Here we provide RD Sharma Class 12 Ex 19.25 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 19.25 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 | 19.25 |
| Category | RD Sharma Solutions |
RD Sharma Class 12 Ex 19.25 Solutions Chapter 19 Indefinite Integrals
Evaluate the following integrals:
Question 1. ∫x cosxdx
Solution:
Given that, I = ∫x cosxdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = x∫cosxdx – ∫(1 × ∫cosxdx)dx + c
= xsinx – ∫sinxdx + c
Hence, I = x sinx + cosx + c
Question 2. ∫log(x + 1)dx
Solution:
Given that, I = ∫log(x + 1)dx
= ∫1 × log(x + 1)dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = log(x + 1)∫1dx – ∫(1/(x + 1) × ∫ 1dx)dx + c
= xlog(x + 1) – ∫(x/(x + 1))dx + c
= x log(x + 1) – ∫(1 – 1/(x + 1))dx + c
Hence, I = x log(x + 1) – x + log(x + 1) + c
Question 3. ∫x3 logxdx
Solution:
Given that, I = ∫ x3 logxdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = logx ∫x3 dx – ∫(1/x × ∫x3 dx)dx + c
= x4/4 logx – ∫x4/4x dx+c
= x4/4 logx – 1/4∫x3 dx + c
= x4/4 logx – 1/4 ∫x4/4 dx + c
I = x4/4 logx – 1/16 x4 + c
Question 4. ∫xex dx
Solution:
Given that I = ∫xex dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = xex – ∫1.ex dx
= xex – ex + c
Hence, I = = xex – ex + c
Question 5. ∫xe2x dx
Solution:
Given that, I = ∫xe2x dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = x∫e2x dx – ∫(1 × ∫ e2x dx) dx + c
= x∫e2x dx – ∫(1 × ∫e2x dx)dx + c
= (xe2x)/2 – ∫(e2x/2)dx + c
= (xe2x)/2 – e2x/4 + c
Hence, I = (x/2 – 1/4) e2x + c
Question 6. ∫x2 e-x dx
Solution:
Given that I = ∫x2 e-x dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = x2 ∫e-x dx – ∫(2x∫e-x dx)dx
= -x2 e-x – ∫(2x)(-e-x)dx
= -x2 e-x + 2∫xe-x dx
= -x2 e-x + 2[x∫e-x dx – ∫(1 × ∫ e-x dx) dx]
= -x2 e-x + 2[x(-e-x) – ∫(-e-x)dx]
= -x2 e-x – 2xe-x + 2∫e-x dx
Hence, I = -x2 e-x – 2xe-x – 2e-x + c
Question 7. ∫ x2cosxdx
Solution:
Given that, I = ∫ x2cosxdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = x2 ∫ cosxdx – ∫(2x)cosxdx)dx
= x2 sinx – 2∫(x)(sinx)dx
= x2 sinx – 2[x∫sinxdx – ∫(1 × ∫sinxdx)dx]
= x2 sinx – 2[x(-cosx) – ∫(-cosx)dx]
= x2 sinx + 2xcosx – 2∫(cosx)dx
Hence, I = x2sinx + 2xcosx – 2sinx + c
Question 8. ∫x2cos2xdx
Solution:
Given that, I = ∫x2cos2xdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = x2 ∫cos2xdx – ∫(2x∫ cos2xdx)dx
= x2 (sin2x)/2 – 2∫x((sin2x)/2)dx
= 1/2 x2 sin2x – ∫xsin2xdx
= 1/2 x2 sin2x – [x∫sin2xdx – ∫ (1∫ sin2xdx)dx]
= 1/2 x2 sin2x – [x((-cos2x)/2) – ∫(-(cos2x)/2)dx]
= 1/2 x2sin2x + x/2 cos2x – 1/2 ∫(cos2x)dx
Hence, I = 1/2 x2 sin2x + x/2 cos2x – 1/4 sin2x + c
Question 9. ∫xsin2xdx
Solution:
Given that, I =∫xsin2xdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = x∫sin2xdx – ∫(1)sin2xdx)dx
