Chhaya Mathematics- Indefinite Integral [Part-3]

WB Board Class 12 Mathematics – Integration (SN Dey) Solutions

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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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1. \( \displaystyle \int \sin x\,\sin 2x\,\cos 3x\,dx \)

Use \( \sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]\).

\[ \sin x\sin2x=\tfrac12(\cos x-\cos3x). \] \[ \text{Integrand } = \tfrac12(\cos x-\cos3x)\cos3x = \tfrac12(\cos x\cos3x - \cos^2 3x). \] \[ \cos x\cos3x=\tfrac12(\cos2x+\cos4x),\qquad \cos^2 3x=\tfrac{1+\cos6x}{2}. \] \[ \text{Integrand }= \tfrac14(\cos2x+\cos4x-1-\cos6x). \] \[ \int = \tfrac14\Big(\tfrac{\sin2x}{2}+\tfrac{\sin4x}{4}-x-\tfrac{\sin6x}{6}\Big)+C. \]

Hence proved. \( \boxed{\dfrac{1}{4}\Big(\dfrac{\sin2x}{2}+\dfrac{\sin4x}{4}-\dfrac{\sin6x}{6}-x\Big)+C} \)

2. \( \displaystyle \int \cos x\,\cos2x\,\cos3x\,dx \)

\( \cos x\cos2x=\tfrac12(\cos x+\cos3x) \Rightarrow \) integrand \(=\tfrac12(\cos x\cos3x + \cos^2 3x).\) \[ \cos x\cos3x=\tfrac12(\cos2x+\cos4x),\qquad \cos^2 3x=\tfrac{1+\cos6x}{2}. \] \[ \text{Integrand }= \tfrac14(\cos2x+\cos4x+1+\cos6x). \] \[ \int = \tfrac14\Big(x+\tfrac{\sin2x}{2}+\tfrac{\sin4x}{4}+\tfrac{\sin6x}{6}\Big)+C. \]

Hence proved. \( \boxed{\dfrac{1}{4}\Big(x+\dfrac{\sin2x}{2}+\dfrac{\sin4x}{4}+\dfrac{\sin6x}{6}\Big)+C} \)

3. \( \displaystyle \int \sin x\,\cos2x\,\cos3x\,dx \)

\( \cos2x\cos3x=\tfrac12(\cos x+\cos5x) \Rightarrow \) integrand \(=\tfrac12(\sin x\cos x+\sin x\cos5x).\) \[ \sin x\cos x=\tfrac12\sin2x,\quad \sin x\cos5x=\tfrac12(\sin6x-\sin4x). \] \[ \text{Integrand }=\tfrac14(\sin2x+\sin6x-\sin4x). \] \[ \int = \tfrac14\Big(-\tfrac{\cos2x}{2}-\tfrac{\cos6x}{6}+\tfrac{\cos4x}{4}\Big)+C. \]

Hence proved. \( \boxed{-\dfrac{1}{4}\Big(\dfrac{\cos2x}{2}-\dfrac{\cos4x}{4}+\dfrac{\cos6x}{6}\Big)+C} \)

4. \( \displaystyle \int \sin x\,\sin2x\,\sin3x\,dx \)

\( \sin2x\sin3x=\tfrac12(\cos x-\cos5x) \Rightarrow \) integrand \(=\tfrac12(\sin x\cos x-\sin x\cos5x).\) \[ \sin x\cos x=\tfrac12\sin2x,\quad \sin x\cos5x=\tfrac12(\sin6x-\sin4x). \] \[ \text{Integrand }=\tfrac14(\sin2x+\sin4x-\sin6x). \] \[ \int=\tfrac14\Big(-\tfrac{\cos2x}{2}-\tfrac{\cos4x}{4}+\tfrac{\cos6x}{6}\Big)+C = -\tfrac{1}{48}(6\cos2x+3\cos4x-2\cos6x)+C. \]

Hence proved. \( \boxed{-\dfrac{1}{48}(6\cos2x+3\cos4x-2\cos6x)+C} \)

