xhxr0JeU1mtZ7NZHrbJSWOa-kKdxoGWR0i7JIgbD5lkqkKmNveqkyAOpueI8a3H-SeeIit3syRVtuN2XteZ4c=s0-d)
= ½∫cos3x dx - ½∫cos7x dx
=
Integration of the powers of tangent and secant functions
- Case 1. Has the form ∫tanmu secnu du. In this case, when m is odd factor out sec u tan u du and change the remaining tangent into secant using the identity, tan2u = sec2u - 1.
- Case 2. Has the form ∫tanmu secnu du. When n is even greater than 2, factor out sec2u du and replace the remaining secant by tangent using the identity, sec2u = 1 + tan2u.
- Case 3. Has the form ∫tanmu secnu du. When m is even the integrand is tangent only use the identity, tan2u = sec2u - 1.
Examples 11:
∫tan3x sec3x dx = ∫tan2x sec2x ∙ secx tanx dx
= ∫(sec2x - 1)sec2x ∙ secx tanx dx
= ∫(sec4x - sec2x)secx tanx dx
Let u = secx, du = secx tanx dx
= ∫ u4 du - u2 du
=![answer for the integral of tan^3xsec^3x dx solution for the integral of tan^3xsec^3x dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_tW86oKjT0VjyVOkCoYJRonZD-asa-2UuUcKSclERIocz47dfSaJKHborZpykAVs1IMgKiUFYNLlmQZ5jl_jEJGmM8lIGS1YTCar_08iMpvtHzCHlhnWrNJ6zOKI_l6rlJd-5GwWFkmjg0Q6ChE4R5nuaMW3vgPlBR60FHSlZHy7Zmj_D3EWCeaJqKbVK76UZwEdo2R0SA_l6p73PROfgjdUjtNnFScPGehuBUELWcXAHHZyPi106GKu-SH4IEepxgFpM2nIaYZgQQ7HASBw0yI=s0-d)
= ∫(sec2x - 1)sec2x ∙ secx tanx dx
= ∫(sec4x - sec2x)secx tanx dx
Let u = secx, du = secx tanx dx
= ∫ u4 du - u2 du
=
Examples 12:
∫tan5x sec4x dx = ∫tan5x sec2x ∙ sec2x dx
= ∫tan5x(1 + tan2x)sec2x dx
= ∫tan5x sec2x dx + ∫tan7x sec2x dx
Let u = tanx, du = sec2x dx
= ∫ u5 du + u7 du
=![answer for the integral of tan^5xsec^4x dx solution for the integral of tan^5xsec^4x dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vzIOVBYBUX_46RWWCfyHi-WSkg6cL05s5JvgjDnU9WypRHXh1Dq8cVKWWOvNVkBvFg3SCSAX1amNWJ9HAZEtq-X2uEqwOo9YIvlVWw1YL6CqTKRHVFjN-3Mt-eIb918fKPVhnA5E5K-cnpMnPosXug_fyZiofys9fjco3FPiG9NKLO8vaZdCBqt6xsKEHYX9omRQRix5mBkE7dB-4ZMpuXFpxgNV5X3lMyq20sblEHK8I0_EFcilyoNedKCJalvmpPBvsEy00H4Vt1DSKz7x8qzw=s0-d)
= ∫tan5x(1 + tan2x)sec2x dx
= ∫tan5x sec2x dx + ∫tan7x sec2x dx
Let u = tanx, du = sec2x dx
= ∫ u5 du + u7 du
=
Examples 13:
∫tan4x dx = ∫tan2x ∙(sec2x - 1) dx
= ∫tan2x sec2x dx - ∫tan2x dx
= ∫tan2x sec2x dx - ∫(sec2x - 1) dx
Let u = tanx, du = sec2x dx
= ∫u2 du - ∫du + ∫dx
=![answer for the integral of tan^4x dx solution for the integral of tan^4x dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vYVHSqe9l5LkQ0dIlHMJAWE6SDNqasNmSIX2tIqmDiFSLGyUWfbdaz7iNJxm7XfeLycjJKHdqVdrVIBtPtuig9nu1AN472crx7iFgoScM161H0voYlBtj5owVGjVrUCpAgwGdaUyEK9DS1b5JqLHB1pJvPXOJ6GHgFzFPMsn8d18oFF9jOIz1tScpyTKEgWdP3UIlvprE-cX9nIsIXnSs=s0-d)
= ∫tan2x sec2x dx - ∫tan2x dx
= ∫tan2x sec2x dx - ∫(sec2x - 1) dx
Let u = tanx, du = sec2x dx
= ∫u2 du - ∫du + ∫dx
=
Integration of the powers of cotangent and Cosecant functions
- Case 1. Has the form ∫cotmu cscnu du. When m is odd, factor out csc u cot u du and change the remaining cotangent into cosecant using the identity, cot2u = csc2u - 1.