= x(-(cos2x)/2) – ∫(-(cos2x)/2)dx
= -x/2 cos2x + 1/2 ∫cos2xdx
= -x/2 cos2x + 1/2(sin2x)/2 + c
Hence, I = -x/2 cos2x + 1/4 sin2x + c
Question 10. ∫(log(logx))/x dx
Solution:
Given that, I = ∫(log(logx))/x dx
= ∫(1/x)(log(logx))dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = loglogx]1/x dx – ∫(1/(xlogx)∫1/x dx)dx
= logx × log(logx) – ∫(1/(xlogx) logx)dx
= logx × log(logx) – ∫1/x dx
= logx × log(logx) – logx + c
Hence, I = logx(loglogx – 1) + c
Question 11. ∫x2 cosxdx
Solution:
Given that I = ∫x2 cosxdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = x2∫ cosxdx – ∫(2x]cosxdx)dx
= x2sinx – 2∫xsinxdx
= x2 sinx – 2[x∫sinxdx – ∫(1]sinxdx)dx]
= x2 sinx – 2[x(-cosx) – ∫(-cosx)dx]
= x2 sinx + 2xcosx – 2∫(cosx)dx
Hence, I = x2 sinx + 2xcosx – 2sinx + c
Question 12. ∫xcosec2xdx
Solution :
Given that, I = ∫xcosec2xdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = x∫cosec2xdx – ∫(∫ cosec2xdx)dx
= -xcotx + ∫cotxdx
= -x cotx + log |sinx| + c
Hence, I = -x cotx + log |sinx| + c
Question 13. ∫xcos2xdx
Solution:
Given that, I = ∫xcos2xdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = x∫cos2xdx – ∫(1∫ cos2xdx)dx
= x∫((cos2x + 1)/2)dx – ∫(∫((1 + cos2x)/2)dx)dx
= x/2 [(sin2x)/2 + x] – 1/2∫(x + (sin2x)/2)dx
= x/4 sin2x + x2/2 – 1/2 × x2/2 – 1/4 (-(cos2x)/2) + c
Hence, I = x/4 sin2x + x2/4 + 1/8 cos2x + c
Question 14. ∫xn logx dx
Solution:
Given that, I = ∫xn logxdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = logx∫xn dx – ∫(1/x ∫xndx)dx
= xn+1/(n + 1) logx – ∫(1/x × xn+1/(n + 1))dx
= xn+1/(n + 1) logx – ∫(xn/(n + 1))dx
Hence, I = xn+1/(n + 1) logx – 1/(n + 1)2 × (xn+1) + c
Question 15. ∫(logx)/xn dx
Solution:
Given that, I = ∫(logx)/xn dx = ∫(logx)(1/xn)dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = logx∫(1/xn)dx – ∫((d(logx))/dx)(∫(1/xn)dx)dx
= logx(x1-n/(1 – n)) – ∫1/x (x1-n/(1 – n))dx
= logx(x1-n/(1 – n)) – ∫(xn/(1 – n))dx
= logx(x1-n/(1 – n)) – (1/(1 – n))(x1-n/(1 – n))
Hence, I = logx(x1-n/(1 – n)) – (x1-n/([1 – n]2)) + c
Question 16. ∫x2 sin2xdx
Solution:
Given that, I = ∫x2 sin2xdx
= ∫x2 ((1 – cos2x)/2)dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= ∫x2/2 dx – ∫((x2 cos2x)/2)dx
= x3/6 – 1/2 [∫x2 cos2xdx]
= x3/6 – 1/2 [x2 ∫cos2xdx – ∫ (2x∫cos2xdx)dx]
= x3/6 – 1/2 (x2(sin2x)/2) + 1/2 × 2∫(x (sin2x)/2)dx
= x3/6 – 1/4 x2sin2x + 1/2 [x ∫sin2xdx – ∫(1∫sin2xdx)dx]
= x3/6 – 1/4 x2 sin2x + 1/2 [x(-(cos2x)/2) – ∫(-(cos2x)/2)dx]
= x3/6 – 1/4 x2 sin2x + 1/2 x(-(cos2x)/2) + 1/4 × (sin2x/2) + c
= x3/6 – 1/4 x2 sin2x – 1/4 x(cos2x) + 1/8 × (sin2x) + c
Hence, I = x3/6 – 1/4 x2 sin2x – 1/4 x(cos2x) + 1/8 × (sin2x) + c
Question 17. 
