WB Board Class 12 Mathematics – Integration (SN Dey) Complete Solutions
Explore the complete step-by-step solutions of the Integration chapter from the SN Dey Mathematics Book for West Bengal Class 12 students. Each problem is solved like a real classroom demonstration — showing every step, applied formula, and concept in a clean, animated layout.
The solutions include exercise -Indefinite Integral from Question 3 to 8(Very Short Type Questions), with proper formulas used and clear mathematical formatting. “Solved by TheMathFellow”
Topics Covered: Integration of standard functions, substitution method, integration by parts, trigonometric transformations, and definite integrals.
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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:
\[y=\int(3x^2+2)\,dx.\]
Split using linearity:
\[y=3\int x^2dx + 2\int dx.\]
Compute:
\[\int x^2dx=\frac{x^3}{3},\quad \int dx = x.\]
So
\[y=3\cdot\frac{x^3}{3} + 2x + C = x^3 + 2x + C.\]
\(\boxed{y=x^3+2x+C}\)
3(ii) \(dy=(ax+b)dx\)
Start: \(dy=(ax+b)dx\). Integrate:
\[y=\int (ax+b)dx = a\int xdx + b\int dx.\]
Compute:
\[\int xdx=\tfrac{x^2}{2},\quad \int dx = x.\]
So
\[y= a\cdot\tfrac{x^2}{2} + b x + C = \tfrac{a}{2}x^2 + bx + C.\]
\(\boxed{y=\tfrac{a}{2}x^2+bx+C}\)
3(iii) \(\dfrac{dy}{dx}=2\sin2x\)
Integrate:
\[y=\int 2\sin2x\,dx = 2\int \sin2x\,dx.\]
Use \(\int \sin(kx)dx = -\tfrac{1}{k}\cos(kx)\):
\[\int \sin2x\,dx = -\tfrac{1}{2}\cos2x.\]
So
\[y=2\cdot\left(-\tfrac{1}{2}\cos2x\right)+C = -\cos2x + C.\]
\(\boxed{y=-\cos2x+C}\)
3(iv) \(\dfrac{dy}{dx}=\sqrt{x},\; y(4)=3\)
Write \(\sqrt{x}=x^{1/2}\). Integrate:
\[y=\int x^{1/2}dx = \frac{x^{3/2}}{3/2} + C = \tfrac{2}{3}x^{3/2} + C.\]
Apply initial condition \(y(4)=3\):
\[3=\tfrac{2}{3}\cdot 4^{3/2} + C.\]
Compute \(4^{3/2}=(\sqrt{4})^3=2^3=8\):
\[3=\tfrac{2}{3}\cdot 8 + C = \tfrac{16}{3} + C \Rightarrow C = 3 - \tfrac{16}{3} = -\tfrac{7}{3}.\]
Final:
\[y=\tfrac{2}{3}x^{3/2} - \tfrac{7}{3}.\]
\(\boxed{y=\tfrac{2}{3}x^{3/2}-\tfrac{7}{3}}\)
4. Integrate each function w.r.t. \(x\).
4(i) \(\displaystyle \int x^{2/3}\,dx\)
Use power rule \(\int x^n dx = \tfrac{x^{n+1}}{n+1}+C\).
Here \(n=\tfrac{2}{3}\Rightarrow n+1=\tfrac{5}{3}\).
\[\int x^{2/3}dx=\frac{x^{5/3}}{5/3}+C=\frac{3}{5}x^{5/3}+C.\]
\(\boxed{\tfrac{3}{5}x^{5/3}+C}\)
4(ii) \(\displaystyle \int \dfrac{2}{4\sqrt{x^3}}\,dx\)
Simplify integrand: \(\dfrac{2}{4\sqrt{x^3}}=\dfrac{1}{2}x^{-3/2}.\)
Integrate using power rule:
\[\int \tfrac12 x^{-3/2}dx = \tfrac12\cdot\frac{x^{-1/2}}{-1/2} = -x^{-1/2} + C = -\dfrac{1}{\sqrt{x}}+C.\]
\(\boxed{-\dfrac{1}{\sqrt{x}}+C}\)
4(iii) \(\displaystyle \int (x^4-2x^2+5)\,dx\)
Integrate termwise:
\[\int x^4dx=\tfrac{x^5}{5},\quad \int -2x^2dx = -\tfrac{2x^3}{3},\quad \int 5dx = 5x.\]
So
\[\int (x^4-2x^2+5)dx = \tfrac{x^5}{5} - \tfrac{2x^3}{3} + 5x + C.\]
\(\boxed{\tfrac{x^5}{5}-\tfrac{2x^3}{3}+5x+C}\)
4(iv) \(\displaystyle \int\Big(x-\dfrac{2}{x}\Big)^3 dx\)
Expand cube: use binomial or compute directly.
