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WB Board Class 12 Mathematics – Integration (SN Dey) Solutions
All solutions verified and handwritten-style explanations prepared by The Math Fellow
Explore complete step-by-step solutions of the Integration chapter from the Chhaya Mathematics (SN Dey) book. Each problem is solved carefully — showing every step, formula, and reasoning — just as explained in a live classroom.
This collection includes Indefinite Integration (Very Short Answer Type Questions/2 Marks) questions with algebraic, trigonometric, exponential, and logarithmic forms — all neatly formatted and watermark-protected with myqpaper.in for authenticity.
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3. Using the definition of integration, find (y) in terms of (x).
3(i) (dfrac{dy}{dx}=3x^2+2)
Given (dfrac{dy}{dx}=3x^2+2). Integrate both sides: [ y=int(3x^2+2),dx = 3int x^2dx + 2int dx = 3cdotfrac{x^3}{3} + 2x + C = x^3 + 2x + C. ]
(boxed{y=x^3+2x+C})
3(ii) (dy=(ax+b)dx)
Integrate: [ y=int (ax+b)dx = aint xdx + bint dx = tfrac{a}{2}x^2 + bx + C. ]
(boxed{y=tfrac{a}{2}x^2+bx+C})
3(iii) (dfrac{dy}{dx}=2sin2x)
Integrate: [ y=int 2sin2x,dx = 2left(-tfrac{1}{2}cos2xright) + C = -cos2x + C. ]
(boxed{y=-cos2x+C})
3(iv) (dfrac{dy}{dx}=sqrt{x},; text{given } y(4)=3)
Write (sqrt{x}=x^{1/2}). Integrate: [ y=int x^{1/2}dx = tfrac{2}{3}x^{3/2} + C. ] Use condition (y(4)=3): [ 3 = tfrac{2}{3}cdot 4^{3/2} + C = tfrac{16}{3} + C Rightarrow C = -tfrac{7}{3}. ] So [ y=tfrac{2}{3}x^{3/2} – tfrac{7}{3}. ]
(boxed{y=tfrac{2}{3}x^{3/2}-tfrac{7}{3}})
4. Integrate each function w.r.t. (x).
4(i) (displaystyle int x^{2/3},dx)
(int x^{n}dx = dfrac{x^{n+1}}{n+1}+C). So (int x^{2/3}dx = dfrac{x^{5/3}}{5/3} + C = dfrac{3}{5}x^{5/3} + C.)
(boxed{tfrac{3}{5}x^{5/3}+C})
4(ii) (displaystyle int dfrac{2}{sqrt[4]{x^3}},dx)
(dfrac{2}{sqrt[4]{x^3}} = 2x^{-3/4}). [ int 2x^{-3/4}dx = 2cdot frac{x^{1/4}}{1/4} + C = 8x^{1/4} + C = 8sqrt[4]{x} + C. ]
(boxed{8sqrt[4]{x}+C})
4(iii) (displaystyle int (x^4-2x^2+5),dx)
Integrate termwise: [ tfrac{x^5}{5} – tfrac{2x^3}{3} + 5x + C. ]
(boxed{tfrac{x^5}{5}-tfrac{2x^3}{3}+5x+C})
4(iv) (displaystyle intBig(x-dfrac{2}{x}Big)^3 dx)
Expand: [ Big(x-tfrac{2}{x}Big)^3 = x^3 -6x + tfrac{12}{x} – tfrac{8}{x^3}. ] Integrate: [ tfrac{x^4}{4} – 3x^2 + 12ln|x| + 4x^{-2} + C. ]
(boxed{tfrac{x^4}{4}-3x^2+12ln|x|+tfrac{4}{x^2}+C})
4(v) (displaystyle int(x+3)(x^2-5),dx)
Multiply: ((x+3)(x^2-5)=x^3+3x^2-5x-15). Integrate: [ tfrac{x^4}{4}+x^3-tfrac{5x^2}{2}-15x + C. ]
(boxed{tfrac{x^4}{4}+x^3-tfrac{5x^2}{2}-15x+C})
4(vi) (displaystyle int dfrac{1}{sqrt{x}}Big(sqrt{x}+dfrac{1}{sqrt{x}}Big)^2 dx)
Expand: ((sqrt{x}+tfrac{1}{sqrt{x}})^2 = x + 2 + tfrac{1}{x}). Multiply by (x^{-1/2}): (x^{1/2}+2x^{-1/2}+x^{-3/2}). Integrate: [ tfrac{2}{3}x^{3/2} + 4sqrt{x} – 2x^{-1/2} + C. ]
(boxed{tfrac{2}{3}x^{3/2}+4sqrt{x}-dfrac{2}{sqrt{x}}+C})
5. Evaluate the following integrals.
5(i) (displaystyle int e^{3log x},dx)
(e^{3ln x}=x^3). So (int x^3 dx = tfrac{x^4}{4} + C.)
(boxed{tfrac{x^4}{4}+C})
5(ii) (displaystyle int dfrac{e^{5x}-2e^{3x}+3}{e^x},dx)
Divide: (=e^{4x}-2e^{2x}+3e^{-x}). Integrate: (tfrac{e^{4x}}{4}-e^{2x}-3e^{-x}+C.)
(boxed{tfrac{e^{4x}}{4}-e^{2x}-3e^{-x}+C})
5(iii) (displaystyle int dfrac{e^{5x}+e^{3x}}{e^x+e^{-x}},dx)
Simplify: (dfrac{e^{3x}(e^{2x}+1)}{e^{-x}(e^{2x}+1)}=e^{4x}.) So integral (=tfrac{e^{4x}}{4}+C.)