Solution:
Given that, l =
Let us assume, x2 = t
2xdx = dt
I = ∫t × et dt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= t∫et dt – ∫(1 × ∫etdt)dt
= tet – ∫et dt
= tet – et + c
= et-1 + c
Hence, I =
(x2 – 1) + c
Question 18. ∫x3 cosx2 dx
Solution:
Given that, I = ∫x3 cosx2 dx
Let us assume x2 = t
2xdx = dt
I = 1/2 ∫tcostdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= 1/2[t∫costdt – ∫(1 × ∫costdt)dt]
= 1/2 [t × sint – ∫sintdt]
= 1/2[tsint + cost] + c
Hence, I = 1/2 [x² sinx2 + cosx2] + c
Question 19. ∫xsinxcosxdx
Solution:
Given that, I = ∫xsinxcosxdx
= ∫x/2(2sinxcosx)dx
= 1/2 ∫xsin2xdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= 1/2 [x∫sin2xdx – ∫(1 × ∫sin2xdx)dx]
= 1/2 [x((-cos2x)/2) – ∫((-cos2x)/2)dx]
= -1/4 xcos2x + 1/4 ∫cos2xdx
Hence, I = -1/4 xcos2x + 1/8 sin2x + c
Question 20. ∫sinx(logcosx)dx
Solution:
Given that, I = ∫sinx(logcosx)dx
Let us considered, cosx = t
-sinxdx = dt
I = -∫ logtdt
= -∫1 × logtdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= -[logt∫dt – ∫(1/t × ∫dt)dt]
= -[tlogt – ∫1/t × tdt]
= -[tlogt-∫ dt]
= -[tlogt – t + c1 ]
= t(1 – logt) + c
Hence, I = cosx(1 – logcosx) + c
Evaluate the following integrals:
Question 21. ∫(logx)2 x dx
Solution:
Given that, I = ∫(logx)2 x dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = (logx)2∫xdx – ∫(2(logx)(1/x) ∫xdx)dx
= x2/2(logx)2 – 2∫(logx)(1/x)(x2/2)dx
= x2/2(logx)2 – ∫x(logx)dx
= x2/2(logx)2 – [logx∫xdx – ∫ (1/x ∫xdx)dx]
= x2/2(logx)2 – [x22/2 logx – ∫(1/x × x2/2)dx]
= x2/2(logx)2 – x2/2 logx + 1/2 ∫xdx
= x2/2(logx)2 – x2/2 logx + 1/4 x2 + c
Hence, I = x2/2 [(logx)2 – logx + 1/2] + c
Question 22. ∫e√x dx
Solution:
Given that, I = ∫ e√x dx
Let us assume, √x = t
x = t2
dx = 2tdt
I = 2∫ et tdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = 2[t∫etdt – ∫(1∫etdt)dt]
= 2[tet – ∫et dt]
= 2[tet – et] + c
= 2et (t – 1) + c
Hence, I = 2e√x(√x – 1) + c
Question 23. ∫(log(x + 2))/((x + 2)2) dx
Solution:
Given that, I = ∫(log(x + 2))/((x + 2)2) dx
Let us assume (1/(x + 2) = t
-1/((x + 2)2) dx = dt
I = -∫log(1/t)dt
= -∫logt-1 dt
= -∫1 × logtdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = logt∫dt – ∫(1/t ∫dt)dt
= tlogt – ∫(1/t × t)dt
= tlogt – ∫dt
= tlogt – t + c
= 1/(x + 2) (log(x + 2)-1 – 1) + c
Hence, I = (-1)/(x + 2) – (log(x + 2))/(x + 2) + c
Question 24. ∫(x + sinx)/(1 + cosx) dx
Solution:
Given that, I = ∫(x + sinx)/(1 + cosx) dx
= ∫x/(2cos2x/2) dx + ∫(2sinx/2 cosx/2)/(2cos2x/2) dx
= 1/2 ∫xsec2x/2 dx + ∫tanx/2 dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= 1/2 [x∫sec2x/2 dx – ∫(1∫ sec²x/2 dx)dx] + ∫tanx/2 dx
= 1/2 [2xtanx/2 – 2∫tanx/2 dx] + ∫tanx/2 dx + c
= xtanx/2 – ∫tanx/2 dx + ∫tanx/2 dx+c
Hence, I = xtanx/2 + c
Question 25. ∫log10xdx
Solution:
Given that, I = ∫log10xdx
= ∫(logx)/(log10) dx
= 1/(log10) ∫1 × logxdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= 1/(log10) [logx∫dx – ∫(1/x ∫dx)dx]
= 1/(log10) [xlogx – ∫(x/x)dx]
= 1/(log10)[xlogx – x]
Hence, I = (x/(log10)) × (logx – 1) + c
Question 26. ∫cos√x dx
Solution:
Given that, I = ∫cos√x dx
Let us assume, √x = t
x = t2
dx = 2tdt
= ∫2tcostdt
I = 2∫tcostdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = 2[t]costdt – ∫(1 ∫costdt)dt]
= 2[tsint – ∫sintdt]
= 2[tsint + cost] + c
Hence, I = 2[√x sin√x + cos√x] + c
Question 27. ∫(xcos-1x)/√(1 – x2) dx
Solution:
Given that, I = ∫(xcos-1x)/√(1 – x2) dx
