some Hard ∫ , really need the help.?

Question

1. ∫cos(lnx)dx \ 2.∫x^7*e^(x^2)dx \ 3.∫(sec^3x+e^sinx)/secx dx \ 4. ∫(x+1)/(9x^2+18x+10)^3/2 dx \ 5. ∫sin^2xcos^2x dx \ 6.∫cot^4*3x dx \ 7.∫tan^-3x sec^4x dx \ 8.∫x*(x-4)^1/3 dx \ 9.∫dx/(x^2+2x+26)^1/2 \ 11.∫(3x^2-8x+13)/(x+3)(x-1)^2 dx \ 12.∫(6x^2-3x+1)/(4x+1)(x^2+1) dx \ 13.∫(6x^2-15x+22)/(x+3)(x^2+2)^2 dx \ 14.∫dx/((x)^1/2+(x)^1/3) dx \ 15. ∫(1/(1+x)^2)*((1-x)/(1+x)^1/3)

Answer 1

in every question, let A= integral (question)

1. let lnx=t ; or,x = e^t ; so dx=(e^t)dt;
so, ietegral becomes :- ∫ cos(t).(e^t)dt = A (say)
now using integration by parts,
A = cos(t).∫ (e^t)dt - ∫[{d(cos(t))/dt}.∫(e^t)dt]dt
= (e^t).cos(t) + (e^t).sin(t) - A
or, 2A = (e^t).cos(t) + (e^t).sin(t)
or, A = [(e^t).cos(t) + (e^t).sin(t)]/2 +c

2. write x^7 as x(x^6) ; or, ((x^2)^3)x
now, put (x^2) = t ; on diff. it we get (2x)dx = dt; or (x.dx) = (dt)/2
now, A = (0.5)∫t^3*e^t(dt) ; which can be solved using integration by parts three times which you can do

3. or, A = ∫(sec^2 x)dx + ∫ (e^sinx)cosx dx
= tanx + B ; where B = ∫ (e^sinx)cosx dx
for B, let sin x = t , (cos x)dx = dt
or, B = ∫ (e^t)dt = e^t + c = e^sin x + c
so, A = tan x + e^sin x + k ; where k = constant

4. here , ( 9x^2 + 18x + 10) can be written as, [ 1 +9(x+1)^2]
now, let [ 1 +9(x+1)^2] = t ;
so, 18(x+1) dx = dt ( on differentiating)
or, (x+1)dx = (dt)/18
so, A = ∫ t^(-3/2) (dt)/18
= [{t^(-1/2)}/(-1/2)]/18 + c
= -[t^(-1/2)]/36 + c , where t = ( 9x^2 + 18x + 10)

5. sin^2xcos^2x = [(2sinxcosx)/2]^2 = [sin^2 (2x)]/4
= [1- cos 4x]/8
so, A = ∫ [1- cos 4x]/8 dx = [ x - (sin 4x)/4]/8 + c

6. cot^4 t = (cot^2 t)(cosec^2 t - 1) = (cot^2 t)(cosec^2 t) -cot^2 t
= (cot^2 t)(cosec^2 t) - (cosec^2 t - 1)
now, A = ∫ [(cot^2 t)(cosec^2 t) - (cosec^2 t - 1)]dt/3 ,
by assuming 3x = t, so dx = dt/3
let B = ∫ (cot^2 t)(cosec^2 t)dt , & D = ∫ (cosec^2 t - 1)]dt
so, A = (B- D)/3
for B, let cot t = u , so (cosec^2 t) dt = - du
B = - ∫ (u^2)du = -(u^3)/3 + c = -(cot^3 t)/3 + c
D = ∫ (cosec^2 t - 1)]dt = -(cot t)- t + m
so, A = -(cot^3 t)/3 +(cot t)+ t + k , where t = 3x

7.tan^-3x sec^4x=[(cos^3 x)/(sin^3 x)]sec^4 x = 1/[(cos x)(sin^3 x)
= 1/[(cot x)(sin^4 x)] = (tan x)(cosec^2 x)(cosec^2 x)
= (1/cot x)(1 + cot^2 x)(cosec^2 x)
now, A = ∫ (1/cot x)(1 + cot^2 x)(cosec^2 x)dx
let cot x = t , so (cosec^2 x)dx = - dt
so, A = -∫(1/t)(1+t^2)dt = -∫(t+1/t)dt = -t^2 - ln t + k; where t=cotx

