23for first put x = atan2θ
for second put 1+lnx = t2
for third
∫dx√(1-sin4x)
=∫dxcosx√(1+sin2x)
= ∫cosxdxcos2x√(1+sin2x)
= ∫cosxdx(1-sin2x)√(1+sin2x)
put sinx = t
integrate the following:
1. sin inverse ( root of x / a+x)
2. root of ( 1+ lnx ) / x lnx
3. 1 / root ( 1 - sin ^ 4 x )
dont give the soln. just give the necess. hints....
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13 Answers
23
1Ques :2
root of ( 1+ lnx ) / x lnx
Let 1 + ln(x) = t^2 => ln(x) = t^2-1
2t.dt/dx = 1/x
or dx = 2.t.x.dt
Now ∫ t .dx / x.ln(x) = ∫t.(2.t.x.dt) / x.(t^2-1)
∫2t^2.dt/(t^2-1) = ? ---(1)
Now ... 2t^2 /(t^2 -1 ) = 2t^2 /(t-1)(t+1) = t / (t-1) + t/(t+1)
t/(t-1) + t/(t+1) = 1 + 1/(t-1) + 1 - 1/(t+1) = 2 + 1/(t-1) - 1/(t+1)
So from (1)
∫ [2 + 1/(t-1) - 1/(t+1)].dt = ∫2.dt + ∫dt/(t-1) -∫dt/(t+1)
Ans = 2t + ln(t-1) -ln(t+1) + C
11. sin inverse ( root of x / a+x)
Let x = a tan2(x)
root x / (x + x) = root atan2(x) / a ( 1+ tan2(x))
root x/a+x = root ( tan2(x) / sec2(x) )
root( x/a+x) = root(sin2(x)) = sin(x)
Hence sin inverse ( sin(x) ) = x
hence integration is simply = x^2/2 + C
6ok a few more
1. sinx / sin4x
2. e^x ( 2-x^2) / (1-x) root of (1-x^2)
3. ln (e/x) /(ln x)^2
1@Akhil ...either i attempt the question till the final line...or i don't....i never write two or three line named as "HINT"
1Ques :1 ∫sin(x)/sin(4x)
sin(x)/sin(4x) = sin(x) / 2.sin(2x).cos(2x) = sin(x)/4.sin(x).cos(x).cos(2x)
sin(x)/sin(4x) = 1/4.cos(x).cos(2x)
4.sin(x)/sin(4x) = 1/cos(x).(2cos2x -1) = 2cos(x)/(2cos2x -1) -1/cos(x)
4.sin(x)/sin(4x) = 2cos(x)/(2cos2x -1) - sec(x)
∫(4.sin(x)/sin(4x)).dx = ∫2cos(x).dx/(2cos2x -1) - ∫sec(x).dx ------------(1)
............................= ∫2cos(x).dx/(1-2sin2x) -- ∫sec(x).dx
............................=Let sin(x) = t
............................= ∫dt / (1-2t^2) - ∫sec(x).dx
...........................= M - N
M = ∫dt / (1-2t^2) = ∫dt/(1-√2t)(1+√2t) = 1/2 [ ∫dt/(1-√2t) + ∫dt/(1+√2t) ]
M ............................................................= 1/2√2 [ -ln ( 1-√2t) + ln(1+√2t) ]
N = ∫sec(x).dx = ln(secx + tanx) + C
M = 1/2√2 [ ln(1+√2sin(x) - ln(1-2√sinx) ]
So from equation (1)
4.∫sin(x)/sin(4x) = M - N = 1/2√2 [ ln(1+√2sin(x) - ln(1-2√sinx) ] - ln(secx + tanx) + C
∫sin(x)/sin(4x) = 1/8√2 [ ln(1+√2sin(x) - ln(1-2√sinx)] - 1/8 ln(secx + tanx) + C
23∫ex ( 2-x2) (1-x) √(1-x2) dx
hint : ∫ex(f(x)+f'(x)) = exf(x) + c
try to convert it into f(x) and f'(x)
( 2-x2) (1-x) √(1-x2)= ( 1+1-x2) (1-x) √(1-x2)
= ( 1) (1-x) √(1-x2)+ ( 1-x2) (1-x) √(1-x2)
= ( 1) (1-x) √(1-x2)+ (√1+x)√(1-x)
d (√1+x)dx√(1-x)
= (√1-x)22√(1+x)(1-x)2
= 1√(1-x2)(1-x)
13rd is similar to 2nd.prob. Put
ln(ex)-t
Den u wil have integral of frm
∫epowerx(f'(x) +f(x) - epower x .f(x)
So ans is
(- e power lnx)/ lnx
Using mobil to anser. Forgiv mistake