Let us assume, t = cos-1x
dt = (-1)/√(1 – x2) dx
Also, cost = x
I = -∫tcostdt
Now, using integration by parts,
So, let
u = t;
du = dt
∫costdt = ∫dv
sint = v
Therefore,
I = -[tsint – ∫sintdt]
= -[tsint + cost] + c
On substituting the value t = cos-1x we get,
I = -[cos-1xsint + x] + c
Hence, I = -[cos-1x√(1 – x²) + x] + c
Question 28. ∫cosec3xdx
Solution:
Given that, I =∫cosec3xdx
=∫cosecx × cosec2xdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= cosecx × ∫cosec2xdx + ∫(cosecxcotx]cosec2xdx)dx
= cosecx × (-cotx) + ∫cosecxcotx(-cotx)dx
= -cosecxcotx – ∫cosecxcot2xdx
= -cosecxcotx – ∫cosecx(cosec2x – 1)dx
= -cosecxcotx – ∫cosec3xdx + ∫cosecxdx
I = -cosecxcotx – I + log|tanx/2| + c1
2l = -cosecxcotx + log|tanx/2| + c1
Hence, I = -1/2cosecxcotx + 1/2 log|tanx/2| + c
Question 29. ∫sec-1√x dx
Solution:
Given that, I = ∫sec-1√x dx
Let us assume, √x = t
x = t2
dx = 2tdt
I = ∫2tsec-1tdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= 2[sec-1t∫tdt – ∫(1/(t√(t2-1))∫tdt)dt]
= 2[t2/2 sec-1 – ∫(t/(2t√(t2 – 1)))dt]
= t2 sec-1t – ∫t/√(t2 – 1) dt
= t2 sec-1t – 1/2∫2t/√(t2 – 1) dt
= t2 sec-1t – 1/2 × 2√(t2 – 1) + c
Hence, I = xsec-1√x – √(x – 1) + c
Question 30. ∫sin-1√x dx
Solution:
Given that, I = ∫sin-1√x dx
Let us assume, x = t
dx = 2tdt
∫sin-1√x dx = ∫sin-1√(t2) 2tdt
= ∫sin-1t2tdt
= sin-1t∫2tdt – (∫(dsin-1t)/dt (∫2tdt)dt
= sin-1t(t2) – ∫1/√(1 – t2) (t2)dt
Now, lets solve ∫1/√(1 – t2) (t2)dt
∫1/√(1 – t2) (t2)dt = ∫(t2 – 1 + 1)/√(1 – t2) dt
= ∫(t2 – 1)/√(1 – t2) dt + ∫1/√(1 – t2) dt
As we know that, value of ∫1/√(1 – t2) dt = sin-1t
So, the remaining integral to evaluate is
∫(t2 – 1)/√(1 – t2) dt= ∫-√(1 – t2) dt
Now, substitute, t = sinu, dt = cosudu, we gte
∫-√(1 – t2) dt = ∫-cos2udu = -∫[(1 + cos2u)/2]du
= -u/2-(sin2u)/4
Now substitute back u = sin-1x and t = √x, we get
= -(sin-1√x)/2 – (sin(2sin-1√x))/4
∫sin-1√x dx = xsin-1√x-(sin-1√x)/2 – (sin(2sin-1√x))/4
sin(2sin-1√x) = 2√x √(1 – x)
Hence, I = xsin-1√x – (sin-1√x)/2 – √(x(1 – x))/2 + c
Question 31. ∫xtan2xdx
Solution:
Given that, I =∫xtan2xdx
= ∫x(sec2x – 1)dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= ∫xsec8xdx – ∫xdx
= [x∫sec2xdx – ∫(1∫sec2xdx)dx] – x2/2
= xtanx – ∫tanxdx – x2/2
Hence, I = xtanx – log|secx| – x2/2 + c
Question 32. ∫ x((sec2x – 1)/(sec2x + 1))dx
Solution:
Given that, I = ∫ x((sec2x – 1)/(sec2x + 1))dx
= ∫x((1 – cos2x)/(1 + cos2x))dx
= ∫x((sec2x)/(cos2x))dx
= ∫xtan2xdx
= ∫x(sec2x – 1)dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= ∫xsec2xdx – ∫dx
= [x∫sec2xdx – ∫(1∫ sec2xdx)dx] – x2/2
= xtanx – ∫tanxdx – x2/2
= xtanx – log|secx| – x2/2 + c
Hence, I = xtanx – log|secx| – x2/2 + c
Question 33. ∫(x + 1)exlog(xex)dx
Solution:
Given that, I = ∫(x + 1)exlog(xex)dx
Let us assume, xex = t
(1 × ex + xex)dx = dt
(x + 1)exdx = dt
I = ∫logtdt
= ∫1 × logtdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= logt∫dt – ∫(1/t∫dt)dt
= tlogt – ∫(1/t × t)dt
= tlogt – ∫dt
= tlogt – t + c
= t(logt – 1) + c
Hence, I = xex (logxex – 1) + c
Question 34. ∫sin-1(3x – 4x3)dx
Solution:
Given that, I = ∫sin-1(3x – 4x3)dx
Let us assume, x = sinθ
dx = cosθdθ
= ∫sin-1(3sinθ – 4sin3θ)cosθdθ
= ∫sin-1(sin3θ)cosθdθ
= ∫3θcosθdθ
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= 3[θ]cosθdθ – ∫(1∫cosθdθ)dθ]
= 3[θsinθ – ∫sinθdθ]
= 3[θsinθ + cosθ] + c
Hence, I = 3[xsin-1x + √(1 – x2)] + c)
Question 35. ∫sin-1(2x/(1 + x2))dx.