5. \( \displaystyle \int \sin^{2}x\,\cos2x\,dx \)

\( \sin^{2}x=\tfrac{1-\cos2x}{2} \Rightarrow \sin^{2}x\cos2x=\tfrac12\cos2x - \tfrac12\cos^22x.\) \[ \cos^22x=\tfrac{1+\cos4x}{2}\Rightarrow \text{integrand }=\tfrac12\cos2x - \tfrac14 - \tfrac14\cos4x. \] \[ \int = \tfrac{\sin2x}{4} - \tfrac{x}{4} - \tfrac{\sin4x}{16} + C = \tfrac14\Big(\sin2x - \tfrac{\sin4x}{4} - x\Big)+C. \]

Hence proved. \( \boxed{\dfrac{1}{4}\Big(\sin2x - \dfrac{\sin4x}{4} - x\Big) + C} \)

6. \( \displaystyle \int 2\cos^{2}2x\,\sin4x\,dx \)

Use \( \sin4x=2\sin2x\cos2x.\) Then integrand \(= 2\cos^{2}2x \cdot 2\sin2x\cos2x = 4\cos^{3}2x\sin2x.\) Put \( t=\cos2x\Rightarrow dt=-2\sin2x\,dx.\) \[ I = 4\int t^{3}\sin2x\,dx = 4\int t^{3}\Big(-\tfrac12dt\Big) = -2\int t^{3}dt = -\tfrac12 t^{4}+C. \] \[ \therefore I = -\tfrac12\cos^{4}2x + C. \]

Hence proved. \( \boxed{-\dfrac{1}{2}\cos^{4}2x + C} \)

7. \( \displaystyle \int \frac{\cos2x - \cos2\alpha}{\cos x - \cos\alpha}\,dx \)

Use \( \cos2\theta=2\cos^{2}\theta-1.\) \[ \cos2x-\cos2\alpha = 2(\cos^{2}x-\cos^{2}\alpha) = 2(\cos x-\cos\alpha)(\cos x+\cos\alpha). \] \[ \therefore \frac{\cos2x-\cos2\alpha}{\cos x-\cos\alpha}=2(\cos x+\cos\alpha). \] \[ I= \int 2(\cos x+\cos\alpha)\,dx = 2\sin x + 2x\cos\alpha + C. \]

Hence proved. \( \boxed{2(\sin x + x\cos\alpha)+C} \)

8. \( \displaystyle \int \frac{\sin x - \cos2x}{1+\sin x}\,dx \)

Use \( \cos2x=1-2\sin^{2}x.\) \[ \sin x - \cos2x = \sin x - (1-2\sin^{2}x) = 2\sin^{2}x + \sin x - 1. \] Factor (put \( s=\sin x \)): \[ 2s^{2}+s-1=(2s-1)(s+1). \] \[ \text{So } \frac{\sin x - \cos2x}{1+\sin x} = \frac{(2s-1)(s+1)}{s+1} = 2s-1 = 2\sin x -1. \] \[ I=\int(2\sin x - 1)\,dx = -2\cos x - x + C. \]

Hence proved. \( \boxed{-(x+2\cos x)+C} \)

9. \( \displaystyle \int \frac{\cos x - \cos2x}{1 - \cos x}\,dx \)

Use \( \cos2x=2\cos^{2}x-1.\) \[ \cos x - \cos2x = \cos x - (2\cos^{2}x -1) = 1 + \cos x - 2\cos^{2}x. \] Put \( c=\cos x\): \(1 + c - 2c^{2}=(1-c)(1+2c).\) \[ \therefore \frac{\cos x - \cos2x}{1-\cos x} = \frac{(1-c)(1+2c)}{1-c}=1+2c=1+2\cos x. \] \[ I=\int(1+2\cos x)\,dx = x + 2\sin x + C. \]

Hence proved. \( \boxed{x+2\sin x + C} \)

10. \( \displaystyle \int \frac{\sin^{8}x - \cos^{8}x}{1 - 2\sin^{2}x\cos^{2}x}\,dx \)

