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

• Last Updated : 30 Apr, 2021

### Question 1. ∫eax cosbx dx

Solution:

We have,

I = ∫eax cosbx dx

Using integration by parts, we get,

I = eax (sinbx)/b − a∫eax (sinbx)/b dx

I = eax (sinbx)/b − (a/b)[−eax (cosbx)/b + a∫eax (cosbx)/b dx]

I = eax (sinbx)/b + (a/b2) eax (cosbx) − (a2/b2)∫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. ∫eax sin(bx+c)dx

Solution:

We have,

I = ∫eax sin(bx+c)dx

Using integration by parts, we get,

I = − eax cos(bx+c)/b + ∫aeax cos(bx+c)/b dx

I = (−1/b) eax cos(bx+c) + (a/b) ∫eax cos(bx+c) dx

I = (−1/b) eax cos(bx+c) + (a/b) [eax sin(bx+c)/b − ∫aeax sin(bx+c)/b dx]

I = (−1/b) eax cos(bx+c) + (a/b2) eax sin(bx+c) − (a2/b2)∫eax sin(bx+c) dx

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

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

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

### Question 3. ∫cos(logx) dx

Solution:

We have,

I = ∫cos(logx) dx

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

=> dx = xdt

=> dx = et dt

So, the equation becomes,

I = ∫et cost dt

Using integration by parts, we get,

I = et sint − ∫et sint dt

I = et sint − [−et cost + ∫et cost dt]

I = et sint + et cost − I + c

2I = et (sint+cost) + c

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

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

### Question 4. ∫e2x cos(3x+4)dx

Solution:

We have,

I = ∫e2x cos(3x+4)dx

Using integration by parts, we get,

I = e2x sin(3x+4)/3 − ∫2e2x sin(3x+4)/3 dx

I = (1/3) e2x sin(3x+4) − (2/3) ∫e2x sin(3x+4) dx

I = (1/3) e2x sin(3x+4) − (2/3) [−e2x cos(3x+4)/3 + ∫2e2x cos(3x+4)/3 dx]

I = (1/3) e2x sin(3x+4) + (2/9) e2x cos(3x+4) − (4/9)∫2e2x cos(3x+4) dx]

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

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

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

### Question 5. ∫e2x sinx cosx dx

Solution:

We have,

I = ∫e2x sinx cosx dx

I = (1/2)∫e2x (2sinx cosx) dx

I = (1/2)∫e2x sin2x dx

Using integration by parts, we get,

I = (1/2)[−e2x (cos2x)/2 + ∫2e2x (cos2x)/2 dx]

2I = −e2x (cos2x)/2 + ∫e2x (cos2x)dx

2I = −e2x (cos2x)/2 + [e2x (sin2x)/2 − ∫2e2x (sin2x)/2 dx]

2I = −e2x (cos2x)/2 + e2x (sin2x)/2 − 2I + c

4I = e2x(sin2x−cos2x)/2 + c

Therefore, I = e2x(sin2x−cos2x)/8 + c

### Question 6. ∫e2x sinx dx

Solution:

We have,

I = ∫e2x sinx dx

Using integration by parts, we get,

I = e2x(−cosx) + ∫2e2x cosx dx

I = −e2x cosx + 2[e2x sinx − 2∫e2x sinx dx]

I = −e2x cosx + 2e2x sinx − 4∫e2x sinx dx

I = e2x(2sinx−cosx) − 4I + c

5I =  e2x(2sinx−cosx) + c

Therefore, I = e2x(2sinx−cosx)/5 + c

### Question 7. ∫e2x sin(3x+1) dx

Solution:

We have,

I = ∫e2x sin(3x+1) dx

Using integration by parts, we get,

I = −e2x cos(3x+1)/3 + ∫2e2x cos(3x+1)/3 dx

I = −(1/3) e2x cos(3x+1) + (2/3) [e2x sin(3x+1)/3 − ∫2e2x sin(3x+1)/3 dx]

I = −(1/3) e2x cos(3x+1) + (2/3) [e2x sin(3x+1)/3 − (2/3)∫e2x sin(3x+1)dx]

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

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

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

### Question 8. ∫ex sin2x dx

Solution:

We have,

I = ∫ex sin2x dx

I = (1/2)∫ex (1−cos2x)dx

I = (1/2)∫ex dx − (1/2)∫ex cos2x dx

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

I = (1/2)∫ex dx − (1/2) I1   . . . . (1)

Using integration by parts in I1, we get,

I1 = ex (sin2x)/2 − ∫ex (sin2x)/2 dx

I1 = ex (sin2x)/2 − (1/2)[−ex (cos2x)/2 + ∫ex (cos2x)/2 dx

I1 = ex (sin2x)/2 + ex (cos2x)/4 − (1/4)∫ex cos2x dx

I1 = ex[2sin2x+cos2x)]/4 − (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)∫ex − (1/2) (1/5) ex[2sin2x+cos2x)] + c

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

### Question 9. ∫(1/x3) sin(logx) dx

Solution:

We have,

I = ∫(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 = ∫(1/e3t) (sint) et dt

I = ∫e−2t sint dt

Using integration by parts, we get,

I = −e−2t cost − ∫2e−2t cost dt

I = −e−2t cost − 2[e−2t sint + ∫2e−2t sint]

I = −e−2t cost − 2e−2t sint − 4∫e−2t sint

I = −e−2t[cost+2sint] − 4I + c

5I = −e−2t[cost+2sint] + c

I = −e−2t[cost+2sint]/5 + c

I = −e−2logx[cos(logx)+2sin(logx)]/5 + c

I = −x-2[cost+2sint]/5 + c

Therefore, I = −[cost+2sint]/5x2 + c

### Question 10. ∫e2x cos2x dx

Solution:

We have,

I = ∫e2x cos2x dx

I = (1/2) ∫e2x (1+cos2x) dx

I = (1/2) ∫e2x dx + (1/2) ∫e2x cos2x dx

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

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

Using integration by parts in I1, we get,

I1 = e2x (sin2x)/2 − ∫2e2x (sin2x)/2 dx

I1 = e2x (sin2x)/2 − ∫e2x sin2x dx

I1 = e2x (sin2x)/2 − [−e2x (cos2x)/2 + ∫2e2x (cos2x)/2 dx]

I1 = e2x (sin2x)/2 + e2x (cos2x)/2 − ∫e2x (cos2x) dx

I1 = e2x(sin2x+cos2x)/2 − I1

2I1 = e2x(sin2x+cos2x)/2

I1 = e2x(sin2x+cos2x)/4   . . . . (2)

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

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

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

### Question 11. ∫e−2x sinx dx

Solution:

We have,

I = ∫e−2x sinx dx

Integrating by parts, we get,

I = −e−2x cosx − ∫2e−2x cosx dx

I = −e−2x cosx − 2[e−2x sinx+∫2e−2x sinx dx]

I = −e−2x cosx − 2e−2x sinx − 4∫e−2x sinx dx

I = −e−2x(cosx+2sinx) − 4I + c

5I = −e−2x(cosx+2sinx) + c

Therefore, I = −e−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) ∫et cost dt

Integrating by parts, we get,

I = (1/3) [et sint − ∫et sint dt]

I = (1/3) [et sint − (−et cost + ∫et cost dt)]

I = (1/3) et sint + (1/3) et cost − (1/3) ∫et cost dt

I = et[sint+cost]/3 − I + c

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

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

Therefore, I =

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