NCERT Solutions Class 12 Maths (integrals) Miscellaneous Exercise
NCERT Solutions Class 12 Maths from class
12th Students will get the answers of
Chapter-7(integrals) Miscellaneous Exercise This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of
NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Q1. Integrate the function 1x−x3
Answer. 1x−x3=1x(1−x2)=1x(1−x)(1+x) Let 1x(1−x)(1+x)=Ax+B(1−x)+C1+x⇒1=A(1−x2)+Bx(1+x)+Cx(1−x)⇒1=A−Ax2+Bx+Bx2+Cx−Cx2 Equating the coefficients of x2 , x, and constant term, we obtain −A+B−C=0B+C=0A=1 On solving these equations, we obtain A=1,B=12, and C=−12 From equation (1), we obtain 1x(1−x)(1+x)=1x+12(1−x)−12(1+x) 1x(1−x)(1+x)=1x+12(1−x)−12(1+x)⇒∫1x(1−x)(1+x)dx=∫1xdx+12∫11−xdx−12∫11+xdx =log|x|−12log|(1−x)|−12log|(1+x)|=log|x|−log∣∣(1−x)12∣∣−log∣∣(1+x)12∣∣=log∣∣
∣∣x(1−x)12(1+x)12∣∣
∣∣+C=log∣∣∣x21−x2∣∣∣+C=12log∣∣∣x21−x2∣∣∣+C
Q2. Integrate the function 1√x+a+√(x+b)
Answer. 1√x+a+√x+b=1√x+a+√x+b×√x+a−√x+b√x+a−√x+b=√x+a−√x+b(x+a)−(x+b)=(√x+a−√x+b)a−b ⇒∫1√x+a−√x+bdx=1a−b∫(√x+a−√x+b)dx=1(a−b)[(x+a)3232−(x+b)3232]=23(a−b)[(x+a)32−(x+b)32]+C
Q3. Integrate the function 1x√ax−x2[ Hint: Put x=at]
Answer. 1x√ax−x2 Let x=at⇒dx=−at2dt ⇒∫1x√ax−x2dx=∫1at√a⋅at−(at)2(−at2dt)=−∫1at⋅1√1t−1t2dt =−1a∫1√t2t−t2t2dt=−1a∫1√t−1dt=−1a[2√t−1]+C=−1a[2√ax−1]+C =−2a(√a−x√x)+C=−2a(√a−xx)+C
Q4. Integrate the function 1x2(x4+1)34
Answer. 1x2(x4+1)34 Multiplying and dividing by x−3 , we obtain x−3x2⋅x−3(x4+1)34=x−3(x4+1)−34x2⋅x−3 =(x4+1)−34x5⋅(x4)34=1x5(x4+1x4)34=1x5(1+1x4)−34 Let 1x4=t⇒−4x5dx=dt⇒1x5dx=−dt4∴∫1x2(x4+1)34dx=∫1x5(1+1x4)−34dx =−14∫(1+t)34dt=−14[(1+t)1414]+C =−14(1+1x4)1414+C=−(1+1x4)14+C
Q5. Integrate the function 1x12+x13⎡⎢⎣ Hint: 1x12+x13=1x13(1+x16), put x=t6⎤⎥⎦
Answer. 1x12+x13=1x13(1+x16) Let x=t6⇒dx=6t5dt ∴∫1x12+x13dx=∫1x13(1+x16)dx=∫6t5t2(1+t)dt=6∫t3(1+t)dt On dividing, we obtain ∫1x12+x13dx=6∫{(t2−t+1)−11+t}dt =6[(t33)−(t22)+t−log|1+t|]=2x12−3x13+6x16−6log(1+x16)+C=2√x−3x13+6x16−6log(1+x16)+C
Q6. Integrate the function 5x(x+1)(x2+9)
