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Class 12 RD Sharma Solutions- Chapter 19 Indefinite Integrals – Exercise 19.12

Last Updated : 03 May, 2022
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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 dt

 I = ∫ 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

= 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 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 = ∫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)t5 dt

= ∫(1 – t2)t5 dt

= ∫(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



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