NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.2
NCERT Solutions Class 12 Maths from class
12th Students will get the answers of
Chapter-7 (Integrals)Exercise 7.2 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.
Exercise 7.2
Q1. Integrate the function 2x1+x2.
Answer. Let 1+x2=t∴2xdx=dt⇒∫2x1+x2dx=∫11dt=log|t|+C=log∣∣1+x2∣∣+C=log(1+x2)+C
Q2. Integrate the function (logx)2x
Answer. Let log |x| = t , ∴1xdx=dt ⇒∫(log|x|)2xdx=∫t2dt=t33+C=(log|x|)33+C
Q3. Integrate the function1x+xlogx
Answer. 1x+xlogx=1x(1+logx) Let 1+logx=t1xdx=dt ⇒∫1x(1+logx)dx=∫1tdt=log|t|+C=log|1+logx|+C
Q4. Integrate the function: sin x ⋅ sin (cos x)
Answer. sin x ⋅ sin (cos x) Let cos x = t ∴ −sin x dx = dt ⇒∫sinx⋅sin(cosx)dx=−∫sintdt=−[−cost]+C=cost+C=cos(cosx)+C
Q5. Integrate the function sin (ax + b ) cos ( ax + b )
Answer. sin(ax+b)cos(ax+b)=2sin(ax+b)cos(ax+b)2=sin2(ax+b)2 Let 2(ax+b)=t∴2adx=dt ⇒∫sin2(αx+b)2dx=12∫sintdt2a=14a[−cost]+C=−14acos2(ax+b)+C
Q6. Integrate the function √ax+b
Answer. Let ax+b=t⇒adx=dt∴dx=1adt ⇒∫(ax+b)12dx=1a∫t12dt=1a(t232)+C=23a(ax+b)32+C
Q7. Integrate the function x√x+2 Answer. Let (x + 2) = t ∴ dx = dt ⇒∫x√x+2dx=∫(t−2)√tdt=∫t32−2t12)dt=∫t32dt−2∫t12dt =t25−2(t1232)+C=25t52−43t32+C=25(x+2)52−43(x+2)32+C
Q8. Integrate the function x√1+2x2
Answer. Let 1+2x2=t∴4xdx=dt ⇒∫x√1+2x2dx=∫√tdt4=14∫t12dt=14(t32)+C=16(1+2x2)32+C
Q9. Integrate the function: (4x+2)√x2+x+1
Answer. Let x2+x+1=t∴(2x+1)dx=dt∫(4x+2)√x2+x+1dx ∫(4x+2)√x2+x+1dx=∫2√tdt=2∫√tdt =2(t3232)+C=43(x2+x+1)32+C
Q10. Integrate the function : 1x−√x
Answer. 1x−√x=1√x(√x−1) Let (√x−1)=t12√xdx=dt ⇒∫1√x(√x−1)dx=∫2tdt=2log|t|+C=2log|√x−1|+C
Q11. Integrate the function : x√x+4,x>0
Answer. Let x + 4 = t ∴ dx = dt ∫x√x+4dx=∫(t−4)√tdt=∫(√t−4√t)dt=t3232−4(1212)+C =23(t)32−8(t)12+C=23t⋅t12−8t12+C=23t12(t−12)+C=23(x+4)12(x+4−12)+C=23√x+4(x−8)+C
Q12. Integrate the function: (x3−1)13x5
Answer. Let x3−1=t∴3x2dx=dt ⇒∫(x3−1)13x5dx=∫(x3−1)13x3⋅x2dx=∫t13(t+1)dt3=13∫t43+t13)dt =13[t733+t4343]+C=13[37t13+34t43]+C=17(x3−1)73+14(x3−1)43+C
Q13. Integrate the function : x2(2+3x3)3
Answer. Let 2+3x3=t∴9x2dx=dt ⇒∫x2(2+3x3)3dx=19∫dt(t)3=19[t−2−2]+C=−118(1t2)+C=−118(2+3x3)2+C
Q14. Integrate the function : 1x(logx)m,x>0,m≠1 Answer. Let log x = t ∴1xdx=dt⇒∫1x(logx)mdx=∫dt(t)m=(logx)1−π(1−m)+C=(logx)1−π(1−m)+C