Answer. Let 5x(x+1)(x2+9)=A(x+1)+Bx+C(x2+9)….(1)⇒5x=A(x2+9)+(Bx+C)(x+1)⇒5x=Ax2+9A+Bx2+Bx+Cx+C Equating the coefficients of x2,x, and constant term, we obtain A+B=0B+C=59A+C=0 On solving these equations, we obtain A=−12,B=12, and C=92 From equation (1), we obtain 5x(x+1)(x2+9)=−12(x+1)+x2+92(x2+9) ∫5x(x+1)(x2+9)dx=∫{−12(x+1)+(x+9)2(x2+9)}dx=−12log|x+1|+12∫xx2+9dx+92∫1x2+9dx =−12log|x+1|+14∫2xx2+9dx+92∫1x2+9dx=−12log|x+1|+14log∣∣x2+9∣∣+92⋅13tan−1x3=−12log|x+1|+14log(x2+9)+32tan−1x3+C
Q7. Integrate the function sinxsin(x−a)
Answer. sinxsin(x−a)Letx−a=tdx=dt∫sinxsin(x−a)dx=∫sin(t+a)sintdt =∫sintcosa+costsinasintdt=∫(cosa+cottsina)dt =tcosa+sinalog|sint|+C1=(x−a)cosa+sinalog|sin(x−a)|+C1=xcosa+sinalog|sin(x−a)|−acosa+C1=sinalog|sin(x−a)|+xcosa+C
Q8. Integrate the function : e5logx−e4logxe3logx−e2logx
Answer. e5logx−e4logxe3logx−e2logx=e4logx(elogx−1)e2logx(elogx−1)=e2logx=elogx2=x2 ∴∫e5logx−e4logxe3logx−e2logxdx=∫x2dx=x33+C
Q9. Integrate the function cosx√4−sin2x
Answer. cosx√4−sin2x Let sinx=tcosxdx=dt ⇒∫cosx√4−sin2xxdx=∫dt√(2)2−(t)2 =sin−1(t2)+C=sin−1(sinx2)+C
Q10. Integrate the function : sin8−cos8x1−2sin2xcos2x
Answer. sin8x−cos8x1−2sin2xcos2x=(sin4x+cos4x)(sin4x−cos4x)sin2x+cos2x−sin2xcos2x−sin2xcos2x =(sin4x+cos4x)(sin2x+cos2x)(sin2x−cos2x)(sin2x−sin2xcos2x)+(cos2x−sin2xcos2x)=(sin4x+cos4x)(sin2x−cos2x)sin2x(1−cos2x)+cos2x(1−sin2x) =−(sin4x+cos4x)(cos2x−sin2x)(sin4x+cos4x)=−cos2x ∴∫sin8x−cos8x1−2sin2xcos2xdx=∫−cos2xdx=−sin2x2+C
Q11. Integrate the function 1cos(x+a)cos(x+b)
Answer. 1cos(x+a)cos(x+b) Multiplying and dividing by sin(a−b), we obtain 1sin(a−b)[sin(a−b)cos(x+a)cos(x+b)]=1sin(a−b)[sin[(x+a)−(x+b)]cos(x+a)cos(x+b)] =1sin(a−b)[sin(x+a)⋅cos(x+b)−cos(x+a)sin(x+b)cos(x+a)cos(x+b)]=1sin(a−b)[sin(x+a)cos(x+a)−sin(x+b)cos(x+b)] =1sin(a−b)[tan(x+a)−tan(x+b)]∫1cos(x+a)cos(x+b)dx=1sin(a−b)∫[tan(x+a)−tan(x+b)]dx =1sin(a−b)[−log|cos(x+a)|+log|cos(x+b)|]+C=1sin(a−b)log∣∣∣cos(x+b)cos(x+a)∣∣∣+C
Q12. Integrate the function x3√1−x8
Answer. x3√1−x8 Let x4=t4x3dx=dt ⇒∫x3√1−x8dx=14∫dt√1−t2=14sin−1t+C=14sin−1(x4)+C
