Class 12 RD Sharma Solutions - Chapter 19 Indefinite Integrals - Exercise 19.25...
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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
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
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