NCERT Solutions Class 12 Maths (integrals) Miscellaneous Exercise

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.
Solutions Class 12 Maths (integrals) Miscellaneous Exercise

Q1. Integrate the function 

Answer. 1xx3=1x(1x2)=1x(1x)(1+x)  Let 1x(1x)(1+x)=Ax+B(1x)+C1+x1=A(1x2)+Bx(1+x)+Cx(1x)1=AAx2+Bx+Bx2+CxCx2 Equating the coefficients of x2 , x, and constant term, we obtain A+BC=0B+C=0A=1 On solving these equations, we obtain A=1,B=12, and C=12 From equation (1), we obtain 1x(1x)(1+x)=1x+12(1x)12(1+x) 1x(1x)(1+x)=1x+12(1x)12(1+x)1x(1x)(1+x)dx=1xdx+1211xdx1211+xdx 

Q2. Integrate the function 

Answer. 1x+a+x+b=1x+a+x+b×x+ax+bx+ax+b=x+ax+b(x+a)(x+b)=(x+ax+b)ab 

Q3. Integrate the function 

Answer. 1xaxx2 Let x=atdx=at2dt 1xaxx2dx=1ataat(at)2(at2dt)=1at11t1t2dt =1a1t2tt2t2dt=1a1t1dt=1a[2t1]+C=1a[2ax1]+C 

Q4. Integrate the function 

Answer. 1x2(x4+1)34 Multiplying and dividing by x3 , we obtain x3x2x3(x4+1)34=x3(x4+1)34x2x3 =(x4+1)34x5(x4)34=1x5(x4+1x4)34=1x5(1+1x4)34  Let 1x4=t4x5dx=dt1x5dx=dt41x2(x4+1)34dx=1x5(1+1x4)34dx =14(1+t)34dt=14[(1+t)1414]+C 

Q5. Integrate the function 

Answer. 1x12+x13=1x13(1+x16) Let x=t6dx=6t5dt 1x12+x13dx=1x13(1+x16)dx=6t5t2(1+t)dt=6t3(1+t)dt  On dividing, we obtain 1x12+x13dx=6{(t2t+1)11+t}dt 

Q6. Integrate the function 

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|+12xx2+9dx+921x2+9dx 

Q7. Integrate the function sinxsin(xa)

Answer. sinxsin(xa)Letxa=tdx=dtsinxsin(xa)dx=sin(t+a)sintdt =sintcosa+costsinasintdt=(cosa+cottsina)dt 

Q8. Integrate the function : e5logxe4logxe3logxe2logx

Answer. e5logxe4logxe3logxe2logx=e4logx(elogx1)e2logx(elogx1)=e2logx=elogx2=x2 

Q9. Integrate the function cosx4sin2x

Answer. cosx4sin2x Let sinx=tcosxdx=dt cosx4sin2xxdx=dt(2)2(t)2 

Q10. Integrate the function : 

Answer. sin8xcos8x12sin2xcos2x=(sin4x+cos4x)(sin4xcos4x)sin2x+cos2xsin2xcos2xsin2xcos2x =(sin4x+cos4x)(sin2x+cos2x)(sin2xcos2x)(sin2xsin2xcos2x)+(cos2xsin2xcos2x)=(sin4x+cos4x)(sin2xcos2x)sin2x(1cos2x)+cos2x(1sin2x) =(sin4x+cos4x)(cos2xsin2x)(sin4x+cos4x)=cos2x 

Q11. Integrate the function 1cos(x+a)cos(x+b)

Answer. 1cos(x+a)cos(x+b) Multiplying and dividing by sin(ab), we obtain  1sin(ab)[sin(ab)cos(x+a)cos(x+b)]=1sin(ab)[sin[(x+a)(x+b)]cos(x+a)cos(x+b)] =1sin(ab)[sin(x+a)cos(x+b)cos(x+a)sin(x+b)cos(x+a)cos(x+b)]=1sin(ab)[sin(x+a)cos(x+a)sin(x+b)cos(x+b)] =1sin(ab)[tan(x+a)tan(x+b)]1cos(x+a)cos(x+b)dx=1sin(ab)[tan(x+a)tan(x+b)]dx 

Q12. Integrate the function x31x8

Answer. x31x8 Let x4=t4x3dx=dt 

Q13. Integrate the function ex(1+ex)(2+ex)

Answer. ex(1+ex)(2+ex) Let ex=texdx=dtex(1+ex)(2+ex)dx=dt(t+1)(t+2) 

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=131x2+1dx131x2+4dx 

Q15. Integrate the function 

Answer. cos3xelogsinx=cos3x×sinx Let cosx=tsinxdx=dt 

Q16. Integrate the function 

Answer. e3logx(x4+1)1=elogx3(x4+1)1=x3(x4+1) Let x4+1=t4x3dx=dte3logx(x4+1)1dx=x3(x4+1)dx 

