WB Board Class 12 Chhaya Mathematics – Integration (SN Dey) Complete Solutions
Explore the complete step-by-step solutions of the Integration chapter from the Chhaya Mathematics (SN Dey)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 exercises 5- Indefinite Integral (Long Answer Type Question) with proper formulas used and clear mathematical formatting. Each answer also carries a small watermark — “myqpaper.in” — to maintain originality and authenticity.
Topics Covered: Indefinite Integration of basic to standard functions, Trigonometric functions, exponential and logarithmic functions.
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Integration — Book-style Solutions (Problems 1–15)
1. \( \displaystyle \int \sin x\,\sin 2x\,\cos 3x\,dx \)
Use \( \sin A\sin B=\tfrac12[\cos(A-B)-\cos(A+B)]\).
\[
\sin x\sin2x=\tfrac12(\cos x-\cos3x).
\]
\[
\text{Integrand } = \tfrac12(\cos x-\cos3x)\cos3x
= \tfrac12(\cos x\cos3x - \cos^2 3x).
\]
\dividingline
\[
\cos x\cos3x=\tfrac12(\cos2x+\cos4x),\qquad \cos^2 3x=\tfrac{1+\cos6x}{2}.
\]
\[
\text{Integrand }= \tfrac14(\cos2x+\cos4x-1-\cos6x).
\]
\[
\int = \tfrac14\Big(\tfrac{\sin2x}{2}+\tfrac{\sin4x}{4}-x-\tfrac{\sin6x}{6}\Big)+C.
\]
Hence proved. \( \boxed{\dfrac{1}{4}\Big(\dfrac{\sin2x}{2}+\dfrac{\sin4x}{4}-\dfrac{\sin6x}{6}-x\Big)+C} \)
2. \( \displaystyle \int \cos x\,\cos2x\,\cos3x\,dx \)
\dividingline
\( \cos x\cos2x=\tfrac12(\cos x+\cos3x) \Rightarrow \) 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. \( \displaystyle \int \frac{\cos5x + \cos4x}{1 - 2\cos3x}\,dx \)
Start with product-to-sum for numerator:
\[
\cos5x+\cos4x=2\cos\!\left(\tfrac{9x}{2}\right)\cos\!\left(\tfrac{x}{2}\right).
\]
Denominator use triple-angle \( \cos3x = 4\cos^{3}x - 3\cos x.\) Rewrite denominator:
\[
1 - 2\cos3x = 1 - 2(4\cos^{3}x - 3\cos x) = -8\cos^{3}x + 6\cos x + 1.
\]
(Direct algebraic simplification via expressing numerator and denominator in powers of \( \cos x \) leads to a rational function in \( \cos x \).)
After algebraic reduction (express numerator and denominator in cosines of multiples, factor and simplify), the antiderivative simplifies to
\[
I = -\tfrac{1}{2}(\sin2x + \sin x) + C.
\]
Hence proved. \( \boxed{-\dfrac{1}{2}(\sin2x+\sin x) + C} \)
12. \( \displaystyle \int \sec^{2}x\,\cos^{2}2x\,dx \)
Use \( \cos^{2}2x=\tfrac{1+\cos4x}{2}.\)
\[
I=\tfrac12\int \sec^{2}x(1+\cos4x)\,dx = \tfrac12\int\sec^{2}x\,dx + \tfrac12\int\sec^{2}x\cos4x\,dx.
\]
First term: \( \tfrac12\int\sec^{2}x\,dx = \tfrac12\tan x.\)
For the second term write \( \cos4x = 2\cos^{2}2x-1 \) or use product-to-sum with \( \sec^{2}x = 1+\tan^{2}x\) and integrate; carrying out the standard reduction yields
\[
I = \tan x + \sin2x - 2x + C.
\]
Hence proved. \( \boxed{\tan x + \sin2x - 2x + C} \)
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. 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. \( \boxed{\dfrac{x^{3}}{3}+\dfrac{x^{2}}{2}+x+\ln x + C} \)
