Chhaya Mathematics- Indefinite Integral [Part-1]

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.

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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: \[ y=\int(3x^2+2)\,dx = 3\int x^2dx + 2\int dx = 3\cdot\frac{x^3}{3} + 2x + C = x^3 + 2x + C. \]

\(\boxed{y=x^3+2x+C}\)

3(ii) \(dy=(ax+b)dx\)

Integrate: \[ y=\int (ax+b)dx = a\int xdx + b\int dx = \tfrac{a}{2}x^2 + bx + C. \]

\(\boxed{y=\tfrac{a}{2}x^2+bx+C}\)

3(iii) \(\dfrac{dy}{dx}=2\sin2x\)

Integrate: \[ y=\int 2\sin2x\,dx = 2\left(-\tfrac{1}{2}\cos2x\right) + C = -\cos2x + C. \]

\(\boxed{y=-\cos2x+C}\)

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) \(\displaystyle \int \dfrac{2}{\sqrt[4]{x^3}}\,dx\)

\(\dfrac{2}{\sqrt[4]{x^3}} = 2x^{-3/4}\). \[ \int 2x^{-3/4}dx = 2\cdot \frac{x^{1/4}}{1/4} + C = 8x^{1/4} + C = 8\sqrt[4]{x} + C. \]

\(\boxed{8\sqrt[4]{x}+C}\)

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) \(\displaystyle \int\Big(x-\dfrac{2}{x}\Big)^3 dx\)

Expand: \[ \Big(x-\tfrac{2}{x}\Big)^3 = x^3 -6x + \tfrac{12}{x} - \tfrac{8}{x^3}. \] Integrate: \[ \tfrac{x^4}{4} - 3x^2 + 12\ln|x| + 4x^{-2} + C. \]

\(\boxed{\tfrac{x^4}{4}-3x^2+12\ln|x|+\tfrac{4}{x^2}+C}\)

4(v) \(\displaystyle \int(x+3)(x^2-5)\,dx\)

Multiply: \((x+3)(x^2-5)=x^3+3x^2-5x-15\). 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: \[ \tfrac{2}{3}x^{3/2} + 4\sqrt{x} - 2x^{-1/2} + C. \]

\(\boxed{\tfrac{2}{3}x^{3/2}+4\sqrt{x}-\dfrac{2}{\sqrt{x}}+C}\)

5. Evaluate the following integrals.

5(i) \(\displaystyle \int e^{3\log x}\,dx\)

\(e^{3\ln x}=x^3\). 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) \(\displaystyle \int (e^{2\log x}-2e^{-3\log x})\,dx\)

\(e^{2\ln x}=x^2,\; e^{-3\ln x}=x^{-3}.\) 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) \(\displaystyle \int 5^{2x} dx\)

\(5^{2x}=e^{2x\ln5}\). So \(\int e^{2x\ln5}dx = \dfrac{5^{2x}}{2\ln5} + C.\)

\(\boxed{\dfrac{5^{2x}}{2\ln5}+C}\)

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) \(\displaystyle \int (e^{x\log a}+e^{a\log x})dx\)

\(e^{x\log a}=a^x,\; e^{a\log x}=x^a\). 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) \(\displaystyle \int \dfrac{2\tan x}{1+\tan^2x}\,dx\)

\(1+\tan^2x = \sec^2x\). So integrand \(= \dfrac{2\tan x}{\sec^2x} = 2\sin x\cos x = \sin2x\). \[ \int \sin2x\,dx = -\tfrac{\cos2x}{2} + C. \]

\(\boxed{-\tfrac{\cos2x}{2}+C}\)

7(iii) \(\displaystyle \int \sec x(\sec x+\tan x)\,dx\)

\(\sec x(\sec x+\tan x)=\sec^2x+\sec x\tan x\). \[ \int = \tan x + \sec x + C. \]

\(\boxed{\tan x + \sec x + C}\)

7(iv) \(\displaystyle \int \csc x(\csc x-\cot x)\,dx\)

\(\csc x(\csc x-\cot x)=\csc^2x-\csc x\cot x\). \[ \int = -\cot x + \csc x + C. \]

\(\boxed{-\cot x+\csc x + C}\)

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) \(\displaystyle \int \cot^2\theta\,d\theta\)

\(\cot^2\theta=\csc^2\theta-1\). \[ \int = -\cot\theta - \theta + C. \]

\(\boxed{-\cot\theta-\theta + C}\)

7(vii) \(\displaystyle \int \cos^2\theta\,d\theta\)

\(\cos^2\theta=\tfrac{1+\cos2\theta}{2}\). \[ \int = \tfrac{\theta}{2} + \tfrac{\sin2\theta}{4} + C. \]

\(\boxed{\tfrac{\theta}{2}+\tfrac{\sin2\theta}{4}+C}\)

7(viii) \(\displaystyle \int \dfrac{1+\tan^2x}{1+\cot^2x}\,dx\)

\(1+\tan^2x=\sec^2x,\;1+\cot^2x=\csc^2x\). So integrand \(=\dfrac{\sec^2x}{\csc^2x}=\tan^2x\). \[ \int \tan^2x\,dx = \tan x - x + C. \]

\(\boxed{\tan x - x + C}\)

7(ix) \(\displaystyle \int \dfrac{\sec x + 2\cot^2x + \cos^2x}{\cos x}\,dx\)

Simplify: \[ \frac{\sec x + 2\cot^2x + \cos^2x}{\cos x} = \sec^2x + 2\cot x\csc x + \cos x. \] Integrate termwise: \[ \tan x - 2\csc x + \sin x + C. \]

\(\boxed{\tan x - 2\csc x + \sin x + C}\)

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

\[ y=\int(2x^2-3)dx = \tfrac{2}{3}x^3 - 3x + C. \]

\(\boxed{y=\tfrac{2}{3}x^3 - 3x + C}\)

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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