Solution:
Given that, I = ∫sin-1(2x/(1 + x2))dx
Let us assume, x = tanθ
dx = sec2θdθ
sin-1(2x/(1 + x2)) = sin-1((2tanθ)/(1 + tan²θ))
= sin-1(sin2θ) = 2θ
∫sin-1(2x/(1 + x2))dx = ∫2θsec2θdθ = 2∫θsec2θdθ
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
2[θ∫sec2θdθ – ∫{(d/dθ θ) ∫sec2θdθ}dθ
= 2[θtanθ – ∫tanθdθ]
= 2[θtanθ + log|cosθ|] + c
= 2[xtan-1x + log|1/√(1 + x2)|] + c
= 2xtan-1x + 2log(1 + x2)1/2 + c
= 2xtan-1x + 2[-1/2 log(1 + x2)] + c
= 2xtan-1x – log(1 + x2) + c
Hence, I = 2xtan-1x – log(1 + x2) + c
Question 36. ∫ tan-1((3x – x3)/(1 – 3x2))dx
Solution:
Given that, I = ∫tan-1((3x – x3)/(1 – 3x2))dx
Let us assume, x = tanθ
dx = sec2θdθ
I = ∫tan-1((3tanθ – tan3θ)/(1 – 3tan2x)) sec2θdθ
=∫tan-1(tan3θ)sec2θdθ
= ∫3θsec2θdθ
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= 3[θ∫ sec2θdθ – ∫(1∫sec2θdθ)dθ]
= 3[θtanθ – ∫tanθdθ]
= 3[θtanθ + logsecθ] + c
= 3[xtan-1x – log√(1 + x2)] + c
Hence, I = 3[xtan-1x – log√(1 + x2)] + c
Question 37. ∫x2sin-1xdx
Solution:
Given that, I = ∫x2sin-1xdx
I = sin-1x∫x2 dx – ∫(1/√(1 – x2) ∫x2 dx)dx
= x3/3 sin-1x – ∫x3/(3√(1 – x2)) dx
I = x3/3 sin-1x – 1/3 I1 + c1 …..(1)
Let I1 = ∫x3/√(1 – x2) dx
Let 1 – x2 = t2
-2xdx = 2tdt
-xdx = tdt
I1 = -∫(1 – t2)tdt/t
= ∫(t2 – 1)dt
= t3/3 – t + c2
= (1 – x2)3/2/3 – (1 – x2)1/2 + c2
Now, put the value of I1 in eq(1), we get
Hence, I = x3/3 sin-1x – 1/9 (1 – x2)3/2 + 1/3 (1 – x2)1/2 + c
Question 38. ∫(sin-1x)/x2dx
Solution:
Given that, I =∫(sin-1x)/x2dx
= ∫(1/x2)(sin-1x)dx
I = [sin-1x∫1/x2dx – ∫(1/√(1 – x2) ∫1/x2dx)dx]
= sin-1×(-1/x) – ∫1/√(1 – x2) (-1/x)dx
I = -1/x sin-1x + ∫1/(x√(1 – x2)) dx
I = -1/x sin-1x + I1 …….(1)
Where,
I1 = ∫1/(x√(1 – x2)) dx
Let 1 – x2 = t2
-2xdx = 2tdt
I1 = ∫x/(x2√(1 – x2)) dx
= -∫tdt/((1 – t2) √t)
= -∫dt/((1 – t2))
= ∫1/(t2 – 1) dt
= 1/2 log|(t – 1)/(t + 1)|
= 1/2 log|(t – 1)/(t + 1)|
= 1/2 log|(√(1 – x2) – 1)/(√(1 – x2) + 1)| + c1
Now, put the value of I1 in eq(1), we get
I = -(sin-1x)/x + 1/2 log|((√(1 – x2) – 1)/(√(1 – x2) + 1))((√(1 – x2) – 1)/(√(1 – x2) – 1))| + c
= -(sin-1x)/x + 1/2 log|(√(1 – x2) – 1)2/(1 – x2 – 1)| + c
= -(sin-1x)/x + 1/2 log|(√(1 – x2) – 1)2/(-x2)| + c
= -(sin-1x)/x + log|(√(1 – x2) – 1)/(-x)| + c
Hence, I = -(sin-1x)/x + log|(1 – √(1 – x2))/x| + c
Question 39. ∫(x2 tan-1x)/(1 + x2) dx
Solution:
Given that, I = ∫(x2 tan-1x)/(1 + x2) dx
Let us assume, tan-1x = t [x = tant]
1/(1 + x2) dx = dt
I = ∫t × tan2tdt
= ∫t(sec2t – 1)dt
= ∫(tsec2t – t)dt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= ∫tsec2tdt – ∫tdt
= [t∫sec2tdt – ∫(1)sec2tdt)dt] – t2/2
= [t × tant – ∫tantdt] – t2/2
= t tant – logsect – t2/2 + c
= xtan-1x – log√(1 + x2) – (tan2x)/2 + c
Hence, I = xtan-1x – 1/2 log|1 + x2| – (tan2x)/2 + c
Question 40. ∫cos-1(4x3 – 3x)dx
Solution:
Given that, I = ∫cos-1(4x3 – 3x)dx
Let us assume, x = cosθ
dx = -sinθdθ
I = -∫cos-1(4cos3θ – 3cosθ)sinθdθ
= – ∫cos-1(cos3θ)sinθdθ
= -∫3θsinθdθ
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= -3[θ]sinθdθ – ∫(1∫sinθdθ)dθ]
= -3[-θcosθ + ∫cosθdθ]
= 3θcosθ – 3sinθ + c
Hence, I = 3xcos-1x – 3√(1 – x2) + c
Evaluate the following integrals:
Question 41. ∫cos-1((1 – x2)/(1 + x2))dx
Solution:
Given that, I = ∫cos-1((1 – x2)/(1 + x2))dx)
Let us considered x = tant
dx = sec²tdt
I = ∫cos-1((1 – tan2t)/(1 + tan2t)) sec2tdt
= ∫cos-1(cos2t)sec2tdt
= ∫2tsec2tdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = 2[t∫sce2tdt – ∫(1∫sec2tdt)dt]
= 2[t × tan2t – ∫tantdt]
= 2[t × tan2t – logsect] + c
= 2[xtan-1x – log√(1 + x2)] + c
Hence, I = 2xtan-1x – log|1 + x2| + c
Question 42. ∫tan-1(2x/(1 – x2))dx
Solution:
Given that, I = ∫tan-1(2x/(1 – x2))dx
Let us considered x = tanθ
dx = sec2θdθ
I = ∫tan-1((2tanθ)/(1 – tan2θ)) sec2θdθ
= ∫tan-1(tan2θ)sec2θdθ
= ∫2θsec2θdθ
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = 2[θ∫sec2θdθ – ∫(1∫ sec2θdθ)dθ]
= 2[θtanθ – ∫tanθdθ]
= 2[θtanθ – logsecθ] + c
= 2[xtan-1x – log√(1 + x2)] + c
Hence, I = 2xtan-1x – log|1 + x2| + c
Question 43. ∫(x + 1)logxdx
Solution:
Given that, I = ∫(x + 1)logxdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = logx∫ (x + 1)dx – ∫(1/x ∫(x + 1)dx)dx
= (x2/2 + x)logx – ∫1/x (x2/2 + x)dx
= (x2/2 + x)logx – 1/2 ∫xdx – ∫dx
= (x + x2/2)logx – 1/2 × x2/2 – x + c
Hence, I = (x + x2/2)logx – 1/2 × x2/2 – x + c
Question 44. ∫ x2 tan-1xdx
Solution:
Given that, I = ∫ x2 tan-1xdx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = tan-1x∫x2 dx – ∫(1/(1 + x2) ∫x2 dx) dx
= tan-1x(x3/3) – 1/3∫x3/(1 + x2) dx
= 1/3 x3 tan-1x – 1/3 ∫(x – x/(1 + x2))dx
= 1/3 x3 tan-1x – 1/3 × x2/2 + 1/3 ∫x/(1 + x2) dx
Hence, I = 1/3 x3tan-1x – 1/6 x2 + 1/6 log|1 + x2| + c
Question 45. ∫(elogx + sinx) cosxdx
Solution:
Given that, I = ∫(elogx + sinx)cosxdx
= ∫(x + sinx)cosxdx
= ∫xcosxdx + ∫sinxcosxdx
= [x∫cosxdx – ∫(1]cosxdx)dx] + 1/2 ∫sin2xdx
= [xsinx – ∫ sinxdx] + 1/2 (-(cos2x)/2) + c
I = xsinx+cosx – 1/4 cos2x + c
= xsinx + cosx – 1/4 [1 – 2sin2x] + c
= xsinx + cosx – 1/4 + 1/2 sin2x + c
= xsinx + cosx – 1/4 + 1/2 sin2x + c
Hence, I = xsinx + cosx + 1/2 sin2x + d [d = c-/4]
Question 46. ∫((xtan-1x))/(1 + x2)3/2 dx
Solution:
Given that, I = ∫((xtan-1x))/(1 + x2)3/2dx
Let us considered tan-1x = t
1/(1 + x2) dx = dt
I = ∫(t tant)/√(1 + tan2t) dt
= ∫(t × tant)/(sect) dt
= ∫t (sint)/(cost) costdt
= ∫tsintdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = [t]sintdt – ∫(1)sintdt)dt]
= [-tcost + ∫costdt]
= [-tcost + sint] + c
= -(tan-1x)/√(1 + x2) + x/√(1 + x2) + c
Hence, I = -(tan-1x)/√(1 + x2) + x/√(1 + x2) + c
Question 47. ∫ tan-1(√x)dx
Solution:
Given that, I = ∫ tan-1(√x)dx
Let us considered x = t2
dx = 2tdt
I = ∫2ttan-1tdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= 2[tan-1)t∫tdt – ∫(1/(1 + t2) ∫tdt)dt]
