Here we provide RD Sharma Class 12 Ex 19.12 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 19.12 Solutions Chapter 19 Indefinite Integrals book pdf download. Now you will get step-by-step solutions to each question.
Textbook | NCERT |
Class | Class 12th |
Subject | Maths |
Chapter | 19 |
Exercise | 19.12 |
Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 19.12 Solutions Chapter 19 Indefinite Integrals
Question 1. ∫sin4x cos3x dx
Solution:
Let I = ∫ sin4x cos3x dx -(i)
Let sinx = t
On differentiating with respect to x:
cosx = dt/dx
cosx dx = dt
dx = dt/cosx
Putting value of dx and sinx in equation (i):
I = ∫ t4 cosxdt/cosx
I = ∫ t4 cos2 x dtI = ∫ t4 (1 – sin2 x) dt
I = ∫ t4 (1 – t2) dt
I = ∫ (t4– t2) dt
I = t5/5 – t7/7 + c
I = sin5/5 – sin7/7 + c
Question 2. ∫ sin5x dx
Solution:
Let I = ∫ sin5x dx
I = ∫sin3xsin2x dx
= ∫sin3x(1 – cos2x)dx
= ∫(sin3x – sin3xcos2x)dx
= ∫[sinxsin2x – sin3xcos2x]dx
= ∫[sinx(1 – cos2x) – sin3xcos2x]dx
= ∫(sinx – sinxcos2x – sin3xcos2x)dx
I = ∫sinx dx – ∫sinxcos2x dx – ∫sin3xcos2x dx
Putting cosx = t and -sinxdx = dt in 2nd and 3rd integral:
I = ∫sinx dx + ∫t2dt + ∫sin2xt3dt/t
= ∫sinx dx + ∫t2 dt + ∫sin2xt2 dt
= ∫sinx dx + ∫t2 dt + ∫(1 – cos2x)t2 dt
Putting value of t:
Question 3. ∫cos5x dx
Solution:
Let I = ∫cos5x dx
I = ∫cos2xcos3x dx
= ∫(1 – sin2x)cos3x dx
= ∫(cos3x−sin2xcos3x)dx
= ∫(cos2xcosx – sin2xcos2xcosx)dx
= ∫[(1 – sin2x)cosx – sin2x(1 – sin2x)cosx]dx
= ∫(cosx – sin2xcosx – sin2xcosx + sin4xcosx)dx
= ∫cosx dx – 2∫sin2xcosx dx + ∫sin4xcosx dx
Putting sinx = t and cosxdx = dt in 2nd and 3rd integral we get:
I = ∫cos dx – 2∫t2dt + ∫t4dt
= sinx – 2t3/3 + t5/5 + c
Putting value of t:
I = = sinx – 2sin3x/3 + cos5x/5 + c
Question 4. ∫sin5xcosx dx
Solution:
Let I = ∫sin5xcosx dx −(i)
Let sinx = t:
On differentiating with respect to x:
-cosx = dt/dx
cosx dx = -dt
Putting cosxdx = -dt and sinx = t in eq (i):
I = ∫t5dt
= t6/6 + c
= sin6x/6 + c
Question 5. ∫sin3xcos6x dx
Solution:
Let I = ∫sin3xcos6x dx −(i)
Let cosx = t
On differentiating both sides w.r.t′x′:
-sinx = dt/dx
sinxdx = -dt
Putting cosx = t and sinxdx = -dt in eq (i):
I = -∫sin2x t6dt
= -∫(1 – cos2x)t6dt
= -∫(1 – t2)t6dt
= -∫(t6 – t8)dt
= -(t7/7 – t9/9) + c
Putting value of t:
I = -(cos7x/7 – cos9x/9) + c
Question 6. ∫cos7x dx
Solution:
Let I = ∫cos7x dx
= ∫cos6xcosx dx
= ∫(cos2x)3cosx dx
= ∫(1 – sin2x)3cosx dx
= ∫(1 – sin6x – 3sin2x + 3sin4x)cosx dx
= ∫(cosx – sin6xcosx – 3sin2xcosx + 3sin4xcosx)dx −(i)
Putting sinx = t and cosx dx = t in 2nd,3rd and 4th integral in (i):
I = ∫cosx dx – ∫t6dt – 3∫t2dt + 3∫t4dt
= sinx – t7/7 - 3t3/3 +3t5/5 + c
Putting value of t:
= sinx – sin7x/7 - 3sin3x/3 +3sin5x/5 + c
Question 7. ∫xcos3x2sinx2dx
Solution:
Let I = ∫xcos3x2sinx2dx −(i)
Let cosx2 = t
On differentiating both sides:
-2xsinx2 = dt/dx
xsinx2 dx = -dt/2
Putting values in (i):
= -t4/8 + c
Putting value of t:
Question 8. ∫sin7x dx
Solution:
Let I = ∫sin7x dx
I = ∫sin6x sinx dx
= ∫(sin2x)3sinx dx
= ∫(1 – cos2x)3sinx dx
= ∫(1 – cos6x – 3cos2x + 3cos4x)sinx dx
I = ∫sinx dx – ∫cos6xsinx dx + 3∫cos4xsinx dx – 3∫cos2xsinx dx
Putting cosx = t and sinx dx = -dt in 2nd,3rd and 4th integral:
I = ∫sinx dx – ∫t6(-dt) + 3∫t4(-dt) – 3∫t2(-dt)
Question 9. ∫sin3xcos5x dx
Solution:
Let I = ∫sin3xcos5x dx −(i)
Let cosx = t
On differentiating both sides: -sinx = dt/dx
sinx dx = -dt
Putting values in (i):
I = ∫sin2xt5(-dt)
= −∫(1 – cos2x)t5 dt
= −∫(1 – t2)t5 dt
= ∫(t7 – t5) dt
= t8/8 – t6/6 + c
Putting value of t:
Question 10. 
Solution:
Let I =
Dividing and multiplying the equation by cos6x:
Let tanx = t, then:
sec2x = dt/dx
sec2x dx = dt
Putting values in eq (ii):
Question 11. 
Solution:
Dividing and multiplying by cos8x:
Let tanx=t,then:
Putting values in ii:
Question 12. 
Solution:
Dividing and multiplying by cos4x:
Let tanx=t,then: sec2xdx = dt Putting values in i:
Putting value of t:
Question 13. 
Solution:
Let tanx=t⟹sec2x dx = dt:
Putting value of t:
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