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. ∫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. 

Solution:

We have,

I = 

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 = 

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