Q13. Integrate the function ex(1+ex)(2+ex)
Answer. ex(1+ex)(2+ex) Let ex=texdx=dt⇒∫ex(1+ex)(2+ex)dx=∫dt(t+1)(t+2) =∫1(t+1)−1(t+2)]dt=log|t+1|−log|t+2|+C=log∣∣t+1t+2∣∣+C=log∣∣1+ex2+ex∣∣+C
Q14. Integrate the function 1(x2+1)(x2+4)
Answer. ∴1(x2+1)(x2+4)=Ax+B(x2+1)+Cx+D(x2+4)⇒1=(Ax+B)(x2+4)+(Cx+D)(x2+1)⇒1=Ax3+4Ax+Bx2+4B+Cx3+Cx+Dx2+D Equating the coefficients of x3,x2,x, and constant term, we obtain A+C=0B+D=04A+C=04B+D=1 On solving these equations, we obtain A=0,B=13,C=0, and D=−13 From equation (1), we obtain 1(x2+1)(x2+4)=13(x2+1)−13(x2+4)∫1(x2+1)(x2+4)dx=13∫1x2+1dx−13∫1x2+4dx =13tan−1x−13⋅12tan−1x2+C=13tan−1x−16tan−1x2+C
Q15. Integrate the function cos3xelogsinx
Answer. cos3xelogsinx=cos3x×sinx Let cosx=t−sinxdx=dt ⇒∫cos3xelogsinxdx=∫cos3xsinxdx=−∫t⋅dt=−t44+C=−cos4x4+C
Q16. Integrate the function e3logx(x4+1)−1
Answer. e3logx(x4+1)−1=elogx3(x4+1)−1=x3(x4+1) Let x4+1=t⇒4x3dx=dt⇒∫e3logx(x4+1)−1dx=∫x3(x4+1)dx =14∫dtt=14log|t|+C=14log∣∣x4+1∣∣+C=14log(x4+1)+C
Q17. Integrate the function : f′(ax+b)[f(ax+b)]n
Answer. f′(ax+b)[f(ax+b)]n Let f(ax+b)=t⇒af′(ax+b)dx=dt⇒∫f′(ax+b)[f(ax+b)]ndx=1a∫tndt =1a[tn+1n+1]=1a(n+1)(f(ax+b))n+1+C
Q18. Integrate the function 1√sin3xsin(x+α)
Answer. 1√sin3xsin(x+α)=1√sin3x(sinxcosα+cosxsinα)=1√sin4xcosα+sin3xcosxsinα=1sin2x√cosα+cotxsinα=csc2x√cosα+cotxsinα Let cosα+cotxsinα=t⇒−csc2xsinαdx=dt∴∫1sin3xsin(x+α)dx=∫csc2x√cosα+cotxsinααdx=−1sinα∫dt√t=−1sinα[2√t]+C =−1sinα[2√cosα+cotxsinα]+C=−2sinα√cosα+cosxsinαsinx+C=−2sinα√sinxcosα+cosxsinαsinx+C=−2sinα√sin(x+α)sinx+C
Q19. Integrate the function sin−1√x−cos−1√xsin−1√x+cos−1√x,x∈[0,1]
Answer. Let I=∫sin−1√x−cos−1√xsin−1√x+cos−1√xdx It is known that, sin−1√x+cos−1√x=π2 ⇒I=∫(π2−cos−1√x)−cos−1√xπ2dx=2ππ](−2cos−1√x2√x)dx =2ππ⋅2∫1⋅dx−4π∫cos−1√xdx =x−4π∫cos−1√xdx....(1) Let I1=∫cos−1√xdx Also, lct√x=t⇒dx=2tdt⇒I1=2∫cos−1t⋅tdt=2[cos−1t⋅t22−∫−1√1−t2⋅t22dt] =t2cos−1t+∫t2√1−t2dt=t2cos−1t−∫1−t2−1√1−t2dt=t2cos−1t−∫√1−t2dt+∫1√1−t2dt =t2cos−1t−t2√1−t2−12sin−1t+sin−1t=t2cos−1t−t2√1−t2+12sin−1t From equation (1), we obtain I=x−4π[t2cost−t2√1−t2+12sin−1t]=x−4π[xcos−1√x−√x2√1−x+12sin−1√x]=x−4π[x(−2−sin−1√x)−√x−λ22+2sin−1√x] =x−2x+4xπsin−1√x+2π√x−x2−2πsin−1√x=−x+2π[(2x−1)sin−1√x]+2π√x−x2+C=2(2x−1)πsin−1√x+2π√x−x2−x+C