8. let x=t+4, so, dx = dt
A = ∫ (t+4)t^1/3 dt = ∫(t^4/3 +4t^1/3)dt =(t^7/3)/(7/3) + 4(t^4/3)/(4/3)
= (3/7)(t^7/3) + 3(t^4/3) + k

9. (x^2+2x+26) = (x+1)^2 + 5^2
A = ∫ dx/[(x+1)^2 + 5^2 ] = arc(sinh) [(x+1)/5]
where , sinh is hyperbolic sine function
& arc(sinh) means inverse hyperbolic sine function.
10. question is not given
11. 3x^2-8x+13 = 2(x-1)^2+(x+3)(x-1)-6(x+3)+32
so, A = ∫ [2/(x+3) +1/(x-1) - 6/(x-1)^2]dx + 32(B)
= 2ln(x+3) + ln(x-1) + 6/(x-1) +32(B)
where B = ∫ dx/(x+3)(x-1)^2
now, 1 = [(x+3) -(x-1)]/4
so, B = (1/4)*∫ [1/(x-1)^2 - 1/(x+3)(x-1)]dx
or, 4B = ∫ [1/(x-1)^2 - (1/4)[(x+3)-(x-1)]/(x+3)(x-1)]dx
or, 4B = ∫1/(x-1)^2 dx - (1/4)[ ∫ 1/(x-1) dx - ∫ 1/(x+3) dx]
32B= 8[-1/(x-1) - (1/4){ ln(x-1) - ln(x+3)} ]
= -8/(x-1) - 2ln(x-1) + 2ln(x+3)
so, A = 2ln(x+3) + ln(x-1) + 6/(x-1) -8/(x-1) - 2ln(x-1) + 2ln(x+3)
A = 4ln(x+3) - ln(x-1) - 2/(x-1) + k

12. similar to 11.
13. similar to 11.

14. put x = t^6, so dx = 6t^5 dt; and solving we will get
A = ∫ [6(t^3)/(1+t)] dt
now again put t= u-1, dt = du
A = 6*∫ [{(u-1)^3}/u] du
= 6*∫ (u^2 - 3u + 3 -1/u) du
= 6[(1/3)u^3 - (3/2)u^2 + 3u - ln(u)]
= 6[(1/3)(1+t)^3 - (3/2)(1+t)^2 + 3(1+t) - ln(1+t)] + k
where t= x^(1/6)

15. A = ∫[(1-x)/(1+x)^7/3] dx
now let (1+x) = t; or, x=(t-1)
so, dx = dt
A = ∫ [(2-t)/t^7/3] dt = ∫ ( 2/t^7/3 - 1/t^4/3 )dt
= 2{t^(-4/3)}/(-4/3) - {t^(-1/3)}/(-1/3) + k
= (-3/2)t^(-4/3) + 3t^(-1/3) + k ; where t = (1+x)

Answer 2

once you celebration until eventually ultimately your hair smells like smoke, you have no theory the place the clothing you're donning got here from, and additionally you do no longer recognize how in the international it have been given to be 3 days later, that's a celebration. I lived in an apt whilst i replace into youthful with my suited pal who replace right into a drug broking at that element. I undergo in ideas dozens and dozens of untamed roaring inebriated and stoned events the place i did no longer have a clue who 80% of the individuals at it have been, for particular the furnishings did no longer thrive in that environment. i do no longer recognize despite if the final straw for me replace into the fool that theory pouring a bottle of vodka into the aquarium replace into an exceedingly good theory reason the "fish could desire to get lit too", or the fool I caught feeling my subconscious fiancee up, a scene particularly equivalent to Jesus clearing the Pharisees out of the temple ensued after that one, i recognize that. besides Zacho, you have exciting, yet for Me, i'm kinda happy it rather is throughout. i might say celebration with the help of Boston. BA: Geeze, it would be an horrific long checklist. **PJ, Amen on your amen, brother

Answer 3

can I give you a hint?

Never post more than 1 question per question

People do not like to get only 2 points for 1 hour of work

be smart!

Answer 4

Dude, you just copied your whole hmwk. Try this online integrator. http://integrals.wolfram.com/index.jsp

Answer 5

You really do have a nerve!

Answer 6

We are not here to do your homework