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Question Number 14014 by tawa tawa last updated on 26/May/17
∫cos^n (x) dx please i need workings.
∫cosn(x)dxpleaseineedworkings.
Answered by mrW1 last updated on 26/May/17
I_n =∫cos^n (x) dx=∫cos^(n−1) (x) dsin (x) using ∫udv=uv−∫vdu =sin (x) cos^(n−1) (x)+(n−1)∫sin^2 (x) cos^(n−2) (x)dx =sin (x) cos^(n−1) (x)+(n−1)∫[1−cos^2 (x)] cos^(n−2) (x)dx =sin (x) cos^(n−1) (x)+(n−1)∫cos^(n−2) (x) dx−(n−1)∫cos^n (x) dx ⇒I_n =sin (x) cos^(n−1) (x)+(n−1)∫cos^(n−2) (x) dx−(n−1)I_n ⇒nI_n =sin (x) cos^(n−1) (x)+(n−1)I_(n−2) ⇒I_n =(1/n)sin (x) cos^(n−1) (x)+((n−1)/n)I_(n−2) ⇒I_(n−2) =(1/(n−2))sin (x) cos^(n−3) (x)+((n−3)/(n−2))I_(n−4) ...... if n=even .... ⇒I_2 =(1/2)sin (x) cos (x)+(1/2)I_0 ⇒I_0 =∫dx=x+C ...... if n=odd .... ⇒I_3 =(1/3)sin (x) cos^2 (x)+(2/3)I_1 ⇒I_1 =∫cos (x) dx=sin (x)+C
In=∫cosn(x)dx=∫cosn−1(x)dsin(x)using∫udv=uv−∫vdu=sin(x)cosn−1(x)+(n−1)∫sin2(x)cosn−2(x)dx=sin(x)cosn−1(x)+(n−1)∫[1−cos2(x)]cosn−2(x)dx=sin(x)cosn−1(x)+(n−1)∫cosn−2(x)dx−(n−1)∫cosn(x)dx⇒In=sin(x)cosn−1(x)+(n−1)∫cosn−2(x)dx−(n−1)In⇒nIn=sin(x)cosn−1(x)+(n−1)In−2⇒In=1nsin(x)cosn−1(x)+n−1nIn−2⇒In−2=1n−2sin(x)cosn−3(x)+n−3n−2In−4……ifn=even….⇒I2=12sin(x)cos(x)+12I0⇒I0=∫dx=x+C……ifn=odd….⇒I3=13sin(x)cos2(x)+23I1⇒I1=∫cos(x)dx=sin(x)+C
Commented by tawa tawa last updated on 26/May/17
God bless you sir.
Godblessyousir.
Commented by mrW1 last updated on 27/May/17
If n is even we get: I_n =∫cos^n (x) dx =(1/n)sin (x) cos^(n−1) (x) +((n−1)/n)∙(1/(n−2))sin (x) cos^(n−3) (x) +((n−1)/n)∙((n−3)/(n−2))∙(1/(n−4))sin (x) cos^(n−5) (x) +...... +((n−1)/n)∙((n−3)/(n−2))∙((n−5)/(n−4))∙∙∙(5/6)∙(1/4)sin (x) cos^3 (x) +((n−1)/n)∙((n−3)/(n−2))∙((n−5)/(n−4))∙∙∙(5/6)∙(3/4)∙(1/2)sin (x) cos (x) +((n−1)/n)∙((n−3)/(n−2))∙((n−5)/(n−4))∙∙∙(5/6)∙(3/4)∙(1/2)x+C =𝚺_(k=1) ^(n/2) (((n−1)(n−3)∙∙∙(n−2k+3)^(=1 if k=1) )/(n(n−2)∙∙∙(n−2k+2)))∙ sin (x) cos^(n−2k+1) (x) +(((n−1)!!)/(n!!))x+C if n is odd we get: I_n =∫cos^n (x) dx =(1/n) sin (x) cos^(n−1) (x) +((n−1)/n)∙(1/(n−2)) sin (x) cos^(n−3) (x) +((n−1)/n)∙((n−3)/(n−2))∙(1/(n−4)) sin (x) cos^(n−5) (x) +...... +((n−1)/n)∙((n−3)/(n−2))∙((n−5)/(n−4))∙∙∙(4/5)∙(1/3) sin (x) cos^2 (x) +((n−1)/n)∙((n−3)/(n−2))∙((n−5)/(n−4))∙∙∙(4/5)∙(2/3)∙(1/1) sin (x) cos^0 (x)+C =𝚺_(k=1) ^((n+1)/2) (((n−1)(n−3)∙∙∙(n−2k+1)^(=1 if k=1) )/(n(n−2)∙∙∙(n−2k+2)))∙ sin (x) cos^(n−2k+1) (x)+C