Q20. Integrate the function √1−√x1+√x
Answer. I=√1−√x1+√xdx Let x=cos2θ⇒dx=−2sinθcosθdθI=∫√1−cosθ1+cosθ(−2sinθcosθ)dθ =−∫√2sin2θ22cos2θ2sin2θdθ =−∫tanθ2⋅2sinθcosθdθ=−2∫sinθ2cosθ2(2sinθ2cosθ2)cosθdθ =−4∫sin2θ2cosθdθ=−4∫sin2θ2⋅(2cos2θ2−1)dθ=−4∫(2sin2θ2cos2θ2−sin2θ2)dθ =−8∫sin2θ2⋅cos2θ2dθ+4∫sin2θ2dθ=−2∫sin2θdθ+4∫sin2θ2dθ=−2∫1−cos2θ2)dθ+4∫1−cosθ2dθ =−2[θ2−sin2θ4]+4[θ2−sinθ2]+C=−θ+sin2θ2+2θ−2sinθ+C=θ+sin2θ2−2sinθ+C =θ+√1−cos2θ⋅cosθ−2√1−cos2θ+C=cos−1√x+√1−x⋅√x−2√1−x+C=−2√1−x+cos−1√x+√x(1−x)+C=−2√1−x+cos−1√x+√x−x2+C
Q21. Integrate the function 2+sin2x1+cos2xex
Answer. I=∫(2+sin2x1+cos2x)ex=∫(2+2sinxcosx2cos2x)ex=∫(1+sinxcosxcos2x)ex=∫(sec2x+tanx)ex Let f(x)=tanx⇒f′(x)=sec2x∴I=∫(f(x)+f′(x)]exdx=exf(x)+C=extanx+C
Q22. Integrate the function x2+x+1(x+1)2(x+2)
Answer. Let x2+x+1(x+1)2(x+2)=A(x+1)+B(x+1)2+C(x+2)...(1)⇒x2+x+1=A(x+1)(x+2)+B(x+2)+C(x2+2x+1)⇒x2+x+1=A(x2+3x+2)+B(x+2)+C(x2+2x+1)⇒x2+x+1=(A+C)x2+(3A+B+2C)x+(2A+2B+C) Equating the coefficients of x2,x, and constant term, we obtain A+C=13A+B+2C=12A+2B+C=1 On solving these equations, we obtain A=−2,B=1, and C=3 From equation (1), we obtain x2+x+1(x+1)2(x+2)=−2(x+1)+3(x+2)+1(x+1)2 ∫x2+x+1(x+1)2(x+2)dx=−2∫1x+1dx+3∫1(x+2)dx+∫1(x+1)2dx=−2log|x+1|+3log|x+2|−1(x+1)+C
Q23. Integrate the function tan−1√1−x1+x
Answer. I=tan−1√1−x1+xdx Let x=cosθ⇒dx=−sinθdθI=∫tan−1√1−cosθ1+cosθ(−sinθdθ) =−∫tan−1√2sin2θ22cos2θ2sinθdθ =−∫tan−1tanθ2⋅sinθdθ=−12∫θ⋅sinθdθ=−12[θ⋅(−cosθ)−∫1⋅(−cosθ)dθ] =+12θcosθ−12sinθ=12cos−1x⋅x−12√1−x2+C=x2cos−1x−12√1−x2+C =12(xcos−1x−√1−x2)+C
Q24. Integrate the function √x2+1[log(x2+1)−2logx]x4
Answer. √x2+1[log(x2+1)−2logx]x4=√x2+1x4[log(x2+1)−logx2]=√x2+1x4[log(x2+1x2)] =√x2+1x4log(1+1x2)=1x3√x2+1x2log(1+1x2)=1x3√1+1x2log(1+1x2) Let 1+1x2=t⇒−2x3dx=dt∴I=∫1x3√1+1x2log(1+1x2)dx=−12∫√tlogtdt=−12∫t12⋅logtdt Integrating by parts, we obtain I=−12[logt⋅∫t12dt−{(ddtlogt)∫t12dt}dt]=−12[logt⋅t3232−∫1t⋅t322dt]=−12[23t32logt−23∫t12dt] =−12[23t32logt−49t32]=−13t32logt+29t32=−13t32[logt−23]=−13(1+1x2)32[log(1+1x2)−23]+C