Q15. Integrate the function : x9−4x2
Answer. Let 9−4x2=t∴−8xdx=dt⇒∫x9−4x2dx=−18∫1tdt =−18log|t|+C=−18log∣∣9−4x2∣∣+C
Q16. Integrate the function : e2x+3
Answer. Let 2x + 3 = t ∴ 2dx = dt ⇒∫e2x+3dx=12∫e′dt=12(et)+C=12e(2x+3)+C
Q17. Integrate the function : xex2
Answer. Let x2=t∴2xdx=dt⇒∫xex2dx=12∫1e′dt =12∫e−tdt=12(e−t−1)+C=−12e−x2+C=−12ex2+C
Q18. Integrate the function : etan−1x1+x2
Answer. Let tan−1x=t11+x2dx=dt ⇒∫etan−1x1+x2dx=∫etdt=et+C=elnn−1x+C
Q19. Integrate the function : e2x−1e2x+1
Answer. e2x−1e2x+1 Dividing numerator and denominator by ex , we obtain (e2x−1)e2x+1)=ex−e−xex+e−xex+1)ex Let ex+e−x=t(ex−e−x)dx=dt ⇒∫e2x−1e2x+1dx=∫ex−e−xex+e−xdx=∫dtt=log|t|+C=log∣∣ex+e−x∣∣+C
Q20. Integrate the function : e2x−e−2xe2x+e−2x
Answer. Let e2x+e−2x=t∴(2e2x−2e−2x)dx=dt⇒2(e2x−e−2x)dx=dt ⇒2(e2x−e−2x)dx=dt⇒∫(e2x−e−2xe2x+e−2x)dx=∫dt2t=12∫1tdt=12log|t|+C=12log∣∣e2x+e−2x∣∣+C
Q21. Integrate the function : tan2(2x−3)
Answer. tan2(2x−3)=sec2(2x−3)−1 Let 2x−3=t∴2dx=dt ⇒∫tan2(2x−3)dx=∫(sec2(2x−3))−1dx=12∫(sec2t)dt−∫1dx=12∫sec2tdt−∫ldx=12tant−x+C=12tan(2x−3)−x+C
Q22. Integrate the function : sec2(7−4x)
Answer. Let 7 − 4x = t ∴ −4dx = dt ∴∫sec2(7−4x)dx=−14∫sec2tdt=−14(tant)+C=−14tan(7−4x)+C
Q23. Integrate the function : sin−1x√1−x2
Answer. Let sin−1x=t ∴1√1−x2dx=dt⇒∫sin−1x√1−x2dx=∫tdt=(sin−1x)22+C=(sin−1x)22+C
Q24. Integrate the function : 2cosx−3sinx6cosx+4sinx
Answer. 2cosx−3sinx6cosx+4sinx=2cosx−3sinx2(3cosx+2sinx) Let 3cosx+2sinx=t (−3sinx+2cosx)dx=dt∫2cosx−3sinx6cosx+4sinxdx=∫dt2t=12∫1tdt=12log|t|+C=12log|2sinx+3cosx|+C
Q25. Integrate the function : 1cos2x(1−tanx)2
Answer. 1cos2x(1−tanx)2=sec2x(1−tanx)2Let(1−tanx)=tz−sec2xdx=dt ⇒∫sec2x(1−tanx)2dx=∫−dtt2=−∫t−2dt=+1t+C=1(1−tanx)+C
Q26. Integrate the function : cos√x√x
Answer. Let √x=t12√xdx=dt ⇒cos√x√xdx=2∫costdt=2sint+C=2sin√x+C
Q27. Integrate the function : √sin2xcos2x
Answer. Let sin2x=t∴2cos2xdx=dt⇒∫√sin2xcos2xdx=12∫√tdt =12(t3232)+C=13t32+C=13(sin2x)32+C
Q28. Integrate the function : cosx√1+sinx
Answer. Let 1 + sin x = t ∴cosxdx=dt⇒∫cosx√1+sinxdx=∫dt√t⇒=t122+C=2√t+C=2√1+sinx+C
Q29. Integrate the function : cot x log sin x
Answer. Let log sin x = t ⇒1sinx⋅cosxdx=dt∴cotxdx=dt⇒cotxlogsinxdx=∫tdt⇒=t22+C=12(logsinx)2+C