Q17. Integrate the function : f(ax+b)[f(ax+b)]n

Answer. f(ax+b)[f(ax+b)]n Let f(ax+b)=taf(ax+b)dx=dtf(ax+b)[f(ax+b)]ndx=1atndt 

Q18. Integrate the function 

Answer. 1sin3xsin(x+α)=1sin3x(sinxcosα+cosxsinα)=1sin4xcosα+sin3xcosxsinα=1sin2xcosα+cotxsinα=csc2xcosα+cotxsinα  Let cosα+cotxsinα=tcsc2xsinαdx=dt1sin3xsin(x+α)dx=csc2xcosα+cotxsinααdx=1sinαdtt=1sinα[2t]+C 

Q19. Integrate the function 

Answer.  Let I=sin1xcos1xsin1x+cos1xdx It is known that, sin1x+cos1x=π2 I=(π2cos1x)cos1xπ2dx=2ππ](2cos1x2x)dx =2ππ21dx4πcos1xdx =x4πcos1xdx....(1)  Let I1=cos1xdx Also, lctx=tdx=2tdtI1=2cos1ttdt=2[cos1tt2211t2t22dt] =t2cos1t+t21t2dt=t2cos1t1t211t2dt=t2cos1t1t2dt+11t2dt =t2cos1tt21t212sin1t+sin1t=t2cos1tt21t2+12sin1t From equation (1), we obtain  I=x4π[t2costt21t2+12sin1t]=x4π[xcos1xx21x+12sin1x]=x4π[x(2sin1x)xλ22+2sin1x] 

Q20. Integrate the function 

Answer. I=1x1+xdx Let x=cos2θdx=2sinθcosθdθI=1cosθ1+cosθ(2sinθcosθ)dθ =2sin2θ22cos2θ2sin2θdθ =tanθ22sinθcosθdθ=2sinθ2cosθ2(2sinθ2cosθ2)cosθdθ =4sin2θ2cosθdθ=4sin2θ2(2cos2θ21)dθ=4(2sin2θ2cos2θ2sin2θ2)dθ =8sin2θ2cos2θ2dθ+4sin2θ2dθ=2sin2θdθ+4sin2θ2dθ=21cos2θ2)dθ+41cosθ2dθ =2[θ2sin2θ4]+4[θ2sinθ2]+C=θ+sin2θ2+2θ2sinθ+C=θ+sin2θ22sinθ+C 

Q21. Integrate the function 

Answer. I=(2+sin2x1+cos2x)ex=(2+2sinxcosx2cos2x)ex=(1+sinxcosxcos2x)ex=(sec2x+tanx)ex 

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 

Q23. Integrate the function tan11x1+x

Answer. I=tan11x1+xdx Let x=cosθdx=sinθdθI=tan11cosθ1+cosθ(sinθdθ) =tan12sin2θ22cos2θ2sinθdθ =tan1tanθ2sinθdθ=12θsinθdθ=12[θ(cosθ)1(cosθ)dθ] =+12θcosθ12sinθ=12cos1xx121x2+C=x2cos1x121x2+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)=1x3x2+1x2log(1+1x2)=1x31+1x2log(1+1x2)  Let 1+1x2=t2x3dx=dtI=1x31+1x2log(1+1x2)dx=12tlogtdt=12t12logtdt Integrating by parts, we obtain I=12[logtt12dt{(ddtlogt)t12dt}dt]=12[logtt3232t1t322dt]=12[23t32logt23t12dt] 

Q25. Evaluate the definite integral 

Answer. I=π2πex(1sinx1cosx)dx=π2πex(12sinx2cosx22sin2x2)dx =x2πex(csc2x22cotx2)dx  Let f(x)=cotx2f(x)=(12csc2x2)=12csc2x2I=ππ2ex(f(x)+f(x)]dx =[exf(x)dx]π2π=[excotx2]π2x=[ex×cotπ2eπ2×cotπ4] 

Q26. Evaluate the definite integral 0π4sinxcosxcos4x+sin4xdx

Answer. I=0π4sinxcosxcos4x+sin4xdx I=0π4(sinxcosx)cos4x(cos4x+sin4x)cos4xdx  I=0x4tanxsec2x1+tan4xdx Let tan2x=t2tanxsec2xdx=dt  When x=0,t=0 and  when x=π4,t=1I=12dt1+t2=12[tan1t]01=12[tan11tan10]=12[π4] 