\[(x-\tfrac{2}{x})^3 = x^3 - 3x^2\cdot\tfrac{2}{x} + 3x\cdot\tfrac{4}{x^2} - \tfrac{8}{x^3} = x^3 -6x + \tfrac{12}{x} - \tfrac{8}{x^3}.\]
Integrate termwise:
\[\int x^3dx=\tfrac{x^4}{4},\; \int -6x dx = -3x^2,\; \int \tfrac{12}{x}dx = 12\ln|x|,\; \int -\tfrac{8}{x^3}dx = 4x^{-2}.\]
So final:
\[\tfrac{x^4}{4}-3x^2+12\ln|x|+4x^{-2}+C.\]
\(\boxed{\tfrac{x^4}{4}-3x^2+12\ln|x|+\tfrac{4}{x^2}+C}\)
4(v) \(\displaystyle \int(x+3)(x^2-5)\,dx\)
Multiply polynomials:
\[(x+3)(x^2-5)=x^3+3x^2-5x-15.\]
Integrate termwise:
\[\int x^3dx=\tfrac{x^4}{4},\; \int 3x^2dx = x^3,\; \int -5x dx = -\tfrac{5x^2}{2},\; \int -15 dx = -15x.\]
Final:
\[\tfrac{x^4}{4}+x^3-\tfrac{5x^2}{2}-15x + C.\]
\(\boxed{\tfrac{x^4}{4}+x^3-\tfrac{5x^2}{2}-15x + C}\)
4(vi) \(\displaystyle \int \dfrac{1}{\sqrt{x}}\Big(\sqrt{x}+\dfrac{1}{\sqrt{x}}\Big)^2 dx\)
First expand inside square:
\[(\sqrt{x}+\tfrac{1}{\sqrt{x}})^2 = x + 2 + \tfrac{1}{x}.\]
Multiply by \(x^{-1/2}\):
\[x^{1/2} + 2x^{-1/2} + x^{-3/2}.\]
Integrate termwise:
\[\int x^{1/2}dx = \tfrac{2}{3}x^{3/2},\; \int 2x^{-1/2}dx = 4x^{1/2},\; \int x^{-3/2}dx = -2x^{-1/2}.\]
So
\[\tfrac{2}{3}x^{3/2} + 4\sqrt{x} - 2x^{-1/2} + C = \tfrac{2}{3}x^{3/2} + 4\sqrt{x} - \dfrac{2}{\sqrt{x}} + C.\]
\(\boxed{\tfrac{2}{3}x^{3/2}+4\sqrt{x}-\dfrac{2}{\sqrt{x}} + C}\)
5. Evaluate the following integrals.
5(i) \(\displaystyle \int e^{3\log x}\,dx\)
Use identity \(e^{\ln x}=x\). So \(e^{3\ln x}=x^3.\)
\[\int e^{3\ln x}\,dx=\int x^3 dx = \tfrac{x^4}{4} + C.\]
\(\boxed{\tfrac{x^4}{4}+C}\)
5(ii) \(\displaystyle \int \dfrac{e^{5x}-2e^{3x}+3}{e^x}\,dx\)
Divide through by \(e^x\):
\[\frac{e^{5x}-2e^{3x}+3}{e^x} = e^{4x} - 2e^{2x} + 3e^{-x}.\]
Integrate termwise:
\[\int e^{4x}dx=\tfrac{e^{4x}}{4},\; \int -2e^{2x}dx = -e^{2x},\; \int 3e^{-x}dx = -3e^{-x}.\]
So
\[\tfrac{e^{4x}}{4} - e^{2x} - 3e^{-x} + C.\]
\(\boxed{\tfrac{e^{4x}}{4}-e^{2x}-3e^{-x}+C}\)
5(iii) \(\displaystyle \int \dfrac{e^{5x}+e^{3x}}{e^x+e^{-x}}\,dx\)
Factor numerator and denominator:
\[\frac{e^{5x}+e^{3x}}{e^x+e^{-x}} = \frac{e^{3x}(e^{2x}+1)}{e^{-x}(e^{2x}+1)} = e^{4x}.\]
Integrate:
\[\int e^{4x}dx = \tfrac{e^{4x}}{4} + C.\]
\(\boxed{\tfrac{e^{4x}}{4}+C}\)
5(iv) \(\displaystyle \int (e^{2\log x}-2e^{-3\log x})\,dx\)
Use \(e^{k\ln x}=x^k\):
\[e^{2\ln x}=x^2,\quad e^{-3\ln x}=x^{-3}.\]
So integrand \(=x^2-2x^{-3}\), integrate:
\[\int x^2dx=\tfrac{x^3}{3},\quad \int -2x^{-3}dx = x^{-2}.\]
Final: \(\tfrac{x^3}{3}+x^{-2}+C.\)
\(\boxed{\tfrac{x^3}{3}+\dfrac{1}{x^2}+C}\)
5(v) \(\displaystyle \int e^{-4x}\,dx\)
\[\int e^{-4x}dx = -\tfrac{1}{4}e^{-4x} + C.\]
\(\boxed{-\dfrac{1}{4}e^{-4x}+C}\)
