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

**Question 1. âˆ«e**^{ax }cosbx dx

^{ax }cosbx dx

**Solution:**

We have,

I = âˆ«e

^{ax}cosbx dxUsing integration by parts, we get,

I = e

^{ax }(sinbx)/b âˆ’ aâˆ«e^{ax }(sinbx)/b dxI = e

^{ax}(sinbx)/b âˆ’ (a/b)[âˆ’e^{ax}(cosbx)/b + aâˆ«e^{ax}(cosbx)/b dx]I = e

^{ax}(sinbx)/b + (a/b^{2}) e^{ax}(cosbx) âˆ’ (a^{2}/b^{2})âˆ«e^{ax}(cosbx) dxI = (e

^{ax}/b^{2}) [b sinbx + a cos bx] + (a^{2}/b^{2}) I + c(a

^{2}+b^{2})I/b^{2}= (e^{ax}/b^{2}) [b sinbx + a cos bx] + c

Therefore, I = e^{ax}[b sinbx + a cos bx]/(a^{2}+b^{2}) + c

**Question 2. âˆ«e**^{ax} sin(bx+c)dx

^{ax}sin(bx+c)dx

**Solution:**

We have,

I = âˆ«e

^{ax}sin(bx+c)dxUsing integration by parts, we get,

I = âˆ’ e

^{ax}cos(bx+c)/b + âˆ«ae^{ax}cos(bx+c)/b dxI = (âˆ’1/b) e

^{ax}cos(bx+c) + (a/b) âˆ«e^{ax}cos(bx+c) dxI = (âˆ’1/b) e

^{ax}cos(bx+c) + (a/b) [e^{ax}sin(bx+c)/b âˆ’ âˆ«ae^{ax}sin(bx+c)/b dx]I = (âˆ’1/b) e

^{ax}cos(bx+c) + (a/b^{2}) e^{ax}sin(bx+c) âˆ’ (a^{2}/b^{2})âˆ«e^{ax}sin(bx+c) dxI = (e

^{ax}/b^{2}) [a sin(bx+c) âˆ’ b cos(bx+c)] âˆ’ (a^{2}/b^{2}) I + c(a

^{2}+b^{2}) I/b^{2}= (e^{ax}/b^{2}) [a sin(bx+c) âˆ’ b cos(bx+c)] + c

Therefore, I = (e^{ax})[a sin(bx+c)âˆ’b cos(bx+c)]/(a^{2}+b^{2}) + c

**Question 3. âˆ«cos(logx) dx**

**Solution:**

We have,

I = âˆ«cos(logx) dx

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

=> dx = xdt

=> dx = e

^{t }dtSo, the equation becomes,

I = âˆ«e

^{t }cost dtUsing integration by parts, we get,

I = e

^{t }sint âˆ’ âˆ«e^{t}sint dtI = e

^{t}sint âˆ’ [âˆ’e^{t}cost + âˆ«e^{t}cost dt]I = e

^{t}sint + e^{t}cost âˆ’ I + c2I = e

^{t }(sint+cost) + cI = e

^{t}(sint+cost)/2 + c

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

**Question 4. âˆ«e**^{2x} cos(3x+4)dx

^{2x}cos(3x+4)dx

**Solution:**

We have,

I = âˆ«e

^{2x}cos(3x+4)dxUsing integration by parts, we get,

I = e

^{2x}sin(3x+4)/3 âˆ’ âˆ«2e^{2x}sin(3x+4)/3 dxI = (1/3) e

^{2x}sin(3x+4) âˆ’ (2/3) âˆ«e^{2x}sin(3x+4) dxI = (1/3) e

^{2x}sin(3x+4) âˆ’ (2/3) [âˆ’e^{2x}cos(3x+4)/3 + âˆ«2e^{2x}cos(3x+4)/3 dx]I = (1/3) e