Q27. Evaluate the definite integral 

Answer.  Let I=5π2cos2xcos2x+4sin2xdxI=π2cos2xcos2x+4(1cos2x)dxI=π2cos2xcos2x+44cos2xdx I=130x243cos2x443cos2xdx I=130π243cos2x43cos2xdx+130π2443cos2xdx I=π6+230π2sec2x1+4tan2xdx....(1) Consider, 0π2sec2x1+4tan2xdx  Let 2tanx=t2sec2xdx=dt When x=0,t=0 and when x=π2,t=0x22sec2x1+4tan2xdx=0dt1+t2 =[tan1t]0=[tan1()tan1(0)]=π2 

Q28. Evaluate the definite integral π6π3sinx+cosxsin2xdx

Answer.  Let I=π6π3sinx+cosxsin2xdxI=π6π3(sinx+cosx)(sin2x)dxdxI=ππ3sinx+cosx(1+12sinxcosx)dx I=π6π3(sinx+cosx)1(sin2x+cos2x2sinxcosx)dxI=π6π3(sinx+cosx)dx1(sinxcosx)2 Let (sinxcosx)=t(sinx+cosx)dx=dt When x =π6,t=(132)x=π3,t=(312)I=3231dt1t2I=31231dt1t2 1 As 11(t)2=11t2, therefore, 11t2 is an even function. It is known that if f(x) is an even function, then aπf(x)dx=20af(x)dx I=2031dt1t2 =[2sin1t]0312 

Q29. Evaluate the definite integral 01dx1+xx

Answer. Let I = 01dx1+xx I=011(1+xx)×(1+x+x)(1+x+x)dx =0d1+x+x1+xxdx=041+xdx+01xdx=[23(1+x)32]01+[23(x)32]01 

Q30. Evaluate the definite integral 0π4sinx+cosx9+16sin2xdx

Answer.  Let I=0π4sinx+cosx9+16sin2xdx Also, let sinxcosx=t(cosx+sinx)dx=dt When x=0,t=1 and when x=π4,t=0 (sinxcosx)2=t2sin2x+cos2x2sinxcosx=t21sin2x=t2sin2x=1t2 I=10dt9+16(1t2)=10dt9+1616t2=10dt2516t2=10dt(5)2(4t)2 

Q31. Evaluate the definite integral 0π2sin2xtan1(sinx)dx

Answer.  Let I=0π2sin2xtan1(sinx)dx=0π22sinxcosxtan1(sinx)dx Also, let sinx=tcosxdx=dt When x=0,t=0 and when x=π2,t=1 I=201ttan1(t)dt....(1) Consider ttan1tdt=tan1ttdt{ddt(tan1t)tdt}dt =tan1tt2211+t2t22dt=t2tan1t212t2+111+t2dt=t2tan1t2121dt+1211+t2dt =t2tan1t212t+12tan1t 01ttan1tdt=[t2tan1t2t2+12tan1t]01=12[π41+π4]=12[π21]=π412 

Q32. Evaluate the definite integral 0πxtanxsecx+tanxdx

Answer. Let I = 0πxtanxsecx+tanxdx....(1) I=0π{(πx)tan(πx)sec(πx)+tan(πx)}dx(0af(x)dx=0ef(ax)dx) I=0π{(πx)tanx(secx+tanx)}dxI=0π(πx)tanxsecx+tanxdx...(2) Adding (1) and (2), we obtain  2I=0ππtanxsecx+tanxdx 2I=π0πsinxcosx1cosx+sinxcosxdx 2I=π0πsinx+111+sinxdx2I=π0π1.dxπ0π11+sinxdx 2I=π[x]0ππ0π1sinxcos2xdx2I=π2π0π(sec2xtanxsecx)dx 2I=π2π[tanxsecx]0x2I=π2π[tanπsecπtan0+sec0]2I=π2π[0(1)0+1] 

Q33. Evaluate the definite integral 14[|x1|+|x2|+|x3|]dx

Answer.  Let I=14[|x1|+|x2|+|x3|]dxI=14|x1|dx+4|x2|dx+4|x3|dxI=I1+I2+I3...(3)  where, I1=4|x1|dx,I2=14|x2|dx, and I3=14|x3|dxI1=14|x1|dx(x1)0 for 1x4 I1=14(x1)dxI1=[x2xx]14I1=[8412+1]=92...(2) I2=14|x2|dxx20 for 2x4 and x20 for 1x2I2=12(2x)dx+21(x2)dx I2=12(2x)dx+24(x2)dxI2=[2xx22]12+[x222x]24I2=[422+12]+[882+4] I2=12+2=52.....(3) I3=14|x3|dxx30 for 3x4 and x30 for 1x3 I3=13(3x)dx+54(x3)dxI3=[3xx22]13+[x223x]34 I3=[9923+12]+[81292+9]I3=[64]+[12]=52 

Q34. Prove the following 

Answer.  Let I=13dxx2(x+1) Also, let 1x2(x+1)=Ax+Bx2+Cx+11=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 =[logx1x+log(x+1)]13=[log(x+1x)1x]13 =log(43)13log(21)+1=log4log3log2+23=log2log3+23 

Q35. Prove the following 01xexdx=1

Answer.  Let I=01xexdx Integrating by parts, we obtain I=x01exdx01{(ddx(x))exdx}dx =[xex]0101exdx=[xex]01[ex]01=ee+1=1 Hence, the given result is proved.