6. Integrate.
6(i) \(\displaystyle \int 5^{2x} dx\)
Rewrite base: \(5^{2x}=e^{2x\ln5}\). Integrate:
\[\int e^{2x\ln5}dx = \dfrac{1}{2\ln5}e^{2x\ln5} + C = \dfrac{5^{2x}}{2\ln5} + C.\]
\(\boxed{\dfrac{5^{2x}}{2\ln5}+C}\)
6(ii) \(\displaystyle \int (x^4+4^x)dx\)
Integrate termwise:
\[\int x^4dx=\tfrac{x^5}{5},\quad \int 4^x dx = \dfrac{4^x}{\ln4}.\]
So \(\tfrac{x^5}{5}+\dfrac{4^x}{\ln4}+C.\)
\(\boxed{\tfrac{x^5}{5}+\dfrac{4^x}{\ln4}+C}\)
6(iii) \(\displaystyle \int (e^{x\log a}+e^{a\log x})dx\)
Interpret logs as natural log: \(e^{x\ln a}=a^x,\; e^{a\ln x}=x^a\).
Integrate:
\[\int a^x dx = \dfrac{a^x}{\ln a},\quad \int x^a dx = \dfrac{x^{a+1}}{a+1}.\]
Final: \(\dfrac{a^x}{\ln a}+\dfrac{x^{a+1}}{a+1}+C.\)
\(\boxed{\dfrac{a^x}{\ln a}+\dfrac{x^{a+1}}{a+1}+C}\)
7. Evaluate the following integrals (trigonometric).
7(i) \(\displaystyle \int \sin^2 2x\,dx\)
Use identity: \(\sin^2 t=\tfrac{1-\cos2t}{2}.\)
\[\sin^2 2x=\tfrac{1-\cos4x}{2}.\]
\[\int \sin^2 2x\,dx = \tfrac{1}{2}\int 1\,dx - \tfrac{1}{2}\int\cos4x\,dx = \tfrac{x}{2} - \tfrac{1}{2}\cdot\tfrac{\sin4x}{4}+C = \tfrac{x}{2} - \tfrac{\sin4x}{8}+C.\]
\(\boxed{\tfrac{x}{2}-\tfrac{\sin4x}{8}+C}\)
7(ii) \(\displaystyle \int \dfrac{2\tan x}{1+\tan^2x}\,dx\)
Note \(1+\tan^2x=\sec^2x\). Simplify:
\[\frac{2\tan x}{1+\tan^2x}=2\tan x\cos^2x = 2\cdot\frac{\sin x}{\cos x}\cdot\cos^2x = 2\sin x\cos x = \sin2x.\]
\[\int \sin2x\,dx = -\tfrac{\cos2x}{2} + C.\]
\(\boxed{-\tfrac{\cos2x}{2}+C}\)
7(iii) \(\displaystyle \int \sec x(\sec x+\tan x)\,dx\)
Expand integrand:
\[\sec x(\sec x+\tan x)=\sec^2x+\sec x\tan x.\]
Split integral:
\[\int \sec^2x\,dx + \int \sec x\tan x\,dx.\]
Recognise derivatives: \(\frac{d}{dx}\tan x=\sec^2x,\; \frac{d}{dx}\sec x=\sec x\tan x.\)
So
\[\int\sec^2x\,dx=\tan x,\quad \int\sec x\tan x\,dx=\sec x.\]
Final: \(\tan x+\sec x + C.\)
\(\boxed{\tan x+\sec x+C}\)
7(iv) \(\displaystyle \int cosec x(cosec x-\cot x)\,dx\)
Expand:
\[cosec x(cosec x-\cot x)=cosec^2x-cosec x\cot x.\]
Split and recognise derivatives:
\[\int cosec^2x\,dx = -\cot x,\quad \int cosec x\cot x\,dx = -cosec x.\]
So overall: \(-\cot x + cosec x + C.\)
\(\boxed{-\cot x+cosec x+C}\)
7(v) \(\displaystyle \int \cos^2 x\,dx\) (radians)
Identity: \(\cos^2x=\tfrac{1+\cos2x}{2}\).