Q25. Evaluate the definite integral ∫ππ2ex(1−sinx1−cosx)dx
Answer. I=∫ππ2ex(1−sinx1−cosx)dx=∫ππ2ex(1−2sinx2cosx22sin2x2)dx =∫πx2ex(csc2x22−cotx2)dx Let f(x)=−cotx2⇒f′(x)=−(−12csc2x2)=12csc2x2∴I=∫π2πex(f(x)+f′(x)]dx =[ex⋅f(x)dx]ππ2=−[ex⋅cotx2]xπ2=−[ex×cotπ2−eπ2×cotπ4] =−[eπ×0−eπ2×1]=eπ2
Q26. Evaluate the definite integral ∫π40sinxcosxcos4x+sin4xdx
Answer. I=∫π40sinxcosxcos4x+sin4xdx ⇒I=∫π40(sinxcosx)cos4x(cos4x+sin4x)cos4xdx ⇒I=∫x40tanxsec2x1+tan4xdx Let tan2x=t⇒2tanxsec2xdx=dt When x=0,t=0 and when x=π4,t=1∴I=12∫dt1+t2=12[tan−1t]10=12[tan−11−tan−10]=12[π4] =π8
Q27. Evaluate the definite integral ∫π20cos2xdxcos2x+4sin2x
Answer. Let I=∫π25cos2xcos2x+4sin2xdx⇒I=∫π2cos2xcos2x+4(1−cos2x)dx⇒I=∫π2cos2xcos2x+4−4cos2xdx ⇒I=−13∫x204−3cos2x−44−3cos2xdx ⇒I=−13∫π204−3cos2x4−3cos2xdx+13∫π2044−3cos2xdx ⇒I=−π6+23∫π02sec2x1+4tan2xdx....(1) Consider, ∫π02sec2x1+4tan2xdx Let 2tanx=t⇒2sec2xdx=dt When x=0,t=0 and when x=π2,t=∞⇒∫x202sec2x1+4tan2xdx=∫∞0dt1+t2 =[tan−1t]∞0=[tan−1(∞)−tan−1(0)]=π2 Therefore, from (1), we obtain I=−π6+23[π2]=π3−π6=π6
Q28. Evaluate the definite integral ∫π3−π6sinx+cosx√sin2xdx
Answer. Let I=∫π3π6sinx+cosx√sin2xdx⇒I=∫π3π6(sinx+cosx)√−(−sin2x)dxdx⇒I=∫π3πsinx+cosx√−(−1+1−2sinxcosx)dx ⇒I=∫π3π6(sinx+cosx)√1−(sin2x+cos2x−2sinxcosx)dx⇒I=∫π3π6(sinx+cosx)dx√1−(sinx−cosx)2 Let (sinx−cosx)=t⇒(sinx+cosx)dx=dt When x =π6,t=(1−√32)x=π3,t=(√3−12)I=∫√3−1−√32dt√1−t2⇒I=∫√3−1−√3−12dt√1−t2 1 As 1√1−(−t)2=1√1−t2, therefore, 1√1−t2 is an even function. It is known that if f(x) is an even function, then ∫π−af(x)dx=2∫a0f(x)dx ⇒I=2∫√3−10dt√1−t2 =[2sin−1t]√3−120 =2sin−1(√3−12)
Q29. Evaluate the definite integral ∫10dx√1+x−√x
Answer. Let I = ∫10dx√1+x−√x I=∫101(√1+x−√x)×(√1+x+√x)(√1+x+√x)dx =∫d0√1+x+√x1+x−xdx=∫40√1+xdx+∫10√xdx=[23(1+x)32]10+[23(x)32]10 =23[(2)32−1]+23[1]=23(2)32=2⋅2√23=4√23
Q30. Evaluate the definite integral ∫π40sinx+cosx9+16sin2xdx