Q36. Prove the following 

Answer.  Let I=01tan1(2x11+xx2)dxI=01tan1(x(1x)1+x(1x))dx I=01[tan1xtan1(1x)]dx...(1)I=01[tan1(1x)tan1(11+x)]dx I=01[tan1(1x)tan1(x)]dxI=01[tan1(1x)tan1(x)]dx...(2)  Adding (1) and (2), we obtain 2I=01(tan1x+tan1(1x)tan1(1x)tan1x)dx 

  Let I=11x17cos4xdx Also, letf(x)=x17cos4xf(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 aaf(x)dx=0I=11x17cos4xdx=0 Hence, the given result is proved.

Q37. Prove the following 

Answer.  Let I=0π2sin3xdxI=0π2sin2xsinxdx=0π2(1cos2x)sinxdx =0π2sinxdx0π2cos2xsinxdx=[cosx]0π2+[cos3x3]0π2=1+13[1]=113=23 Hence, the given result is proved.

Q38. Prove the following 

Answer.  Let I=0π42tan3xdxI=20π4tan2xtanxdx=20π(sec2x1)tanxdx=20π4sec2xtanxdx20πtanxdx =2[tan2x2]0π4+2[logcosx]0π4=1+2[logcosπ4logcos0]0π4=1+2[log12log1]=1log2log1=1log2 Hence, the given result is proved.

Q39. Prove the following 

Answer.  Let I=01sin1xdxI=01sin1x1dx Integrating by parts, we obtain  I=[sin1xx]010111x2xdx=[xsin1x]01+121x21(2x)1x2dx Let 1x2=t2xdx=dt  When x=0,t=1 and when x=1,t=0I=[xsin1x]01+12dtt=[xsin1x]01+12[2t]10=sin1(1)+[1]=π21 Hence, the given result is proved.

Q40. Evaluate 01e23xdx as a limit of a sum.

Answer.  Let I=01e23xdx It is known that, 0f(x)dx=(ba)limn1n[f(a)+f(a+h)++f(a+(n1)h)]  Where, h=ban Here, a=0,b=1, and f(x)=e23xh=10n=1n 01e23xdx=(10)limn1n[f(0)+f(0+h)++f(0+(n1)h)]=limn1n[e2+e23h+e23(n1)h] =limn1n[e2{1+e3h+e6h+e9h+e3(n1)k}]=limh1n[e2{1(e3h)n1(e3h)}] =limn1n[e2{1e3n1e3n}]=limn1n[e2(1e3)1e3n]=e2(e31)limn1n[1e331] =e2(e31)limn(13)[3ne3n1]=e2(e31)3limn[3ne3n] 

Q41. Choose the correct answer dxex+ex is equal to  (A) tan1(ex)+C (B) tan1(ex)+C(C)log(exex)+C (D) log(ex+ex)+C

Answer.  Let I=dxex+exdx=exe2x+1dx Also, let ex=texdx=dtI=dt1+t2=tan1t+C 

Q42. Choose the correct answer cos2x(sinx+cosx)2dx is equal to 

Answer.  Let I=cos2x(cosx+sinx)2I=cos2xsin2x(cosx+sinx)2dx=(cosx+sinx)(cosxsinx)(cosx+sinx)2dx =cosxsinxcos+sinxdx Let cosx+sinx=t(cosxsinx)dx=dt 

Q43. Choose the correct answer Iff(a+bx)=f(x), then abxf(x)dx is equal to 

Answer. Let I=sbxf(x)dx ….(1) I=ab(a+bx)f(a+bx)dxI=ab(a+bx)f(x)dx(abf(x)dx=abf(a+bx)dx) I=(a+b)abf(x)dxI[Using(1)]I+I=(a+b)cbf(x)dx 

Q44. Choose the correct answer The value of 01tan1(2x11+xx2)dx is 

Answer.  Let I=01tan1(2x11+xx2)dxI=01tan1(x(1x)1+x(1x))dx I=01[tan1xtan1(1x)]dx...(1)I=01[tan1(1x)tan1(11+x)]dx I=01[tan1(1x)tan1(x)]dxI=01[tan1(1x)tan1(x)]dx...(2)  Adding (1) and (2), we obtain 2I=01(tan1x+tan1(1x)tan1(1x)tan1x)dx