(boxed{tfrac{e^{4x}}{4}+C})
5(iv) (displaystyle int (e^{2log x}-2e^{-3log x}),dx)
(e^{2ln x}=x^2,; e^{-3ln x}=x^{-3}.) So integrand (=x^2-2x^{-3}). Integrate: (tfrac{x^3}{3}+x^{-2}+C.)
(boxed{tfrac{x^3}{3}+dfrac{1}{x^2}+C})
5(v) (displaystyle int e^{-4x},dx)
(int e^{-4x}dx = -tfrac{1}{4}e^{-4x} + C.)
(boxed{-tfrac{1}{4}e^{-4x}+C})
6. Integrate.
6(i) (displaystyle int 5^{2x} dx)
(5^{2x}=e^{2xln5}). So (int e^{2xln5}dx = dfrac{5^{2x}}{2ln5} + C.)
(boxed{dfrac{5^{2x}}{2ln5}+C})
6(ii) (displaystyle int (x^4+4^x)dx)
(int x^4dx=tfrac{x^5}{5},; int 4^x dx=dfrac{4^x}{ln4}.) So (tfrac{x^5}{5}+dfrac{4^x}{ln4}+C.)
(boxed{tfrac{x^5}{5}+dfrac{4^x}{ln4}+C})
6(iii) (displaystyle int (e^{xlog a}+e^{alog x})dx)
(e^{xlog a}=a^x,; e^{alog x}=x^a). Integrate: (dfrac{a^x}{ln a} + dfrac{x^{a+1}}{a+1} + C.)
(boxed{dfrac{a^x}{ln a}+dfrac{x^{a+1}}{a+1}+C})
7. Evaluate the following integrals (trigonometric).
7(i) (displaystyle int sin^2 2x,dx)
(sin^2 t = tfrac{1-cos2t}{2}). (sin^2 2x=tfrac{1-cos4x}{2}). [ int sin^2 2x,dx = tfrac{x}{2} – tfrac{sin4x}{8} + C. ]
(boxed{tfrac{x}{2}-tfrac{sin4x}{8}+C})
7(ii) (displaystyle int dfrac{2tan x}{1+tan^2x},dx)
(1+tan^2x = sec^2x). So integrand (= dfrac{2tan x}{sec^2x} = 2sin xcos x = sin2x). [ int sin2x,dx = -tfrac{cos2x}{2} + C. ]
(boxed{-tfrac{cos2x}{2}+C})
7(iii) (displaystyle int sec x(sec x+tan x),dx)
(sec x(sec x+tan x)=sec^2x+sec xtan x). [ int = tan x + sec x + C. ]
(boxed{tan x + sec x + C})
7(iv) (displaystyle int csc x(csc x-cot x),dx)
(csc x(csc x-cot x)=csc^2x-csc xcot x). [ int = -cot x + csc x + C. ]
(boxed{-cot x+csc x + C})
7(v) (displaystyle int cos x^circ,dx)
Convert degrees: (x^circ = dfrac{pi x}{180}). Let (k=dfrac{pi}{180}). [ int cos(kx),dx = frac{sin(kx)}{k} + C Rightarrow frac{180}{pi}sin x^circ + C. ]
(boxed{dfrac{180}{pi}sin x^circ + C})
7(vi) (displaystyle int cot^2theta,dtheta)
(cot^2theta=csc^2theta-1). [ int = -cottheta – theta + C. ]
(boxed{-cottheta-theta + C})
7(vii) (displaystyle int cos^2theta,dtheta)
(cos^2theta=tfrac{1+cos2theta}{2}). [ int = tfrac{theta}{2} + tfrac{sin2theta}{4} + C. ]
(boxed{tfrac{theta}{2}+tfrac{sin2theta}{4}+C})
7(viii) (displaystyle int dfrac{1+tan^2x}{1+cot^2x},dx)
(1+tan^2x=sec^2x,;1+cot^2x=csc^2x). So integrand (=dfrac{sec^2x}{csc^2x}=tan^2x). [ int tan^2x,dx = tan x – x + C. ]
(boxed{tan x – x + C})
7(ix) (displaystyle int dfrac{sec x + 2cot^2x + cos^2x}{cos x},dx)
Simplify: [ frac{sec x + 2cot^2x + cos^2x}{cos x} = sec^2x + 2cot xcsc x + cos x. ] Integrate termwise: [ tan x – 2csc x + sin x + C. ]
(boxed{tan x – 2csc x + sin x + C})
7(x) (displaystyle int dfrac{sec^2x}{csc^2x},dx)
(dfrac{sec^2x}{csc^2x}=tan^2x). So [ int tan^2x,dx = tan x – x + C. ]
(boxed{tan x – x + C})
8. If (dfrac{dy}{dx}=2x^2-3) at any point ((x,y)) on a curve, find the family of curves.
8. Integrate
[ y=int(2x^2-3)dx = tfrac{2}{3}x^3 – 3x + C. ]
(boxed{y=tfrac{2}{3}x^3 – 3x + C})
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Next Topics in Indefinite Integration
Indefinite Integral – Very Short Answer Type [2 Marks] Indefinite Integral – Short Answer Type [4 Marks] Indefinite Integral – Long Answer Type [5 Marks]
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