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.

Textbook	NCERT
Class	Class 12th
Subject	Maths
Chapter	19
Exercise	19.12
Category	RD Sharma Solutions

Table of Contents

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 = ∫ t⁴cos^xdt/cosx
I = ∫ t⁴cos²x dt
I = ∫ t⁴(1 – sin²x) 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 + ∫t²dt + ∫sin²xt²dt
= ∫sinx dx + ∫t²dt + ∫(1 – cos²x)t²dt
$= -cosx+ \frac{t^3}{3} + \int (1-t^2)t^2dt\\ = -cosx + \frac{t^3}{3} + \int(t^2-t^4)dt\\ -cosx+\frac{t^3}{3}+\frac{t^3}{3}-\frac{t^5}{5}\\ -cosx+\frac{2t^3}{3}+\frac{t^5}{5}+c$ 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

xsinx²dx = -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)t⁵dt

= ⁻∫(1 – t²)t⁵dt

= ∫(t⁷– 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: $\implies I=\frac{1}{2}tan^2x+log|tanx|+c$

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.

RD Sharma Class 12 Ex 19.12 Solutions Chapter 19 Indefinite Integrals

Question 1. ∫sin4x cos3x dx

Question 2. ∫ sin5x dx

Question 3. ∫cos5x dx

Question 4. ∫sin5xcosx dx

Question 5. ∫sin3xcos6x dx

Question 6. ∫cos7x dx

Question 7. ∫xcos3x2sinx2dx

Question 8. ∫sin7x dx

Question 9. ∫sin3xcos5x dx

Question 10.

Question 11.

Question 12.

Question 13.

Related Posts

Leave a Comment Cancel Reply

Question 1. ∫sin⁴x cos³x dx

Question 2. ∫ sin⁵x dx

Question 3. ∫cos⁵x dx

Question 4. ∫sin⁵xcosx dx

Question 5. ∫sin³xcos⁶x dx

Question 6. ∫cos⁷x dx

Question 7. ∫xcos³x²sinx²dx

Question 8. ∫sin⁷x dx

Question 9. ∫sin³xcos⁵x dx

Question 10. $\int \frac{1}{sin^4xcos^2x}dx$

Question 11. $\int \frac{1}{sin^3xcos^5x}dx$

Question 12. $\int \frac{1}{sin^3xcosx}dx$

Question 13. $\int \frac{1}{sinxcos^3x}dx$