Answer. Let I=∫π40sinx+cosx9+16sin2xdx Also, let sinx−cosx=t⇒(cosx+sinx)dx=dt When x=0,t=−1 and when x=π4,t=0 ⇒(sinx−cosx)2=t2⇒sin2x+cos2x−2sinxcosx=t2⇒1−sin2x=t2⇒sin2x=1−t2 ∴I=∫0−1dt9+16(1−t2)=∫0−1dt9+16−16t2=∫0−1dt25−16t2=∫0−1dt(5)2−(4t)2 =14[12(5)log∣∣5+4t5−4t∣∣]0−1=140[log(1)−log∣∣19∣∣]−1=140log9
Q31. Evaluate the definite integral ∫π20sin2xtan−1(sinx)dx
Answer. Let I=∫π20sin2xtan−1(sinx)dx=∫π202sinxcosxtan−1(sinx)dx Also, let sinx=t⇒cosxdx=dt When x=0,t=0 and when x=π2,t=1 ⇒I=2∫10ttan−1(t)dt....(1) Consider ∫t⋅tan−1tdt=tan−1t⋅∫tdt−∫{ddt(tan−1t)∫tdt}dt =tan−1t⋅t22−∫11+t2⋅t22dt=t2tan−1t2−12∫t2+1−11+t2dt=t2tan−1t2−12∫1dt+12∫11+t2dt =t2tan−1t2−12⋅t+12tan−1t ⇒∫10t⋅tan−1tdt=[t2⋅tan−1t2−t2+12tan−1t]10=12[π4−1+π4]=12[π2−1]=π4−12 From equation (1), we obtain I=2[π4−12]=π2−1
Q32. Evaluate the definite integral ∫π0xtanxsecx+tanxdx
Answer. Let I = ∫π0xtanxsecx+tanxdx....(1) I=∫π0{(π−x)tan(π−x)sec(π−x)+tan(π−x)}dx(∫a0f(x)dx=∫e0f(a−x)dx) ⇒I=∫π0{−(π−x)tanx−(secx+tanx)}dx⇒I=∫π0(π−x)tanxsecx+tanxdx...(2) Adding (1) and (2), we obtain 2I=∫π0πtanxsecx+tanxdx ⇒2I=π∫π0sinxcosx1cosx+sinxcosxdx ⇒2I=π∫π0sinx+1−11+sinxdx⇒2I=π∫π01.dx−π∫π011+sinxdx ⇒2I=π[x]π0−π∫π01−sinxcos2xdx⇒2I=π2−π∫π0(sec2x−tanxsecx)dx ⇒2I=π2−π[tanx−secx]x0⇒2I=π2−π[tanπ−secπ−tan0+sec0]⇒2I=π2−π[0−(−1)−0+1] ⇒2I=π2−2π⇒2I=π(π−2)⇒I=π2(π−2)
Q33. Evaluate the definite integral ∫41[|x−1|+|x−2|+|x−3|]dx
Answer. Let I=∫41[|x−1|+|x−2|+|x−3|]dx⇒I=∫41|x−1|dx+∫4|x−2|dx+∫4|x−3|dxI=I1+I2+I3...(3) where, I1=∫4|x−1|dx,I2=∫41|x−2|dx, and I3=∫41|x−3|dxI1=∫41|x−1|dx(x−1)≥0 for 1≤x≤4 ∴I1=∫41(x−1)dx⇒I1=[x2x−x]41⇒I1=[8−4−12+1]=92...(2) I2=∫41|x−2|dxx−2≥0 for 2≤x≤4 and x−2≤0 for 1≤x≤2∴I2=∫21(2−x)dx+∫12(x−2)dx ∴I2=∫21(2−x)dx+∫42(x−2)dx⇒I2=[2x−x22]21+[x22−2x]42⇒I2=[4−2−2+12]+[8−8−2+4] ⇒I2=12+2=52.....(3) I3=∫41|x−3|dxx−3≥0 for 3≤x≤4 and x−3≤0 for 1≤x≤3 ∴I3=∫31(3−x)dx+∫45(x−3)dx⇒I3=[3x−x22]31+[x22−3x]43 ⇒I3=[9−92−3+12]+[8−12−92+9]⇒I3=[6−4]+[12]=52 From equations (1),(2),(3), and (4), we obtain I=92+52+52=192
Q34. Prove the following ∫31dxx2(x+1)=23+log23
