 |
VOOZH | about |
|
VOOZH | about |
In this article, we will delve into the solutions for Exercise 19.25 | Set 2 from Chapter 19 of the RD Sharma Class 12 Mathematics textbook. This chapter focuses on indefinite integrals a crucial concept in calculus that forms the foundation for understanding more advanced topics in the integration. These solutions aim to provide students with a comprehensive understanding of the problems and methodologies required to solve them effectively helping to solidify their knowledge and prepare for board exams.
The Indefinite integrals also known as antiderivatives are functions that reverse the process of differentiation. They are used to determine the original function from its derivative. The general form of an indefinite integral is expressed as ∫f(x)dx where f(x) is the integrand, and the result includes a constant of the integration denoted as C. Indefinite integrals play a vital role in calculus especially in finding the areas under curves and solving differential equations.
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
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
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
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
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
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
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
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
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
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
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
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
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
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)
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
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
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
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
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
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
Read More:
Exercise 19.25 | Set 2 typically deals with integrating rational functions where the denominator is of the form x⁴ - 1. The key points to remember are:
