Class 12 RD Sharma Solutions Chapter 19 Indefinite Integrals Exercise 19.27 (original) (raw)

Last Updated : 3 Sep, 2024

**Question 1. ∫e ax **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 = e ax [b sinbx + a cos bx]/(a 2 +b 2 ) + c

**Question 2. ∫e ax 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 = (e ax )[a sin(bx+c)−b cos(bx+c)]/(a 2 +b 2 ) + 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. ∫e 2x 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 = e 2x [2 cos(3x+4)−3 sin(3x+4)]/13 + c

**Question 5. ∫e 2x 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 = e 2x (sin2x−cos2x)/8 + c

**Question 6. ∫e 2x **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 = e 2x (2sinx−cosx)/5 + c

**Question 7. ∫e 2x 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 = e 2x [2sin(3x+1)−3cos(3x+1)]/13 + c

**Question 8. ∫e x sin 2 x 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 = e x /2 − e x [2sin2x+cos2x)]/10 + c

**Question 9. ∫(1/x 3 ) 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]/5x 2 + c

**Question 10. ∫e 2x cos 2 x 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) e 2x + (1/8) [e 2x (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. ∫x^2 e^{x^3} cosx^3 dx

**Solution:

We have,

I = ∫x^2 e^{x^3} cosx^3 dx

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 = \frac{e^{x^3}(sinx^3+cosx^3)}{6} + c

Summary

Exercise 19.27 in RD Sharma's Class 12 Chapter 19 on Indefinite Integrals focuses on integrating rational functions where the denominator is a cubic polynomial (third-degree). The key techniques used in solving these integrals include:

These problems require a good understanding of algebraic manipulation, factorization of polynomials, and various integration techniques. Students should be comfortable with complex fractions and identifying the most efficient method for each problem.