Factor: \[ \sin^{8}x - \cos^{8}x = (\sin^{4}x - \cos^{4}x)(\sin^{4}x + \cos^{4}x). \] Now \( \sin^{4}-\cos^{4}=(\sin^{2}-\cos^{2})(\sin^{2}+\cos^{2})=\sin^{2}-\cos^{2}.\) Also \( \sin^{4}+\cos^{4} = 1 - 2\sin^{2}x\cos^{2}x.\) \[ \therefore \frac{\sin^{8}x - \cos^{8}x}{1-2\sin^{2}x\cos^{2}x} = \sin^{2}x - \cos^{2}x = -\cos2x. \] \[ I = \int -\cos2x\,dx = -\tfrac12\sin2x + C. \]

Hence proved. \( \boxed{-\dfrac{1}{2}\sin2x + C} \)

11. \( \displaystyle \int \frac{\cos5x + \cos4x}{1 - 2\cos3x}\,dx \)

\[ \begin{aligned} I &= \int \frac{\cos5x+\cos4x}{1-2\cos3x}\,dx \\[6pt] &= \int \frac{\sin3x\big(\cos5x+\cos4x\big)}{\sin3x\big(1-2\cos3x\big)}\,dx \\[6pt] \cos5x+\cos4x &= 2\cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)\\[6pt] \Rightarrow I &= \int \frac{\sin3x\cdot 2\cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)} {\sin3x - 2\sin3x\cos3x}\,dx \\[8pt] \sin3x &= 2\sin\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{3x}{2}\right),\qquad \sin6x = 2\sin3x\cos3x\\[6pt] \sin3x - \sin6x &= \sin3x - 2\sin3x\cos3x = \sin3x(1-2\cos3x)\\[6pt] \sin3x-\sin6x &= -2\cos\!\left(\tfrac{9x}{2}\right)\sin\!\left(\tfrac{3x}{2}\right) \quad\text{(use } \sin A-\sin B = 2\cos\!\tfrac{A+B}{2}\sin\!\tfrac{A-B}{2})\\[8pt] \Rightarrow I &= \int \frac{2\cdot 2\sin\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{3x}{2}\right) \cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)} {-2\cos\!\left(\tfrac{9x}{2}\right)\sin\!\left(\tfrac{3x}{2}\right)}\,dx \\[8pt] &= \int -2\cos\!\left(\tfrac{3x}{2}\right)\cos\!\left(\tfrac{x}{2}\right)\,dx \\[8pt] -2\cos A\cos B &= -\big(\cos(A+B)+\cos(A-B)\big)\\[6pt] \Rightarrow I &= -\int\big(\cos2x+\cos x\big)\,dx \\[6pt] &= -\Big(\tfrac{\sin2x}{2} + \sin x\Big) + C \end{aligned} \]

\[ \boxed{\;-\,\Big(\tfrac{\sin2x}{2}+\sin x\Big) + C\;} \]

12. \( \displaystyle \int \sec^{2}x\,\cos^{2}2x\,dx \)

\[ \begin{aligned} I &= \int \sec^{2}x\,\cos^{2}2x\,dx \\[6pt] &= \int \frac{\cos^{2}2x}{\cos^{2}x}\,dx \\[8pt] \text{Use } \cos2x &= 2\cos^{2}x - 1 \\[6pt] \cos^{2}2x &= (2\cos^{2}x-1)^{2} = 4\cos^{4}x - 4\cos^{2}x + 1 \\[8pt] \therefore \frac{\cos^{2}2x}{\cos^{2}x} &= \frac{4\cos^{4}x - 4\cos^{2}x + 1}{\cos^{2}x} = 4\cos^{2}x - 4 + \frac{1}{\cos^{2}x} \\[8pt] I &= \int \big(4\cos^{2}x - 4 + \sec^{2}x\big)\,dx \\[8pt] &= 4\int \cos^{2}x\,dx - 4\int dx + \int \sec^{2}x\,dx \\[6pt] \cos^{2}x &= \tfrac{1+\cos2x}{2} \;\Rightarrow\; \int \cos^{2}x\,dx = \tfrac{x}{2}+\tfrac{\sin2x}{4} \\[8pt] \Rightarrow I &= 4\Big(\tfrac{x}{2}+\tfrac{\sin2x}{4}\Big) - 4x + \tan x + C \\[8pt] &= 2x + \sin2x - 4x + \tan x + C \\[6pt] &= -2x + \sin2x + \tan x + C \end{aligned} \]

\[ \boxed{\; -2x + \sin2x + \tan x + C \;} \]

