RD Sharma Class 12 Ex 19.12 Solutions Chapter 19 Indefinite Integrals

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.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter19
Exercise19.12
CategoryRD 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 = ∫ tcosxdt/cosx
 I = ∫ tcosx dt

 I = ∫ t(1 – sinx) dt

 I = ∫ t(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 + ∫tdt + ∫sin2xtdt​

= ∫sinx dx + ∫tdt + ∫(1 – cos2x)tdt

= -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

= sin6​x/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 cosx= t

On differentiating both sides:

 -2xsinx= dt/dx

​ xsinxdx = -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 = ∫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)tdt

∫(1 – t2)tdt

= ∫(t– t5) dt

= t8/8 – t6/6​ + 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:

sec2x = dt/dx

sec2x 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

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