Ifnisevenweget:In=∫cosn(x)dx=1nsin(x)cosn−1(x)+n−1n⋅1n−2sin(x)cosn−3(x)+n−1n⋅n−3n−2⋅1n−4sin(x)cosn−5(x)+……+n−1n⋅n−3n−2⋅n−5n−4⋅⋅⋅56⋅14sin(x)cos3(x)+n−1n⋅n−3n−2⋅n−5n−4⋅⋅⋅56⋅34⋅12sin(x)cos(x)+n−1n⋅n−3n−2⋅n−5n−4⋅⋅⋅56⋅34⋅12x+C=∑n2k=1(n−1)(n−3)⋅⋅⋅(n−2k+3)=1ifk=1n(n−2)⋅⋅⋅(n−2k+2)⋅sin(x)cosn−2k+1(x)+(n−1)!!n!!x+Cifnisoddweget:In=∫cosn(x)dx=1nsin(x)cosn−1(x)+n−1n⋅1n−2sin(x)cosn−3(x)+n−1n⋅n−3n−2⋅1n−4sin(x)cosn−5(x)+……+n−1n⋅n−3n−2⋅n−5n−4⋅⋅⋅45⋅13sin(x)cos2(x)+n−1n⋅n−3n−2⋅n−5n−4⋅⋅⋅45⋅23⋅11sin(x)cos0(x)+C=∑n+12k=1(n−1)(n−3)⋅⋅⋅(n−2k+1)=1ifk=1n(n−2)⋅⋅⋅(n−2k+2)⋅sin(x)cosn−2k+1(x)+C
Commented by tawa tawa last updated on 27/May/17
God bless you sir.
Godblessyousir.
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/May/17
I_n =∫cos^2 x.cos^(n−2) xdx= =∫cos^(n−2) xdx−∫sin^2 x.cos^(n−2) xdx= =I_(n−2) −[((−1)/(n−1))sinx.cos^(n−1) x+(1/(n−1))∫cosx.cos^(n−1) xdx= =I_(n−2) +(1/(n−1))sinx.cos^(n−1) x−(1/(n−1))I_n +C. ⇒I_n +(1/(n−1))I_n =I_(n−2) +(1/(n−1))sinx.cos^(n−1) x+C ⇒ (n/(n−1))I_n =I_(n−2) +(1/(n−1))sinx.cos^(n−1) x+C ⇒I_n =((n−1)/n).I_(n−2) +(1/n)sinx.cos^(n−1) x+C. 0)I_0 =∫dx=x+C. 1)I_1 =∫cosxdx=sinx+C 2)I_2 =∫cos^2 xdx=(1/2)x+(1/4)sin2x+C 3)n>2⇒formula given above. example:I_3 I_3 =∫cos^3 xdx=∫cosx(1−sin^2 x)dx= =sinx−(1/3)sin^3 x+C I_3 =((3−1)/3)I_1 +(1/3)sinx.cos^2 x=(2/3)I_1 +(1/3)sinx.cos^2 x= =(2/3)sinx+(1/3)sinx(1−sin^2 x)=sinx−(1/3)sin^3 x+C
In=∫cos2x.cosn−2xdx==∫cosn−2xdx−∫sin2x.cosn−2xdx==In−2−[−1n−1sinx.cosn−1x+1n−1∫cosx.cosn−1xdx==In−2+1n−1sinx.cosn−1x−1n−1In+C.⇒In+1n−1In=In−2+1n−1sinx.cosn−1x+C⇒nn−1In=In−2+1n−1sinx.cosn−1x+C⇒In=n−1n.In−2+1nsinx.cosn−1x+C.0)I0=∫dx=x+C.1)I1=∫cosxdx=sinx+C2)I2=∫cos2xdx=12x+14sin2x+C3)n>2⇒formulagivenabove.example:I3I3=∫cos3xdx=∫cosx(1−sin2x)dx==sinx−13sin3x+CI3=3−13I1+13sinx.cos2x=23I1+13sinx.cos2x==23sinx+13sinx(1−sin2x)=sinx−13sin3x+C
Commented by tawa tawa last updated on 27/May/17
God bless you sir.
Godblessyousir.

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