13. \( \displaystyle \int \frac{\cos^{4}x}{\sin^{2}x}\,dx \)

Write \( \cos^{4}x=(1-\sin^{2}x)^{2}=1-2\sin^{2}x+\sin^{4}x.\) \[ \frac{\cos^{4}x}{\sin^{2}x} = \frac{1}{\sin^{2}x} - 2 + \sin^{2}x = \csc^{2}x - 2 + \sin^{2}x. \] Integrate termwise: \[ \int \csc^{2}x\,dx = -\cot x,\quad \int -2\,dx = -2x,\quad \int \sin^{2}x\,dx = \tfrac{x}{2} - \tfrac{\sin2x}{4}. \] \[ I = -\cot x - 2x + \tfrac{x}{2} - \tfrac{\sin2x}{4} + C = -\cot x - \tfrac{3x}{2} - \tfrac{\sin2x}{4} + C. \]

Hence proved. \( \boxed{-\cot x - \dfrac{3x}{2} - \dfrac{\sin2x}{4} + C} \)

14. \( \displaystyle \int \sin(mx)\sin(nx)\,dx \quad (m^{2}\ne n^{2}) \)

Use identity \( \sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)].\) \[ \sin(mx)\sin(nx)=\tfrac12[\cos((m-n)x)-\cos((m+n)x)]. \] Integrate: \[ I=\tfrac12\Big(\frac{\sin((m-n)x)}{m-n}-\frac{\sin((m+n)x)}{m+n}\Big)+C. \]

Hence proved. \( \boxed{\tfrac12\Big(\dfrac{\sin((m-n)x)}{m-n}-\dfrac{\sin((m+n)x)}{m+n}\Big)+C} \)

15 i. Find \( f(x) \) such that \( f'(x)=\dfrac{x^{4}-1}{x(x-1)} \)

Divide polynomial: \[ \frac{x^{4}-1}{x(x-1)} = \frac{x^{4}-1}{x^{2}-x} = x^{2}+x+1 + \frac{1}{x}. \] (Check: multiply back \(x(x-1)(x^{2}+x+1+\tfrac1x) = x^{4}-1\).) Integrate: \[ f(x)=\int\Big(x^{2}+x+1+\frac{1}{x}\Big)dx = \frac{x^{3}}{3}+\frac{x^{2}}{2}+x+\ln x + C. \]

Hence proved. \( \boxed{\dfrac{x^{3}}{3}+\dfrac{x^{2}}{2}+x+\ln x + C} \)

15(ii). Find \( f(x) \) such that \( f'(x) = 2^{2x-1} + 2^{1-2x} \).

\[ \begin{aligned} f(x) &= \int \big(2^{2x-1} + 2^{1-2x}\big)\,dx \\[8pt] &= \tfrac{1}{2}\int 2^{2x}\,dx + 2\int 2^{-2x}\,dx \\[8pt] &= \tfrac{1}{2}\cdot\frac{2^{2x}}{2\ln2} + 2\cdot\frac{2^{-2x}}{-2\ln2} + C \\[8pt] &= \frac{2^{2x}}{4\ln2} - \frac{2^{-2x}}{\ln2} + C \\[8pt] &= \frac{1}{4\ln2}\big(2^{2x} - 4\cdot2^{-2x}\big) + C \end{aligned} \]

\[ \boxed{\, f(x) = \dfrac{2^{2x}}{4\ln2} - \dfrac{2^{-2x}}{\ln2} + C \,} \]

Next Topics in Indefinite Integration(Click to View)

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