x<\/sup> dx
\n= \\(\\int \\frac{d t}{(1+t)(2+t)}\\)
\nLet \\(\\frac{1}{(1+t)(2+t)}=\\frac{\\mathrm{A}}{(1+t)}+\\frac{\\mathrm{B}}{(2+t)}\\)
\n\u2234 1 = A(2 + t) + B(1 + t) … (1)
\nPut t = – 2 in (1), we get A = 1
\nPut t = – 2 in (1), we get 1 = – B \u2234B = – 1
\n\\(\\frac{1}{(1+t)(2+t)}=\\frac{1}{(1+t)}-\\frac{1}{(2+t)}\\)
\nI = \\(\\int \\frac{1}{1+t} d t-\\int \\frac{1}{2+t} d t\\)
\n= \\(\\log |1+t|-\\log |2+t|+C\\)
\n= \\(\\log \\left|\\frac{1+t}{2+t}\\right|+\\mathrm{C}=\\log \\left(\\frac{1+e^{x}}{2+e^{x}}\\right)+\\mathrm{C}\\)<\/p>\n<\/p>\n
Question 14.
\n\\(\\frac{1}{\\left(x^{2}+1\\right)\\left(x^{2}+4\\right)}\\)
\nSolution:
\n<\/p>\n
Question 15.
\n\\(\\cos ^{3} x e^{\\log \\sin x}\\)
\nSolution:
\n<\/p>\n
Question 16.
\n\\(e^{3 \\log x}\\left(x^{4}+1\\right)^{-1}\\)
\nSolution:
\n<\/p>\n
Question 17.
\n\\(f^{\\prime}(a x+b)[f(a x+b)]^{n}\\)
\nSolution:
\n<\/p>\n
Question 18.
\n\\(\\frac{1}{\\sqrt{\\sin ^{3} x \\sin (x+\\alpha)}}\\)
\nSolution:
\n<\/p>\n
Question 19.
\n\\(\\frac{\\sin ^{-1} \\sqrt{x}-\\cos ^{-1} \\sqrt{x}}{\\sin ^{-1} \\sqrt{x}+\\cos ^{-1} \\sqrt{x}}, x \\in[0,1]\\)
\nSolution:
\n<\/p>\n
Question 20.
\n\\(\\sqrt{\\frac{1-\\sqrt{x}}{1+\\sqrt{x}}}\\)
\nSolution:
\n<\/p>\n
Question 21.
\n\\(\\frac{2+\\sin 2 x}{1+\\cos 2 x} e^{x}\\)
\nSolution:
\n<\/p>\n
Question 22.
\n\\(\\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\\)
\nSolution:
\nLet \\(\\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\\frac{\\mathrm{A}}{(x+1)}+\\frac{\\mathrm{B}}{(x+1)^{2}}+\\frac{\\mathrm{C}}{(x+2)}\\)
\n\u21d2 x\u00b2 + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)\u00b2 … (1)
\nPut x = – 1 in (1), we get B = 1
\nPut x = – 2 in (1), we get C = 3
\nEquating the coefficients of x2, we get
\nA + C = 1 \u2234 A = – 2
\n<\/p>\n
<\/p>\n
Question 23.
\n\\(\\tan ^{-1} \\sqrt{\\frac{1-x}{1+x}}\\)
\nSolution:
\n<\/p>\n
Question 24.
\n\\(\\frac{\\sqrt{x^{2}+1}\\left[\\log \\left(x^{2}+1\\right)-2 \\log x\\right]}{x^{4}}\\)
\nSolution:
\n<\/p>\n
Question 25.
\n\\(\\int_{\\frac{\\pi}{2}}^{\\pi} e^{x}\\left(\\frac{1-\\sin x}{1-\\cos x}\\right) d x\\)
\nSolution:
\n<\/p>\n
Question 26.
\n\\(\\int_{0}^{\\frac{\\pi}{4}} \\frac{\\sin x \\cos x}{\\cos ^{4} x+\\sin ^{4} x} d x\\)
\nSolution:
\n<\/p>\n
Question 27.
\n\\(\\int_{0}^{\\frac{\\pi}{2}} \\frac{\\cos ^{2} x d x}{\\cos ^{2} x+4 \\sin ^{2} x}\\)
\nSolution:
\nLet I = \\(\\int_{0}^{\\frac{\\pi}{2}} \\frac{\\cos ^{2} x d x}{\\cos ^{2} x+4 \\sin ^{2} x}\\)
\nDividing the Nr. and Dr. by cos\u00b2x, we get
\n
\nwhere t\u00b2 = y
\n\u21d2 1 = A(1 + 4y) + B(1 + y) … (1)
\nPut y = – 1 in (1), we get 1 = – 3A \u2234 A = \\(\\frac { – 1 }{ 3 }\\)
\nEquating the coefficients of y, we get
\n4A + B = 0
\n\u2234 B = \\(\\frac { 4 }{ 3 }\\)
\n<\/p>\n
Question 28.
\n\\(\\int_{\\frac{\\pi}{6}}^{\\frac{\\pi}{3}} \\frac{\\sin x+\\cos x}{\\sqrt{\\sin 2 x}} d x\\)
\nSolution:
\n<\/p>\n
<\/p>\n
Question 29.
\n\\(\\int_{0}^{1} \\frac{d x}{\\sqrt{1+x}-\\sqrt{x}}\\)
\nSolution:
\n<\/p>\n
Question 30.
\n\\(\\int_{0}^{\\frac{\\pi}{4}} \\frac{\\sin x+\\cos x}{9+16 \\sin 2 x} d x\\)
\nSolution:
\n<\/p>\n
Question 31.
\n\\(\\int_{0}^{\\frac{\\pi}{2}} \\sin 2 x \\tan ^{-1}(\\sin x) d x\\)
\nSolution:
\n<\/p>\n
Question 32.
