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# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.27

• Last Updated : 30 Apr, 2021

### Question 1. âˆ«eax cosbx dx

Solution:

We have,

I = âˆ«eax cosbx dx

Using integration by parts, we get,

I = eax (sinbx)/b âˆ’ aâˆ«eax (sinbx)/b dx

I = eax (sinbx)/b âˆ’ (a/b)[âˆ’eax (cosbx)/b + aâˆ«eax (cosbx)/b dx]

I = eax (sinbx)/b + (a/b2) eax (cosbx) âˆ’ (a2/b2)âˆ«eax (cosbx) dx

I = (eax/b2) [b sinbx + a cos bx] + (a2/b2) I + c

(a2+b2)I/b2 = (eax/b2) [b sinbx + a cos bx] + c

Therefore, I = eax [b sinbx + a cos bx]/(a2+b2) + c

### Question 2. âˆ«eax sin(bx+c)dx

Solution:

We have,

I = âˆ«eax sin(bx+c)dx

Using integration by parts, we get,

I = âˆ’ eax cos(bx+c)/b + âˆ«aeax cos(bx+c)/b dx

I = (âˆ’1/b) eax cos(bx+c) + (a/b) âˆ«eax cos(bx+c) dx

I = (âˆ’1/b) eax cos(bx+c) + (a/b) [eax sin(bx+c)/b âˆ’ âˆ«aeax sin(bx+c)/b dx]

I = (âˆ’1/b) eax cos(bx+c) + (a/b2) eax sin(bx+c) âˆ’ (a2/b2)âˆ«eax sin(bx+c) dx

I = (eax/b2) [a sin(bx+c) âˆ’ b cos(bx+c)] âˆ’ (a2/b2) I + c

(a2+b2) I/b2 = (eax/b2) [a sin(bx+c) âˆ’ b cos(bx+c)] + c

Therefore, I = (eax)[a sin(bx+c)âˆ’b cos(bx+c)]/(a2+b2) + c

### Question 3. âˆ«cos(logx) dx

Solution:

We have,

I = âˆ«cos(logx) dx

Let log x = t, so we get, (1/x)dx = dt

=> dx = xdt

=> dx = et dt

So, the equation becomes,

I = âˆ«et cost dt

Using integration by parts, we get,

I = et sint âˆ’ âˆ«et sint dt

I = et sint âˆ’ [âˆ’et cost + âˆ«et cost dt]

I = et sint + et cost âˆ’ I + c

2I = et (sint+cost) + c

I = et (sint+cost)/2 + c

Therefore, I = x[cos(logx) + sin(logx)]/2 + c

### Question 4. âˆ«e2x cos(3x+4)dx

Solution:

We have,

I = âˆ«e2x cos(3x+4)dx

Using integration by parts, we get,

I = e2x sin(3x+4)/3 âˆ’ âˆ«2e2x sin(3x+4)/3 dx

I = (1/3) e2x sin(3x+4) âˆ’ (2/3) âˆ«e2x sin(3x+4) dx

I = (1/3) e2x sin(3x+4) âˆ’ (2/3) [âˆ’e2x cos(3x+4)/3 + âˆ«2e2x cos(3x+4)/3 dx]

I = (1/3) e2x sin(3x+4) + (2/9) e2x cos(3x+4) âˆ’ (4/9)âˆ«2e2x cos(3x+4) dx]

I = (e2x/9) [2 cos(3x+4)âˆ’3 sin(3x+4)] âˆ’ (4/9)I + c

(13/9)I = (e2x/9) [2 cos(3x+4)âˆ’3 sin(3x+4)] + c

Therefore, I = e2x[2 cos(3x+4)âˆ’3 sin(3x+4)]/13 + c

### Question 5. âˆ«e2x sinx cosx dx

Solution:

We have,

I = âˆ«e2x sinx cosx dx

I = (1/2)âˆ«e2x (2sinx cosx) dx

I = (1/2)âˆ«e2x sin2x dx

Using integration by parts, we get,

I = (1/2)[âˆ’e2x (cos2x)/2 + âˆ«2e2x (cos2x)/2 dx]