Q30. Integrate the function : sinx1+cosx
Answer. Let 1 + cos x = t ∴ −sin x dx = dt ⇒∫sinx1+cosxdx=∫−dtt=−log|t|+C=−log|1+cosx|+C
Q31. Integrate the function : sinx(1+cosx)2
Answer. Let 1 + cos x = t ∴ −sin x dx = dt ⇒∫sinx(1+cosx)2dx=∫−dtt2=−∫t−2dt=1t+C=11+cosx+C
Q32. Integrate the function : 11+cotx
Answer. Let I=∫11+cotxdx=∫11+cosxsinxdx=∫sinxsinx+cosxdx=12∫2sinxsinx+cosxdx =12∫(sinx+cosx)+(sinx−cosx)(sinx+cosx)dx=12∫1dx+12∫sinx−cosxsinx+cosxdx=12(x)+12∫sinx−cosxsinx+cosxdx Let sin x + cos x = t ⇒ (cos x − sin x) dx = dt ∴I=x2+12∫−(dt)t=x2−12log|t|+C=x2−12|logsinx+cosx|+C
Q33. Integrate the function : 11−tanx Answer. Let I=∫11−tanxdx=∫11−sinxcosxdx=∫cosxcosx−sinxdx=12∫2cosxcosx−sinxdx =12∫(cosx−sinx)+(cosx+sinx)(cosx−sinx)dx=12∫1dx+12∫cosx+sinxcosx−sinxdx=x2+12∫cosx+sinxcosx−sinxdx Put cos x − sin x = t ⇒ (−sin x − cos x) dx = dt ∴I=x2+12∫−(dt)t=x2−12log|t|+C=x2−12log|cosx−sinx|+C
Q34. Integrate the function : √tanxsinxcosx
Answer. Let I=∫√tanxsinxcosxdx=∫√tanxxcosxsinxcosx×cosxdx=∫√tanxtanxcos2xdx=∫sec2xdx√tanx Let tanx=t⇒sec2xdx=dt∴I=∫dt√t=2√t+C=2√tanx+C
Q35. Integrate the function : (1+logx)2x
Answer. (x+1)(x+logx)2x=(x+1x)(x+logx)2=(1+1x)(x+logx)2 Let (x+logx)=t…(1+1x)dx=dt ⇒(1+1x)(x+logx)2dx=∫t2dt=t33+C=13(x+logx)3+C
Q36. Integrate the function : (x+1)(x+logx)2x
Answer. (x+1)(x+logx)2x=(x+1x)(x+logx)2=(1+1x)(x+logx)2 Let (x+logx)=t…(1+1x)dx=dt ⇒(1+1x)(x+logx)2dx=∫t2dt=t33+C=13(x+logx)3+C
Q37. Integrate the function : x3sin(tan−1x4)1+x8
Answer. Let x4=t∴4x3dx=dt ⇒∫x3sin(tan−1x4)1+x8dx=14∫sin(tan−1t)1+t2dt....(1) Let tan−1t=u11+t2dt=du From (1), we obtain ∫x3sin(tan−1x4)dx1+x8=14∫sinudu=14(−cosu)+C =−14cos(tan−1t)+C=−14cos(tan−1x4)+C
Q38. Choose the correct answer : ∫10x9+10xloge10dxx10+10x equals (A) 10x−x10+C (B) 10x+x10+C (C) (10x−x10)−1+C (D) log(10x+x10)+C
Answer. Let x10+10x=t∴(10x9+10xloge10)dx=dt ⇒∫10x9+10xloge10x10+10xdx=∫dtt=logt+C=log(10x+x10)+C Hence, the correct Answer is D.
Q39. Choose the correct answer : ∫dxsin2xcos2x equals (A) tanx+cotx+C (B) tanx−cotx+C (C) tanxcotx+C (D) tanx−cot2x+C
Answer. Let I=∫dxsin2xcos2x=∫1sin2xcos2xdx=∫sin2x+cos2xsin2xcos2xdx =∫sin2xsin2xcos2xdx+∫cos2xsin2xcos2xdx=∫sec2xdx+∫csc2xdx=tanx−cotx+C Hence, the correct Answer is B .