\n\\(\\int_{0}^{\\pi} \\frac{x \\tan x}{\\sec x+\\tan x} d x\\)
\nSolution:
\n<\/p>\n
Question 33.
\n\\(\\left.\\int_{1}^{4}|| x-1|+| x-2|+| x-3 \\mid\\right] d x\\)
\nSolution:
\n<\/p>\n
<\/p>\n
Question 34.
\n\\(\\int_{1}^{3} \\frac{d x}{x^{2}(x+1)}=\\frac{2}{3}+\\log \\frac{2}{3}\\)
\nSolution:
\nLet \\(\\frac{1}{x^{2}(x+1)}=\\frac{A}{x}+\\frac{B}{x^{2}}+\\frac{C}{x+1}\\)
\n1 = Ax(x + 1) + B(x + 1) + C(x\u00b2) … (1)
\nPut x = 0 in (1), we get B = 1
\nPut x = – 1 in (1), we get C = 1
\nEquating the coefficients of x\u00b2, we get
\nA + C = 0 \u2234 A = – 1
\n<\/p>\n
Question 35.
\n\\(\\int_{0}^{1} x e^{x} d x=1\\)
\nSolution:
\n<\/p>\n
Question 36.
\n\\(\\int_{-1}^{1} x^{17} \\cos ^{4} x d x=0\\)
\nSolution:
\nLet f(x) = x17<\/sup>cos4<\/sup>x
\nf(- x) = (- x)17<\/sup>cos< sup>4(-x)
\n\u2234 f(x) is an odd function.
\n\u2234 \\(\\int_{-1}^{1} x^{17} \\cos ^{4} x \\cdot d x=0\\)
\n\\(\\int_{-a}^{a} f(x) d x=0, \\text { if } f(x) \\text { is odd }\\)<\/p>\nQuestion 37.
\n\\(\\int_{0}^{\\frac{\\pi}{2}} \\sin ^{3} x d x=\\frac{2}{3}\\)
\nSolution:
\n<\/p>\n
Question 38.
\n\\(\\int_{0}^{\\frac{\\pi}{4}} 2 \\tan ^{3} x d x=1-\\log 2\\)
\nSolution:
\n<\/p>\n
Question 39.
\n\\(\\int_{0}^{1} \\sin ^{-1} x d x=\\frac{\\pi}{2}-1\\)
\nSolution
\n\u222bsin-1<\/sup> x dx = x sin-1<\/sup> x + \\(\\sqrt{1-x^{2}}\\)
\n(Refer in-tegrals of inverse trigonometric functions)
\n\u2234 \\(\\int_{0}^{1} \\sin ^{-1} x d x=\\left[x \\sin ^{-1} x+\\sqrt{1-x^{2}}\\right]_{0}^{1}\\)
\n= \\(\\left(\\frac{\\pi}{2}+0\\right)-(0+1)=\\frac{\\pi}{2}-1\\)<\/p>\nQuestion 40.
\nEvaluate \\(\\int_{0}^{1} e^{2-3 x} d x\\) as a limit of a sum.
\nSolution:
\nLet I = \\(\\int_{0}^{1} e^{2-3 x} d x\\)
\nHere f(x) = e2-3x<\/sup>, a = 0, b = 1
\nnh = b – a = 1 – 0 = 1
\nf(0 + h) = f(h) = e2-3h<\/sup>
\nf(0 + 2h) = f(2h) = e2-6h<\/sup>
\n……………………………….
\nf(0 + (n – 1)h = f((n – 1)h) = e2-3(n-1)h<\/sup>
\n<\/p>\nQuestion 41.
\n\\(\\int \\frac{d x}{e^{x}+e^{-x}}\\) is equal to
\na. tan-1<\/sup>(ex<\/sup>) + C
\nb. tan-1<\/sup>(e-x<\/sup>) + C
\nc. log(ex<\/sup> – e-x<\/sup>) + C
\nd. log(ex<\/sup> + x-x<\/sup>) + C
\nSolution:
\n<\/p>\n<\/p>\n
Question 42.
\n\\(\\int \\frac{\\cos 2 x}{(\\sin x+\\cos x)^{2}} d x\\) is equal to
\na. \\(\\frac{-1}{\\sin x+\\cos x}+\\mathrm{C}\\)
\nb. log|sin x + cos x| + C
\nc. log|sin x – cos x| + C
\nd. \\(\\frac{1}{(\\sin x+\\cos x)^{2}}\\)
\nSolution:
\n<\/p>\n
Question 43.
\nIf \\(f(a+b-x)=f(x), \\text { then } \\int_{a}^{b} x f(x) d x\\) is equal to
\na. \\(\\frac{a+b}{2} \\int_{a}^{b} f(b-x) d x\\)
\nb. \\(\\frac{a+b}{2} \\int_{a}^{b} f(b+x) d x\\)
\nc. \\(\\frac{b-a}{2} \\int_{a}^{b} f(x) d x\\)
\nd. \\(\\frac{a+b}{2} \\int_{a}^{b} f(x) d x\\)
\nSolution:
\n<\/p>\n
Question 44.
\nThe value of \\(\\int_{0}^{1} \\tan ^{-1}\\left(\\frac{2 x-1}{1+x-x^{2}}\\right) d x\\) is
\na. 1
\nb. 0
\nc. – 1
\nd. \\(\\frac{\\pi}{4}\\)
\nSolution:
\n<\/p>\n","protected":false},"excerpt":{"rendered":"
These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise Class 12 Maths Chapter 7 Miscellaneous Exercise Solutions Question 1. Solution: Let 1 = A(1 – x)(l + x) + Bx(l …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise - MCQ Questions<\/title>\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\t\n\t\n\t\n