Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.27

Question 1. ∫e^ax cosbx dx

Solution:

We have,

I = ∫e^ax cosbx dx 

Using integration by parts, we get,

I = e^ax (sinbx)/b − a∫e^ax (sinbx)/b dx

I = e^ax (sinbx)/b − (a/b)[−e^ax (cosbx)/b + a∫e^ax (cosbx)/b dx]

I = e^ax (sinbx)/b + (a/b²) e^ax (cosbx) − (a²/b²)∫e^ax (cosbx) dx

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

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

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

Question 2. ∫e^ax sin(bx+c)dx

Solution:

We have,

I = ∫e^ax sin(bx+c)dx

Using integration by parts, we get,

I = − e^ax cos(bx+c)/b + ∫ae^ax cos(bx+c)/b dx

I = (−1/b) e^ax cos(bx+c) + (a/b) ∫e^ax cos(bx+c) dx

I = (−1/b) e^ax cos(bx+c) + (a/b) [e^ax sin(bx+c)/b − ∫ae^ax sin(bx+c)/b dx] 

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

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

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

Therefore, I = (e^ax)[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 = e^t dt

So, the equation becomes,

I = ∫e^t cost dt

Using integration by parts, we get,

I = e^t sint − ∫e^t sint dt

I = e^t sint − [−e^t cost + ∫e^t cost dt]

I = e^t sint + e^t cost − I + c

2I = e^t (sint+cost) + c

I = e^t (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 = ∫e^2x cos(3x+4)dx 

Using integration by parts, we get,

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

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

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

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

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

(13/9)I = (e^2x/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 = ∫e^2x sinx cosx dx

I = (1/2)∫e^2x (2sinx cosx) dx

I = (1/2)∫e^2x sin2x dx

Using integration by parts, we get,

I = (1/2)[−e^2x (cos2x)/2 + ∫2e^2x (cos2x)/2 dx]

2I = −e^2x (cos2x)/2 + ∫e^2x (cos2x)dx

2I = −e^2x (cos2x)/2 + [e^2x (sin2x)/2 − ∫2e^2x (sin2x)/2 dx]

2I = −e^2x (cos2x)/2 + e^2x (sin2x)/2 − 2I + c

4I = e^2x(sin2x−cos2x)/2 + c

Therefore, I = e^2x(sin2x−cos2x)/8 + c

Question 6. ∫e^2x sinx dx

Solution:

We have,

I = ∫e^2x sinx dx

Using integration by parts, we get,

I = e^2x(−cosx) + ∫2e^2x cosx dx

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

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

I = e^2x(2sinx−cosx) − 4I + c

5I =  e^2x(2sinx−cosx) + c

Therefore, I = e^2x(2sinx−cosx)/5 + c 

Question 7. ∫e^2x sin(3x+1) dx

Solution:

We have,

I = ∫e^2x sin(3x+1) dx

Using integration by parts, we get,

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

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

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

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

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

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

Question 8. ∫e^x sin²x dx

Solution:

We have,

I = ∫e^x sin²x dx

I = (1/2)∫e^x (1−cos2x)dx

I = (1/2)∫e^x dx − (1/2)∫e^x cos2x dx

Let I₁ = ∫e^x cos2x dx. So, our equation becomes,

I = (1/2)∫e^x dx − (1/2) I₁   . . . . (1)

Using integration by parts in I₁, we get,

I₁ = e^x (sin2x)/2 − ∫e^x (sin2x)/2 dx

I₁ = e^x (sin2x)/2 − (1/2)[−e^x (cos2x)/2 + ∫e^x (cos2x)/2 dx

I₁ = e^x (sin2x)/2 + e^x (cos2x)/4 − (1/4)∫e^x cos2x dx

I₁ = e^x[2sin2x+cos2x)]/4 − (1/4) I₁

(5/4)I₁ = e^x[2sin2x+cos2x)]/4

I₁ = e^x[2sin2x+cos2x)]/5  . . . . (2)

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

I = (1/2)∫e^x − (1/2) (1/5) e^x[2sin2x+cos2x)] + c

Therefore, I = e^x/2 − e^x[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 = e^t dt

So, the equation becomes,

I = ∫(1/e^3t) (sint) e^t 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. ∫e^2x cos²x dx 

Solution:

We have,

I = ∫e^2x cos²x dx 

I = (1/2) ∫e^2x (1+cos2x) dx

I = (1/2) ∫e^2x dx + (1/2) ∫e^2x cos2x dx

Let I1 = ∫e^2x cos2x dx. So, our equation becomes,

I = (1/2)∫e^2x dx + (1/2) I₁   . . . . (1)

Using integration by parts in I₁, we get,

I₁ = e^2x (sin2x)/2 − ∫2e^2x (sin2x)/2 dx

I₁ = e^2x (sin2x)/2 − ∫e^2x sin2x dx

I₁ = e^2x (sin2x)/2 − [−e^2x (cos2x)/2 + ∫2e^2x (cos2x)/2 dx] 

I₁ = e^2x (sin2x)/2 + e^2x (cos2x)/2 − ∫e^2x (cos2x) dx

 I₁ = e^2x(sin2x+cos2x)/2 − I₁

2I₁ = e^2x(sin2x+cos2x)/2 

I₁ = e^2x(sin2x+cos2x)/4   . . . . (2)

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

I = (1/2)∫e^2x dx + (1/2) (1/4) [e^2x(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) ∫e^t cost dt

Integrating by parts, we get,

I = (1/3) [e^t sint − ∫e^t sint dt]

I = (1/3) [e^t sint − (−e^t cost + ∫e^t cost dt)]

I = (1/3) e^t sint + (1/3) e^t cost − (1/3) ∫e^t cost dt

I = e^t[sint+cost]/3 − I + c

2I = e^t[sint+cost]/3 + c  

I = e^t[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:

• Partial fraction decomposition
• Factoring the denominator
• Recognizing standard integral forms
• Logarithmic integration
• Long division of polynomials (when degree of numerator ≥ degree of denominator)

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.