= 2[t2/2 tan-1t – ∫t2/2(1 + t2)dt]
= t2 tan-1t – ∫(t2 + 1 – 1)/(1 + t2)dt
= t2 tan-1t – ∫(1 – 1/(1 + t2))dt
= t2 tan-1t – t + tan-1t + c
= (t2 + 1) tan-1t – t + c
Hence, I = (x + 1)tan-1√x – √x + c
Question 48. ∫x3 tan-1xdx
Solution:
Given that, I = ∫x3 tan-1xdx
= tan-1x∫x3dx – (∫(dtan-1x)/dx (∫x3 dx)dx)
= tan-1x x4/4 – (∫1/(1 + x2) (x4/4)dx)
= tan-1x x4/4 – (∫1/(1 + x2) (x4/4)dx)
= tan-1x x4/4 – (∫1/(1 + x2) (x4/4)dx)
∫ 1/(1 + x2) (x4/4)dx = 1/4 [∫1/(1 + x2) dx + (x2 – 1)dx]
∫ 1/(1 + x2) (x4/4)dx = 1/4 [tan-1x + x3/3 – x]
Hence, I = x4/4 tan-1x – 1/4 [tan-1x + x3/3 – x] + c
Question 49. ∫xsinxcos2xdx
Solution:
Given that, I = ∫xsinxcos2xdx
= 1/2 ∫x(2sinxcos2x)dx
= 1/2 ∫x(sin(x + 2x) – sin(2x – x))dx
= 1/2 ∫x(sin3x – sinx)dx
= 1/2[x](sin3x – sinx)dx – ∫ (1)(sin3x – sinx)dx)dx]
= 1/2 [x((-cos3x)/3 + cosx) – ∫(-(cos3x)/3 + cosx)dx]
Hence, I = 1/2 [-x (cos3x)/3 + xcosx + 1/9 sin3x – sinx] + c
Question 50. ∫(tan-1x2)xdx
Solution:
Given that, I = ∫(tan-1x2)xdx
Let us considered x2 = t
2xdx = dt
I = 1/2∫tan-1tdt
= 1/2∫1tan-1tdt
= 1/2 [tan-1t∫dt – (∫1/(1 + t2)∫dt)dt]
= 1/2 [t × tan-1t – ∫t/(1 + t2) dt]
= 1/2 t × tan-1t – 1/4∫2t/(1 + t2) dt
= 1/2 t × tan-1t – 1/4 log|1 + t2| + c
Hence, I = 1/2 x2 tan-1x2 – 1/4 log|1 + x4| + c
Question 51. ∫xdx/√(1 – x2)
Solution:
Given that, I = ∫xdx/√(1 – x2)
Let first function be sin-1x and second function be x/√(1 – x2).
Now, first we find the integral of the second function,
∫xdx/√(1 – x2)
Now, put t = 1 – x2
Then dt = -2xdx
Therefore,
∫ xdx/√(1 – x2) = -1/2 ∫dt/√t = -√t = -√(1 – x2)
Hence,
∫(xsin-1x)/√(1 – x2) dx
= (sin-1x)(-√(1 – x2) – ∫1/√(1 – x2) * (-√(1 – x2))dx
= -√(1 – x2) sin-1x + x + c
= x – √(1 – x2) sin-1x + c
Question 52. ∫sin3√x dx
Solution:
Given that, I = ∫sin3√x dx
Let us considered √x = t
x = t2
dx = 2tdt
I = 2∫ tsin3tdt
= 2∫t((3sint – sin3t)/4)dt
= 1/2 ∫t(3sint – sin3t)dt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = 1/2 [t(-3cost + 1/3 cos3t) – ∫(-3cost + (cos3t)/3)dt]
= 1/2 [(-9tcost + tcos3t)/3 – {-3sint + (sin3t)/9}] + c
= 1/2 [(-9tcost + tcos3t)/3 + (27sint – 3sin3t)/9] + c
= 1/18[-27tcost + 3tcos3t + 27sint – 3sin3t] + c
Hence, I = 1/18[3√x cos3√x + 27sin√x – 27√x cos√x – 3sin3√x] + c
Question 53. ∫ xsin3xdx
Solution:
Given that, I = ∫ xsin3xdx
= ∫x((3sinx – sin3x)/4)dx
= 1/4 ∫x(3sinx – sin3x)dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
= 1/4 [x∫ (3sinx – sin3x)dx – ∫(1)(3sinx – sin3x)dx)dx]
= 1/4 [x(-3cosx + (cos3x)/3) – ∫(-3cosx + (cos3x)/3)dx]
= 1/4 [-3xcosx + (xcos3x)/3 + 3sinx – (sin3x)/9] + c
Hence, I = 1/36[3xcos3x – 27xcosx + 27sinx – sin3x] + c
Question 54. ∫cos3√x dx
Solution:
Given that, I = ∫cos3√x dx
Let us considered x = t²
dx = 2tdt
= 2∫tcos3tdt
= 2∫t((3cost + cos3t)/4)dt
= 1/2 ∫t(3cost + cos3t)dt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = 1/2 [t(3sint + 1/3 sin3t) + ∫(1 × 3sint + (sin3t)/3)dt]
