Class 12 RD Sharma Solutions Chapter 19 Indefinite Integrals Exercise 19.12 (original) (raw)
Last Updated : 3 Sep, 2024
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
= -cosx+ \frac{t^3}{3} + \int (1-t^2)t^2dt\\ = -cosx + \frac{t^3}{3} + \int(t^2-t^4)dt\\ -cosx+\frac{t^3}{3}+\frac{t^3}{3}-\frac{t^5}{5}\\ -cosx+\frac{2t^3}{3}+\frac{t^5}{5}+c Putting value of t: \\ I=-cosx+\frac{2cos^3}{3}+\frac{cos^5}{5}+c
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):
I=\int t^3\frac{-dt}{2}\\
= -t4/8 + c
Putting value of t:
I=-\frac{1}{8}cos^4x^2+c
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)
=\ -cosx+\frac{t^7}{7}-\frac{3}{5}t^5+\frac{3}{3}t^3+c\\ =\ -cosx+\frac{cos^x}{7}-\frac{3}{5}cos^5x+cos^3x+c\\ \implies -cosx+cos^3x-\frac{3}{5}cos^5x+\frac{1}{7}cos^7x+c
Question 9.∫sin3xcos5x dx
**Solution:
_Let I = ∫sin 3 xcos 5 x dx −(i)
_Let cosx = t
_On differentiating both sides: -sinx = dt/dx
_sinx dx = -dt
_Putting values in (i):
_I = ∫sin 2 xt 5 (-dt)
_= _− ∫(1 - cos 2 x)t 5 _dt
_= _− ∫(1 - t 2 )t 5 _dt
_= ∫(t 7 _- t 5 ) dt
_= t 8 /8 - t 6 /6 + c
_Putting value of t:
\implies \frac{1}{8}cos^8x-\frac{1}{6}cos^6x+c
Question 10. \int \frac{1}{sin^4xcos^2x}dx
**Solution:
Let I = \int \frac{1}{sin^4xcos^2x}dx\quad -(i)
Dividing and multiplying the equation by cos6x:
I=\int \frac{\frac{1}{cos^6x}}{\frac{sin^4xcos^2x}{cos^6x}}\\ =\ \int \frac{sec^6x}{tan^4x}dx\\ =\ \int \frac{sec^4xsec^2x}{tan^4x}dx\\ =\ \int \frac{(sec^2x)^2sec^2x}{sec^2x}dx\\ =\ \int \frac{(1+tan^2x)^2sec^2x}{tan^4x}dx\\ I=\int \frac{(1+tan^4x+2tan^2x)sec^2x}{tan^4x}dx\quad -(ii)\\
_Let tanx = t, then:
_sec 2 x = dt/dx
_sec 2 x dx = dt
_Putting values in eq (ii):
\\ I=\int \frac{1+t^4+2t^2}{t^4}dt\\ =\ \int (\frac{1}{t^4}+1+\frac{2}{t^2})dt\\ =\ -\frac{1}{3t^3}+t-\frac{2}{t}+c\\ =\ -\frac{1}{3tan^3x}+tanx-\frac{2}{tanx}+c\\ \implies -\frac{1}{3}cot^3x-2cotx+tanx+c
Question 11.\int \frac{1}{sin^3xcos^5x}dx
**Solution:
Let\ I=\int \frac{1}{sin^3xcos^5x}dx Dividing and multiplying by cos8x: \\ =\ \int \frac{\frac{1}{cos^8x}}{\frac{sin^3xcos^5x}{cos^8x}}dx\\ =\ \int \frac{sec^8x}{tan^3x}dx\\ =\ \int \frac{(sec^2x)^3}{tan^3x}sec^2xdx\\ =\ \int \frac{(1+tan^2x)^3}{tan^3x}sec^2xdx\\ =\ \int \frac{(1+tan^6x+3tan^2x+3tan^4x)sec^2x}{tan^3x}dx Let tanx=t,then: \\ sec^2x=\frac{dt}{dx}\\ \implies sec^2xdx=dt Putting values in ii: \\ I=\int \frac{1+t^6+3t^4+3t^2}{t^3}dt\\ =\ \int (\frac{1}{t^3}+t^3+3t+\frac{3}{t})dt\\ =\ \frac{1}{2t^2}+\frac{t^4}{4}+\frac{3t^2}{2}+3logt+c\\ \implies I=\frac{-1}{2tan^2x}+3log|tanx|+\frac{3}{2}tan^2x+\frac{1}{4}tan^4x+c
Question 12.\int \frac{1}{sin^3xcosx}dx
**Solution:
Let\ I=\int \frac{1}{sin^3xcosx}dx Dividing and multiplying by cos4x: \\ I=\int \frac{\frac{1}{cos^4x}}{\frac{sin^3xcosx}{cos^4x}}dx\\ I=\int \frac{sec^4x}{tan^3x}dx\\ I=\int \frac{sec^2xsec^2x}{tan^3x}dx\\ =\ \int \frac{1+tan^2x}{tan^3x}dx \quad -i Let tanx=t,then: sec2xdx = dt Putting values in i: \\ I=\int \frac{1+t^2}{t^3}dt\\ I=\int (\frac{1}{t^3}+\frac{1}{t})dt\\ =\ -\frac{1}{2t^2}+log|t|+c Putting value of t: \\ \implies -\frac{1}{2tan^2x}+log|tanx|+c
Question 13.\int \frac{1}{sinxcos^3x}dx
**Solution:
Let\ I=\int \frac{1}{sinxcos^3x}dx\\ \frac{1}{sinxcos^3x}= \frac{sin^2x+cos^2x}{sinxcos^3x}\\ =\ \frac{sinx}{cos^2x}+\frac{1}{sinxcosx}\\ =\ tanxsec^2x+\frac{\frac{1}{cos^2x}}{\frac{sinxcosx}{cos^2x}}\\ =\ tanxsec^2x+\frac{sec^2x}{tanx} \implies \int \frac{1}{sinxcos^3x}dx= \int tanxsec^2xdx +\int \frac{sec^2x}{tanx}dx Let tanx=t⟹sec2x dx = dt: I= \int tdt+\int \frac{1}{t}dt\\ =\ \frac{t^2}{2}+log|t|+c Putting value of t: \implies I=\frac{1}{2}tan^2x+log|tanx|+c
Practice Questions
1. ∫ (2x + 1) / (x² - 1) dx
2. ∫ (x² + 1) / (x - 2) dx
3. ∫ (x³ + 2x² - x + 3) / (x² + 1) dx
4. ∫ (x⁴ + 1) / (x² + x) dx
5. ∫ (2x³ - 3x² + 4x - 1) / (x² - x + 1) dx
6. ∫ (x² + 2x + 3) / (x³ - 1) dx
7. ∫ (x³ - 2x² + x - 3) / (x² + 2x + 1) dx
8. ∫ (2x⁴ + 3x³ - x² + 2x - 1) / (x³ + x) dx
9. ∫ (x⁵ + 1) / (x³ - x) dx
10. ∫ (x⁴ - 2x³ + 3x² - 2x + 1) / (x³ + x² - x - 1) dx
Summary
Exercise 19.12 in RD Sharma's Class 12 Chapter on Indefinite Integrals focuses on integrating rational functions where the degree of the numerator is greater than or equal to the degree of the denominator. The key techniques used in this exercise include:
1. Long division of polynomials
2. Partial fraction decomposition
3. Integration of simpler rational functions
4. Recognition of standard integral forms
The problems typically involve dividing the numerator by the denominator, then integrating the resulting polynomial and the remaining fraction separately.