^{2x}sin(3x+4) + (2/9) e^{2x }cos(3x+4) âˆ’ (4/9)âˆ«2e^{2x}cos(3x+4) dx]I = (e

^{2x}/9) [2 cos(3x+4)âˆ’3 sin(3x+4)] âˆ’ (4/9)I + c(13/9)I = (e

^{2x}/9) [2 cos(3x+4)âˆ’3 sin(3x+4)] + c

Therefore, I = e^{2x}[2 cos(3x+4)âˆ’3 sin(3x+4)]/13 + c

**Question 5. âˆ«e**^{2x} sinx cosx dx

^{2x}sinx cosx dx

**Solution:**

We have,

I = âˆ«e

^{2x}sinx cosx dxI = (1/2)âˆ«e

^{2x}(2sinx cosx) dxI = (1/2)âˆ«e

^{2x}sin2x dxUsing integration by parts, we get,

I = (1/2)[âˆ’e

^{2x }(cos2x)/2 + âˆ«2e^{2x}(cos2x)/2 dx]2I = âˆ’e

^{2x}(cos2x)/2 + âˆ«e^{2x}(cos2x)dx2I = âˆ’e

^{2x}(cos2x)/2 + [e^{2x }(sin2x)/2 âˆ’ âˆ«2e^{2x}(sin2x)/2 dx]2I = âˆ’e

^{2x}(cos2x)/2 + e^{2x}(sin2x)/2 âˆ’ 2I + c4I = e

^{2x}(sin2xâˆ’cos2x)/2 + c

Therefore, I = e^{2x}(sin2xâˆ’cos2x)/8 + c

**Question 6. âˆ«e**^{2x }sinx dx

^{2x }sinx dx

**Solution:**

We have,

I = âˆ«e

^{2x}sinx dxUsing integration by parts, we get,

I = e

^{2x}(âˆ’cosx) + âˆ«2e^{2x }cosx dxI = âˆ’e

^{2x}cosx + 2[e^{2x}sinx âˆ’ 2âˆ«e^{2x }sinx dx]I = âˆ’e

^{2x}cosx + 2e^{2x}sinx âˆ’ 4âˆ«e^{2x}sinx dxI = e

^{2x}(2sinxâˆ’cosx) âˆ’ 4I + c5I = e

^{2x}(2sinxâˆ’cosx) + c

Therefore, I = e^{2x}(2sinxâˆ’cosx)/5 + c

**Question 7. âˆ«e**^{2x} sin(3x+1) dx

^{2x}sin(3x+1) dx

**Solution:**

We have,

I = âˆ«e

^{2x}sin(3x+1) dxUsing integration by parts, we get,

I = âˆ’e

^{2x}cos(3x+1)/3 + âˆ«2e^{2x}cos(3x+1)/3 dxI = âˆ’(1/3) e

^{2x}cos(3x+1) + (2/3) [e^{2x}sin(3x+1)/3 âˆ’ âˆ«2e^{2x}sin(3x+1)/3 dx]I = âˆ’(1/3) e

^{2x}cos(3x+1) + (2/3) [e^{2x}sin(3x+1)/3 âˆ’ (2/3)âˆ«e^{2x}sin(3x+1)dx]I = âˆ’(1/3) e

^{2x}cos(3x+1) + (2/9) e^{2x}sin(3x+1) âˆ’ (4/9)I + c(13/9)I = e

^{2x}[2sin(3x+1)âˆ’3cos(3x+1)]/9 + c

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

**Question 8. âˆ«e**^{x} sin^{2}x dx

^{x}sin

^{2}x dx

**Solution:**

We have,

I = âˆ«e

^{x}sin^{2}x dxI = (1/2)âˆ«e

^{x }(1âˆ’cos2x)dxI = (1/2)âˆ«e

^{x}dx âˆ’ (1/2)âˆ«e^{x }cos2x dxLet I

_{1}= âˆ«e^{x}cos2x dx. So, our equation becomes,I = (1/2)âˆ«e

^{x }dx âˆ’ (1/2) I_{1}. . . . (1)Using integration by parts in I

_{1}, we get,I

_{1}= e^{x }(sin2x)/2 âˆ’ âˆ«e^{x }(sin2x)/2 dxI

_{1}= e^{x}(sin2x)/2 âˆ’ (1/2)[âˆ’e^{x }(cos2x)/2 + âˆ«e^{x}(cos2x)/2 dxI

_{1}= e^{x}(sin2x)/2 + e^{x}(cos2x)/4 âˆ’ (1/4)âˆ«e^{x}cos2x dxI

_{1}= e^{x}[2sin2x+cos2x)]/4 âˆ’ (1/4) I_{1}(5/4)I

_{1}= e^{x}[2sin2x+cos2x)]/4I

_{1}= e^{x}[2sin2x+cos2x)]/5 . . . . (2)Putting (2) in (1), we get,

I = (1/2)âˆ«e

^{x}âˆ’ (1/2) (1/5) e^{x}[2sin2x+cos2x)] + c

Therefore, I = e^{x}/2 âˆ’ e^{x}[2sin2x+cos2x)]/10 + c

**Question 9. âˆ«(1/x**^{3}) sin(logx) dx

^{3}) sin(logx) dx

**Solution:**

We have,

I = âˆ«(1/x

^{3}) sin(logx) dxLet logx = t, so we have, (1/x)dx = dt

=> dx = xdt

=> dx = e

^{t }dtSo, the equation becomes,

I = âˆ«(1/e

^{3t}) (sint) e^{t }dtI = âˆ«e

^{âˆ’2t }sint dtUsing integration by parts, we get,

I = âˆ’e

^{âˆ’2t}cost âˆ’ âˆ«2e^{âˆ’2t}cost dtI = âˆ’e

^{âˆ’2t}cost âˆ’ 2[e^{âˆ’2t }sint + âˆ«2e^{âˆ’2t }sint]I = âˆ’e

^{âˆ’2t}cost âˆ’ 2e^{âˆ’2t}sint âˆ’ 4âˆ«e^{âˆ’2t }sintI = âˆ’e

^{âˆ’2t}[cost+2sint] âˆ’ 4I + c5I = âˆ’e

^{âˆ’2t}[cost+2sint] + cI = âˆ’e

^{âˆ’2t}[cost+2sint]/5 + cI = âˆ’e

^{âˆ’2logx}[cos(logx)+2sin(logx)]/5 + cI = âˆ’x

^{-2}[cost+2sint]/5 + c

Therefore, I = âˆ’[cost+2sint]/5x^{2}+ c

**Question 10. âˆ«e**^{2x} cos^{2}x dx

^{2x}cos

^{2}x dx

**Solution:**

We have,

I = âˆ«e

^{2x }cos^{2}x dxI = (1/2) âˆ«e

^{2x}(1+cos2x) dxI = (1/2) âˆ«e

^{2x}dx + (1/2) âˆ«e^{2x}cos2x dxLet I1 = âˆ«e

^{2x}cos2x dx. So, our equation becomes,I = (1/2)âˆ«e

^{2x}dx + (1/2) I_{1}. . . . (1)Using integration by parts in I

_{1}, we get,I

_{1}= e^{2x}(sin2x)/2 âˆ’ âˆ«2e^{2x }(sin2x)/2 dxI

_{1}= e^{2x}(sin2x)/2 âˆ’ âˆ«e^{2x }sin2x dxI

_{1}= e^{2x}(sin2x)/2 âˆ’ [âˆ’e^{2x}(cos2x)/2 + âˆ«2e^{2x}(cos2x)/2 dx]I

_{1}= e^{2x}(sin2x)/2 + e^{2x}(cos2x)/2 âˆ’ âˆ«e^{2x}(cos2x) dxI

_{1}= e^{2x}(sin2x+cos2x)/2 âˆ’ I_{1}2I

_{1}= e^{2x}(sin2x+cos2x)/2I

_{1}= e^{2x}(sin2x+cos2x)/4 . . . . (2)Putting (2) in (1), we get,

I = (1/2)âˆ«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. âˆ«e**^{âˆ’2x} sinx dx

^{âˆ’2x}sinx dx

**Solution:**

We have,

I = âˆ«e

^{âˆ’2x}sinx dxIntegrating by parts, we get,

I = âˆ’e

^{âˆ’2x }cosx âˆ’ âˆ«2e^{âˆ’2x }cosx dxI = âˆ’e

^{âˆ’2x}cosx âˆ’ 2[e^{âˆ’2x}sinx+âˆ«2e^{âˆ’2x}sinx dx]I = âˆ’e

^{âˆ’2x}cosx âˆ’ 2e^{âˆ’2x}sinx âˆ’ 4âˆ«e^{âˆ’2x}sinx dxI = âˆ’e

^{âˆ’2x}(cosx+2sinx) âˆ’ 4I + c5I = âˆ’e

^{âˆ’2x}(cosx+2sinx) + c

Therefore, I = âˆ’e^{âˆ’2x}(cosx+2sinx)/5 + c

**Question 12. **

**Solution:**

We have,

I =

Let x

^{3}= t, so we have, 3x^{2}dx = dtSo, the equation becomes,

I = (1/3) âˆ«e

^{t}cost dtIntegrating by parts, we get,

I = (1/3) [e

^{t}sint âˆ’ âˆ«e^{t}sint dt]I = (1/3) [e

^{t}sint âˆ’ (âˆ’e^{t}cost + âˆ«e^{t}cost dt)]I = (1/3) e

^{t}sint + (1/3) e^{t}cost âˆ’ (1/3) âˆ«e^{t}cost dtI = e

^{t}[sint+cost]/3 âˆ’ I + c2I = e

^{t}[sint+cost]/3 + cI = e

^{t}[sint+cost]/6 + c

Therefore, I =

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