# RD Sharma Class 12 Ex 19.27 Solutions Chapter 19 Indefinite Integrals

Here we provide RD Sharma Class 12 Ex 19.27 Solutions Chapter 19 Indefinite Integrals for English medium students, Which will very helpful for every student in their exams. Students can download the latest Sharma Class 12 Ex 19.27 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 19.27 Solutions Chapter 19 Indefinite Integrals

### 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 = edt

So, the equation becomes,

I = ∫ecost dt

Using integration by parts, we get,

I = esint − ∫et sint dt

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

I = et sint + et cost − I + c

2I = e(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)∫e(1−cos2x)dx

I = (1/2)∫ex dx − (1/2)∫ecos2x dx

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

I = (1/2)∫edx − (1/2) I1   . . . . (1)

Using integration by parts in I1, we get,

I1 = e(sin2x)/2 − ∫e(sin2x)/2 dx

I1 = ex (sin2x)/2 − (1/2)[−e(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 = edt

So, the equation becomes,

I = ∫(1/e3t) (sint) edt

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 = I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

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