- Case 2. Has the form ∫cotmu cscnu du. When n is even greater than 2, factor out csc2u du and replace the remaining cosecant by cotangent using the identity, csc2u = 1 + cot2u.
Examples 14:
∫cot32x csc32x dx = ∫cot22x csc22x ∙ (cot2x csc2x dx)
= ∫(csc22x - 1)csc22x ∙ cot2x csc2x dx
= ∫(csc42x - csc22x)cot2x csc2x dx
Let u = csc2x, du = -2csc2x cot2x dx
= -½∫u4 du + ½∫u2 du
=![answer for the integral of cot^3(2x)csc^3(2x) solution for the integral of cot^3(2x)csc^3(2x)](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_syjkHmVHvfj6Fp4TbTxsH-ZS03dZ4gw8DsQkpiuGX-UeVsbpsQkKnTSwdBdOgitnvCto_0447oMT9aWxyjlqDPgBTpBezZ_NQ_NXuWN6SsHjBCO5jzC-kbrnkb2TLFRtOAhrP6GZ6IQdCUxDApJ5TU4FD2AtYAStXP-8ZUW1XzmVXB8X8Nx0y47pfp7DGuqe72nx_7yXdgxKOhZBVxIXqgcVtiYMG4KEFbQ5MkR-4GMScbTGdbbe6q0QGdfDQe5IA9pzF2k5Z9W4XJmSm2dhffA4GTCtM=s0-d)
= ∫(csc22x - 1)csc22x ∙ cot2x csc2x dx
= ∫(csc42x - csc22x)cot2x csc2x dx
Let u = csc2x, du = -2csc2x cot2x dx
= -½∫u4 du + ½∫u2 du
=
Examples 15:
∫cot4x csc4x dx = ∫cot4x csc2x ∙ csc2x dx
= ∫cot4x(1 + cot2x) csc2x dx
= ∫(cot4x + cot6x) csc2x dx
Let u = cotx, du = -csc2x dx
= -∫u4 du - ∫u6 du
=![answer for the integral of cot^4xcsc^x dx solution for the integral of cot^4xcsc^x dx](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_upbPytzJeATsqzdDg21H3POKChzO-7D7Dq_6eN5ZIhWu3co7WDxvYFnk1EczrNWgWNW76vJoO2jvXWQH7uqiU1MFVE7eeklKXygjhTYyI837l1KZKzekjC8UdenpnyAb6Zt8hsJrzcLN_ZOUuNaOtuowJY-YXo20GH_WWGoP5esIvV0Q-Xa1jF7PYLhnE6yTPlt-__uxuRVSHzxycpV5I2S9a6MWodwWn0zW0tzzhuA0NchJlhWyu642vSQ1v8YREFRG8X_6uewrd4_pH4QqaAsjM=s0-d)
= ∫cot4x(1 + cot2x) csc2x dx
= ∫(cot4x + cot6x) csc2x dx
Let u = cotx, du = -csc2x dx
= -∫u4 du - ∫u6 du
=
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