\[\int\cos^2x\,dx = \tfrac{1}{2}\int 1\,dx + \tfrac{1}{2}\int\cos2x\,dx = \tfrac{x}{2} + \tfrac{1}{2}\cdot\tfrac{\sin2x}{2} + C = \tfrac{x}{2}+\tfrac{\sin2x}{4}+C.\]
\(\boxed{\tfrac{x}{2}+\tfrac{\sin2x}{4}+C}\)
7(vi) \(\displaystyle \int \cot^2\theta\,d\theta\)
Rewrite: \(\cot^2\theta=\csc^2\theta-1\).
\[\int\cot^2\theta\,d\theta = \int\csc^2\theta d\theta - \int 1 d\theta = -\cot\theta - \theta + C.\]
\(\boxed{-\cot\theta-\theta + C}\)
7(vii) \(\displaystyle \int \cos^2\theta\,d\theta\)
Same as (v) with variable \(\theta\):
\[\int\cos^2\theta\,d\theta = \tfrac{\theta}{2} + \tfrac{\sin2\theta}{4} + C.\]
\(\boxed{\tfrac{\theta}{2}+\tfrac{\sin2\theta}{4}+C}\)
7(viii) \(\displaystyle \int \dfrac{1+\tan^2x}{1+\cot^2x}\,dx\)
Simplify using identities: \(1+\tan^2x=\sec^2x,\;1+\cot^2x=\csc^2x\).
\[\frac{1+\tan^2x}{1+\cot^2x}=\frac{\sec^2x}{\csc^2x}=\tan^2x.\]
Integrate:
\[\int\tan^2x\,dx = \int(\sec^2x-1)dx = \tan x - x + C.\]
\(\boxed{\tan x - x + C}\)
7(ix) \(\displaystyle \int \dfrac{\sec x + 2\cot^2x + \cos^2x}{\cos x}\,dx\)
Split the fraction:
\[\frac{\sec x}{\cos x} + 2\frac{\cot^2x}{\cos x} + \frac{\cos^2x}{\cos x}.\]
Simplify each term:
\[\frac{\sec x}{\cos x}=\sec^2x,\quad \frac{\cot^2x}{\cos x}=\frac{\cos x}{\sin^2x},\quad \frac{\cos^2x}{\cos x}=\cos x.\]
So integrand becomes \(\sec^2x + 2\dfrac{\cos x}{\sin^2x} + \cos x\). Integrate termwise:
\[\int\sec^2x\,dx=\tan x.\]
For middle term put \(u=\sin x\Rightarrow du=\cos x dx\):
\[2\int \dfrac{\cos x}{\sin^2x}dx = 2\int u^{-2}du = 2(-u^{-1}) = -2\csc x.\]
Last term: \(\int\cos x dx = \sin x.\) Combine:
\[\tan x - 2\csc x + \sin x + C.\]
\(\boxed{\tan x - 2\csc x + \sin x + C}\)
7(x) \(\displaystyle \int \dfrac{\sec^2x}{\cos x\cdot\sec^2x}\,dx\)
Cancel \(\sec^2x\):
\[\frac{\sec^2x}{\cos x\cdot\sec^2x} = \frac{1}{\cos x} = \sec x.\]
So
\[\int \sec x dx = \ln|\sec x + \tan x| + C.\]
\(\boxed{\ln|\sec x + \tan x| + C}\)
8. If \(\dfrac{dy}{dx}=2x^2-3\) at any point \((x,y)\) on a curve, find the equation of the family of curves.
8. Integrate
Integrate RHS:
\[y=\int(2x^2-3)dx = 2\int x^2dx - 3\int dx = 2\cdot\tfrac{x^3}{3} - 3x + C = \tfrac{2}{3}x^3 - 3x + C.\]
\(\boxed{y=\tfrac{2}{3}x^3 - 3x + C}\)
