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# P3 Integration help plz... watch

1. Hi Im stuck on a few questions now as i approach the end of the chapter can someone guide me through these:

1 . Find integral in terms of x:

∫ x/root(x+1) dx u^2 = x + 1

2. Use the sub u = x - 1 ∫ x/root(x-1) dx

3. Use the sub u= sinx to evaluate ∫ cos2xcosx dx (has limits pi/2 n pi/6)
2. (Original post by phiyahless)
Hi Im stuck on a few questions now as i approach the end of the chapter can someone guide me through these:

1 . Find integral in terms of x:

2. Use the sub u = x - 1 ∫ x/root(x-1) dx

3. Use the sub u= sinx to evaluate ∫ cos2xcosx dx (has limits pi/2 n pi/6)

1 . Find integral in terms of x:

∫ x/root(x+1) dx u^2 = x + 1

rearrange your substitution to give x=u^2 - 1 (put in top)
square root your sub. and let the denominator be u
rearrange your substition in terms of u (x=u^1-1) and differentiage with respect to u>>>dx = 2u du
sub all back into integral to get integral u^2 -1 / u * 2u du

integrating gives 2u^3 / 3 -2u
resubbing u gives...2(x+1)^3 - 2(x+1) + C

2. Use the sub u = x - 1 ∫ x/root(x-1) dx
same method as for (1)

numerator = u+1
denominator = u^0.5 and dx = 1.du
resubbing gives integ u/u^0.5 du = integ u^0.5 du = 2u^1.5 /3
resub u gives integ = 2/3 * (x-1) ^1.5

3. Use the sub u= sinx to evaluate ∫ cos2xcosx dx (has limits pi/2 n pi/6)

du / dx = cos x > dx = du / cos x.

resub into original integral and the cos x on the bottom cancels with the one ni the integral.

now use identity for cos 2x to get 1-2sin^2 x. you know sin x = u> sub this in:

integral = u-2/3 u ^ 3 between limits

BUT...you got to change the limits to u, or change the equation into x from and put the angle limits in...for convenience i will convery it to U limits.

at pi / 3 u = sin pi / 3 = (rt 3) / 2 and at pi/6 u = sin pi / 6 = 0.5

put these limits into the integral worked out above...to get:

-((5+2 (rt2) )/12)

answers may be wrong... i'm really tired...awwww.

goodnight people!
3. (Original post by phiyahless)
Find integral in terms of x:

1. ∫ x/root(x+1) dx u^2 = x + 1
2. Use the sub u = x - 1 ∫ x/root(x-1) dx
3. Use the sub u = sinx to evaluate ∫ cos2xcosx dx (has limits pi/2 n pi/6)
1.) u^2 = x + 1 --> x = u^2 - 1 --> dx/du = 2u --> dx = 2u du
∫ x/root(x+1) dx
= ∫ (u^2 - 1)/u . 2u du
= ∫ 2(u^2 - 1) du
= ∫ 2u^2 - 2 du
= (2u^3)/3 - 2u + k
= [2(x + 1)^(3/2)]/3 - 2Sqrt(x + 1) + k

2.) u = x - 1 --> du/dx = 1 --> dx = du
∫ x/root(x-1) dx
= ∫ (u + 1)/(Sqrtu) du
= ∫ u^(1/2) + u^(-1/2) du
= [2u^(3/2)]/3 + 2u^(1/2) + k
= [2(x - 1)^(3/2)]/3 + 2(x - 1)^(1/2) + k

3.) u = sinx --> du/dx = cosx --> dx = 1/cosx du
When x = Pi/2: u = 1
When x = Pi/6: u = 0.5

∫ cos2xcosx dx
= ∫ (Limits 1 and 0.5) cos2xcosx/cosx du
= ∫ (Limits 1 and 0.5) cos2x du
= ∫ (Limits 1 and 0.5) 1 - 2sin^2x du
= ∫ (Limits 1 and 0.5) 1 - 2u^2 du
= [u - (2u^3)/3] (Limits 1 and 0.5)
= (1 - 2/3) - (0.5 - 1/12)
= 12/12 - 8/12 - 6/12 + 1/12
= -1/12
4. thanks a lot people!!
5. firstly, this is blatantly an attempt to make people do ure homework for u :O those questions are, like, the easiest in the integration by substitution chapter...

anyway, here is my solution. i dont think i fully agree with either of the two above... so dnno...

1. u^2 = x + 1, x = u^2 + 1
dx/du = 2u
∫ (u² - 1)/u * 2u du
∫2u² - 2 du
2/3u³ - 2u
2/3(x+1)^3/2 - 2(x+1)^1/2

2. u = x-1, x = u + 1, du/dx = 1, du = dx
∫u^1/2 + u^-1/2 du
2/3u^3/2 + 2u^1/2
2/3(x-1)^3/2 + 2(x-1)^1/2

3. du/dx = cox x, cos 2x = 1 - 2sin²x = 1 - 2u²
∫1-2u²du
[u-2/3u³]
[sin x - 2/3sin ³ x] limits
(sin pi/2 - 2/3(sin pi/2)³) - (sin pi/6 - 2/3sin³pi/6)
(i dont have a calculator on me)

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