Answer. Let I=∫31dxx2(x+1) Also, let 1x2(x+1)=Ax+Bx2+Cx+1⇒1=Ax(x+1)+B(x+1)+C(x2)⇒1=Ax2+Ax+Bx+B+Cx2 Equating the coefficients of x2,x, and constant term, we obtain A+C=0A+B=0B=1 On solving these equations, we obtain A=−1,C=1, and B=1 ∴1x2(x+1)=−1x+1x2+1(x+1)⇒I=∫3{−1x+1x2+1(x+1)}dx =[−logx−1x+log(x+1)]31=[log(x+1x)−1x]31 =log(43)−13−log(21)+1=log4−log3−log2+23=log2−log3+23 =log(23)+23 Hence, the given result is proved.
Q35. Prove the following ∫10xexdx=1
Answer. Let I=∫10xexdx Integrating by parts, we obtain I=x∫10exdx−∫10{(ddx(x))∫exdx}dx =[xex]10−∫10exdx=[xex]10−[ex]10=e−e+1=1 Hence, the given result is proved.
Q36. Prove the following ∫1−1x17cos4xdx=0
Answer. Let I=∫10tan−1(2x−11+x−x2)dx⇒I=∫10tan−1(x−(1−x)1+x(1−x))dx ⇒I=∫10[tan−1x−tan−1(1−x)]dx...(1)⇒I=∫10[tan−1(1−x)−tan−1(1−1+x)]dx ⇒I=∫10[tan−1(1−x)−tan−1(x)]dx⇒I=∫10[tan−1(1−x)−tan−1(x)]dx...(2) Adding (1) and (2), we obtain 2I=∫10(tan−1x+tan−1(1−x)−tan−1(1−x)−tan−1x)dx ⇒2I=0⇒I=0 Hence, the correct Answer is B Let I=∫1−1x17cos4xdx Also, letf(x)=x17cos4x⇒f(−x)=(−x)17cos4(−x)=−x17cos4x=−f(x) Therefore, f(x) is an odd function. It is known that if f(x) is an odd function, then ∫a−af(x)dx=0∴I=∫11x17cos4xdx=0 Hence, the given result is proved.
Q37. Prove the following ∫π20sin3xdx=23
Answer. Let I=∫π20sin3xdxI=∫π20sin2x⋅sinxdx=∫π20(1−cos2x)sinxdx =∫π20sinxdx−∫π20cos2x⋅sinxdx=[−cosx]π20+[cos3x3]π20=1+13[−1]=1−13=23 Hence, the given result is proved.
Q38. Prove the following ∫π402tan3xdx=1−log2
Answer. Let I=∫π402tan3xdxI=2∫π40tan2xtanxdx=2∫π0(sec2x−1)tanxdx=2∫π40sec2xtanxdx−2∫π0tanxdx =2[tan2x2]π40+2[logcosx]π40=1+2[logcosπ4−logcos0]π40=1+2[log1√2−log1]=1−log2−log1=1−log2 Hence, the given result is proved.
Q39. Prove the following ∫10sin−1xdx=π2−1
Answer. Let I=∫10sin−1xdx⇒I=∫10sin−1x⋅1⋅dx Integrating by parts, we obtain I=[sin−1x⋅x]10−∫101√1−x2⋅xdx=[xsin−1x]10+12∫1√1−x2(−2x)√1−x2dx Let 1−x2=t−2xdx=dt When x=0,t=1 and when x=1,t=0I=[xsin−1x]10+12∫dt√t=[xsin−1x]10+12[2√t]01=sin−1(1)+[−√1]=π2−1 Hence, the given result is proved.