= 1/2 [t((9sint + sin3t)/3) + 3cost(cos3t)/9] + c
= 1/18[27tsint + 3tsin3t + 9cost + cos3t] + c
Hence, I = 1/18[27√x sin√x + 3√x sin3√x + 9cos√x + cos3√x] + c
Question 55. ∫xcos3xdx
Solution:
Given that, I = ∫xcos3xdx
= ∫x((3cosx + cos3x)/4)dx
= 1/4 ∫x(3cosx + cos3x)dx
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = 1/4 [x∫(3cosx + cos3x)dx – ∫(1)(3cosx + cos3x)dx)dx]
= 1/4 [x(3sinx + (sin3x)/3) – ∫ (3sinx + (sin3x)/3)dx]
= 1/4 [3xsinx + (xsin3x)/3 + 3cosx + (cos3x)/9] + c
Hence, I = (3xsinx)/4 + (xsin3x)/12 + (3cosx)/4 + (cos3x)/36 + c
Question 56. ∫tan-1√((1 – x)/(1 + x))
Solution:
Given that, I = ∫tan-1√((1 – x)/(1 + x))
Let us considered x = cosθ
dx = -sinθdθ
I = ∫ tan-1(tanθ/2)(-sinθ)dθ
=-1/2 ∫θsinθdθ
Let θ = u and sinθdθ = v
So that sinθ = ∫vdθ
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = -1/2 (-θcosθ – ∫-cosθdθ)
= -1/2(-θcosθ + sinθ)+c
= -1/2 (-θcosθ + √(1 – cos2θ)) + c
= -1/2 (-xcos-1x + √(1 – x2)) + c
Question 57. ∫sin-1√(x/(a + x)) dx
Solution:
Given that, I = ∫sin-1√(x/(a + x)) dx
Let us considered x = atan2θ
dx = 2atanθsec2θdθ
I = ∫(sin-1√((atan2θ)/(a + atan2θ))(2atanθsec2θ)dθ
= ∫ (sin-1√((tan2θ)/(sec2θ)))(2atanθsec2θ)dθ
= ∫ sin-1(sinθ)(2atanθsec2θ)dθ
= ∫ 2θatanθsec2θdθ
= 2a∣θ(tanθsec2θ)dθ)
= ∫2θatanθsec2θdθ
= 2a∫θ(tanθsec2θ)dθ
= 2a[θ]tanθsec2θdθ – ∫(∫tanθsec2θdθ)dθ]
= 2a[θ (tan2θ)/2 – ∫(tan2θ)/2 dθ]
= aθtan2θ – 2a/2∫(sec2θ – 1)dθ
= aθtan2θ – atanθ + aθ + c
= a(tan-1√(x/a)) x/a – a√(x/a) + atan-1√(x/a) + c
Hence, I = xtan-1√(x/a) – √ax + atan-1√(x/a) + c
Question 58. ∫(x3 sin-1x²)/√(1 – x4) dx
Solution:
Given that, I = ∫(x3 sin-1x²)/√(1 – x4) dx
Let us considered sin-1x² = t
(1/√(1 – x4)(2x)dx = dt
I = ∫(x² sin-1x²)/√(1 – x4) xdx
= ∫(sint)t dt/2
= 1/2∫tsintdt
= 1/2 [t∫sintdt – ∫(1∫sintdt)dt]
= 1/2 [t(-cost)dt – ∫(1∫(-cost))dt]
= 1/2[-tcost + sint] + c
Hence, I = 1/2 [x2 – √(1 – x4) sin(-1)x2] + c
Question 59. ∫(x2 sin-1x)/(1 – x2)3/2 dx
Solution:
Given that, I = ∫(x2 sin-1x)/(1 – x2)3/2dx
Let us considered sin-1x = t
(1/√(1 – x2) dx = dt
I = ∫(sin2t × t)/((1 – sin2t)) dt
= ∫(tsin2t)/(cos2t) dt
= ∫t × tan2tdt
= ∫t(sec2t – 1)dt
= ∫tsec2tdt – t2/2 + c
= t∫sec2tdt – ∫(1∫sec2tdt)dt – t2/2 + c
= t × tant – ∫tantdt – t2/2 + c
= t × tant – logsect – t2/2 + c
Hence, I = x/√(1 – x2) sin-1x + log|1 – x2| – 1/2 (sin-1x)2 + c
Question 60. ∫cos-1(1 – x2/ 1 + x2) dx
Solution:
Given that, I = ∫cos-1(1 – x2/ 1 + x2) dx
Let us considered, x = tant
dx = sec2tdt
I = ∫cos-1(1 – tan2t/ 1 + tan2t) sec2tdt
= ∫ 2t sec2tdt
Using integration by parts,
∫u v dx = v∫ u dx – ∫{d/dx(v) × ∫u dx}dx + c
We get
I = 2[t∫sec2tdt – ∫(1 ∫sec2tdt)dt]
= 2[t tan2t – ∫tant dt]
= 2[t tan2t – log sect] + c
= 2[x tan2x – log √1 + x2] + c
Hence, I = 2[xtan2x – log √1 + x2] + c
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.