2I = âˆ’e2x (cos2x)/2 + âˆ«e2x (cos2x)dx

2I = âˆ’e2x (cos2x)/2 + [e2x (sin2x)/2 âˆ’ âˆ«2e2x (sin2x)/2 dx]

2I = âˆ’e2x (cos2x)/2 + e2x (sin2x)/2 âˆ’ 2I + c

4I = e2x(sin2xâˆ’cos2x)/2 + c

Therefore, I = e2x(sin2xâˆ’cos2x)/8 + c

### Question 6. âˆ«e2x sinx dx

Solution:

We have,

I = âˆ«e2x sinx dx

Using integration by parts, we get,

I = e2x(âˆ’cosx) + âˆ«2e2x cosx dx

I = âˆ’e2x cosx + 2[e2x sinx âˆ’ 2âˆ«e2x sinx dx]

I = âˆ’e2x cosx + 2e2x sinx âˆ’ 4âˆ«e2x sinx dx

I = e2x(2sinxâˆ’cosx) âˆ’ 4I + c

5I =  e2x(2sinxâˆ’cosx) + c

Therefore, I = e2x(2sinxâˆ’cosx)/5 + c

### Question 7. âˆ«e2x sin(3x+1) dx

Solution:

We have,

I = âˆ«e2x sin(3x+1) dx

Using integration by parts, we get,

I = âˆ’e2x cos(3x+1)/3 + âˆ«2e2x cos(3x+1)/3 dx

I = âˆ’(1/3) e2x cos(3x+1) + (2/3) [e2x sin(3x+1)/3 âˆ’ âˆ«2e2x sin(3x+1)/3 dx]

I = âˆ’(1/3) e2x cos(3x+1) + (2/3) [e2x sin(3x+1)/3 âˆ’ (2/3)âˆ«e2x sin(3x+1)dx]

I = âˆ’(1/3) e2x cos(3x+1) + (2/9) e2x sin(3x+1) âˆ’ (4/9)I + c

(13/9)I = e2x [2sin(3x+1)âˆ’3cos(3x+1)]/9 + c

Therefore, I = e2x [2sin(3x+1)âˆ’3cos(3x+1)]/13 + c

### Question 8. âˆ«ex sin2x dx

Solution:

We have,

I = âˆ«ex sin2x dx

I = (1/2)âˆ«ex (1âˆ’cos2x)dx

I = (1/2)âˆ«ex dx âˆ’ (1/2)âˆ«ex cos2x dx

Let I1 = âˆ«ex cos2x dx. So, our equation becomes,

I = (1/2)âˆ«ex dx âˆ’ (1/2) I1   . . . . (1)

Using integration by parts in I1, we get,

I1 = ex (sin2x)/2 âˆ’ âˆ«ex (sin2x)/2 dx

I1 = ex (sin2x)/2 âˆ’ (1/2)[âˆ’ex (cos2x)/2 + âˆ«ex (cos2x)/2 dx

I1 = ex (sin2x)/2 + ex (cos2x)/4 âˆ’ (1/4)âˆ«ex cos2x dx

I1 = ex[2sin2x+cos2x)]/4 âˆ’ (1/4) I1

(5/4)I1 = ex[2sin2x+cos2x)]/4

I1 = ex[2sin2x+cos2x)]/5  . . . . (2)

Putting (2) in (1), we get,

I = (1/2)âˆ«ex âˆ’ (1/2) (1/5) ex[2sin2x+cos2x)] + c

Therefore, I = ex/2 âˆ’ ex[2sin2x+cos2x)]/10 + c

### Question 9. âˆ«(1/x3) sin(logx) dx

Solution:

We have,

I = âˆ«(1/x3) sin(logx) dx

Let logx = t, so we have, (1/x)dx = dt

=> dx = xdt

=> dx = et dt

So, the equation becomes,

I = âˆ«(1/e3t) (sint) et dt

I = âˆ«eâˆ’2t sint dt

Using integration by parts, we get,

I = âˆ’eâˆ’2t cost âˆ’ âˆ«2eâˆ’2t cost dt

I = âˆ’eâˆ’2t cost âˆ’ 2[eâˆ’2t sint + âˆ«2eâˆ’2t sint]

I = âˆ’eâˆ’2t cost âˆ’ 2eâˆ’2t sint âˆ’ 4âˆ«eâˆ’2t sint

I = âˆ’eâˆ’2t[cost+2sint] âˆ’ 4I + c

5I = âˆ’eâˆ’2t[cost+2sint] + c

I = âˆ’eâˆ’2t[cost+2sint]/5 + c

I = âˆ’eâˆ’2logx[cos(logx)+2sin(logx)]/5 + c

I = âˆ’x-2[cost+2sint]/5 + c

Therefore, I = âˆ’[cost+2sint]/5x2 + c

### Question 10. âˆ«e2x cos2x dx

Solution:

We have,

I = âˆ«e2x cos2x dx

I = (1/2) âˆ«e2x (1+cos2x) dx

I = (1/2) âˆ«e2x dx + (1/2) âˆ«e2x cos2x dx

Let I1 = âˆ«e2x cos2x dx. So, our equation becomes,

I = (1/2)âˆ«e2x dx + (1/2) I1   . . . . (1)

Using integration by parts in I1, we get,

I1 = e2x (sin2x)/2 âˆ’ âˆ«2e2x (sin2x)/2 dx

I1 = e2x (sin2x)/2 âˆ’ âˆ«e2x sin2x dx

I1 = e2x (sin2x)/2 âˆ’ [âˆ’e2x (cos2x)/2 + âˆ«2e2x (cos2x)/2 dx]

I1 = e2x (sin2x)/2 + e2x (cos2x)/2 âˆ’ âˆ«e2x (cos2x) dx

I1 = e2x(sin2x+cos2x)/2 âˆ’ I1

2I1 = e2x(sin2x+cos2x)/2

I1 = e2x(sin2x+cos2x)/4   . . . . (2)

Putting (2) in (1), we get,

I = (1/2)âˆ«e2x dx + (1/2) (1/4) [e2x(sin2x+cos2x)]

Therefore, I = (1/4) e2x + (1/8) [e2x(sin2x+cos2x)] + c

### Question 11. âˆ«eâˆ’2x sinx dx

Solution:

We have,

I = âˆ«eâˆ’2x sinx dx

Integrating by parts, we get,

I = âˆ’eâˆ’2x cosx âˆ’ âˆ«2eâˆ’2x cosx dx

I = âˆ’eâˆ’2x cosx âˆ’ 2[eâˆ’2x sinx+âˆ«2eâˆ’2x sinx dx]

I = âˆ’eâˆ’2x cosx âˆ’ 2eâˆ’2x sinx âˆ’ 4âˆ«eâˆ’2x sinx dx

I = âˆ’eâˆ’2x(cosx+2sinx) âˆ’ 4I + c

5I = âˆ’eâˆ’2x(cosx+2sinx) + c

Therefore, I = âˆ’eâˆ’2x(cosx+2sinx)/5 + c

### Question 12.

Solution:

We have,

I =

Let x3 = t, so we have, 3x2dx = dt

So, the equation becomes,

I = (1/3) âˆ«et cost dt

Integrating by parts, we get,

I = (1/3) [et sint âˆ’ âˆ«et sint dt]

I = (1/3) [et sint âˆ’ (âˆ’et cost + âˆ«et cost dt)]

I = (1/3) et sint + (1/3) et cost âˆ’ (1/3) âˆ«et cost dt

I = et[sint+cost]/3 âˆ’ I + c

2I = et[sint+cost]/3 + c

I = et[sint+cost]/6 + c

Therefore, I =

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