Q40. Evaluate ∫10e2−3xdx as a limit of a sum.
Answer. Let I=∫10e2−3xdx It is known that, ∫∞0f(x)dx=(b−a)limn→∞1n[f(a)+f(a+h)+…+f(a+(n−1)h)] Where, h=b−an Here, a=0,b=1, and f(x)=e2−3x⇒h=1−0n=1n ∴∫10e2−3xdx=(1−0)limn→∞1n[f(0)+f(0+h)+…+f(0+(n−1)h)]=limn→∞1n[e2+e2−3h+…e2−3(n−1)h] =limn→∞1n[e2{1+e−3h+e−6h+e−9h+…e−3(n−1)k}]=limh→∞1n[e2{1−(e−3h)n1−(e−3h)}] =limn→∞1n[e2{1−e−3n1−e−3n}]=limn→∞1n[e2(1−e−3)1−e3n]=e2(e−3−1)limn→∞1n[1e−33−1] =e2(e−3−1)limn→∞(−13)[−3ne3n−1]=−e2(e−3−1)3limn→∞[−3ne−3n] =−e2(e−3−1)3(1)[limn→∞xex−1]=−e−1+e23=13(e2−1e)
Q41. Choose the correct answer ∫dxex+e−x is equal to (A) tan−1(ex)+C (B) tan−1(ex)+C(C)log(ex−e−x)+C (D) log(ex+e−x)+C
Answer. Let I=∫dxex+e−xdx=∫exe2x+1dx Also, let ex=t⇒exdx=dt∴I=∫dt1+t2=tan−1t+C =tan−1(ex)+C Hence, the correct Answer is A.
Q42. Choose the correct answer ∫cos2x(sinx+cosx)2dx is equal to (A) −1sinx+cosx+C (B) log|sinx+cosx|+C(C)log|sinx−cosx|+C(D)1(sinx+cosx)2
Answer. Let I=cos2x(cosx+sinx)2I=∫cos2x−sin2x(cosx+sinx)2dx=∫(cosx+sinx)(cosx−sinx)(cosx+sinx)2dx =∫cosx−sinxcos+sinxdx Let cosx+sinx=t⇒(cosx−sinx)dx=dt ∴I=∫dtt=log|t|+C=log|cosx+sinx|+C Hence, the correct Answer is B
Q43. Choose the correct answer Iff(a+b−x)=f(x), then ∫baxf(x)dx is equal to (A) a+b2∫baf(b−x)dx (B) a+b2∫baf(b+x)dx(C)b−a2∫baf(x)dx (D) a+b2∫baf(x)dx
Answer. Let I=∫bsxf(x)dx ….(1) I=∫ba(a+b−x)f(a+b−x)dx⇒I=∫ba(a+b−x)f(x)dx(∫baf(x)dx=∫baf(a+b−x)dx) ⇒I=(a+b)∫baf(x)dx−I[Using(1)]⇒I+I=(a+b)∫bcf(x)dx ⇒2I=(a+b)∫baf(x)dx⇒I=(a+b2)∫baf(x)dx Hence, the correct Answer is D.
Q44. Choose the correct answer The value of ∫10tan−1(2x−11+x−x2)dx is (A) 1 (B) 0 (C) −1 (D) π4
Answer. Let I=∫10tan−1(2x−11+x−x2)dx⇒I=∫10tan−1(x−(1−x)1+x(1−x))dx ⇒I=∫10[tan−1x−tan−1(1−x)]dx...(1)⇒I=∫10[tan−1(1−x)−tan−1(1−1+x)]dx ⇒I=∫10[tan−1(1−x)−tan−1(x)]dx⇒I=∫10[tan−1(1−x)−tan−1(x)]dx...(2) Adding (1) and (2), we obtain 2I=∫10(tan−1x+tan−1(1−x)−tan−1(1−x)−tan−1x)dx ⇒2I=0⇒I=0 Hence, the correct Answer is B
 Miscellaneous Exercise){kind=link}