CBSE Class 12 Mathematics Integration Worksheet Set 04

CBSE Class 12 Mathematics Integration Worksheet (2). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Integration (Indefinite Integrals)

Question. \( I = \int \frac{\cos(2x)-\cos(2\alpha)}{\cos x-\cos \alpha} \, dx \)
Answer: \( I = \int \frac{\cos(2x)-\cos(2\alpha)}{\cos x-\cos \alpha} \, dx \)
\( = \int \frac{(2\cos^2 x-1)-(2\cos^2 \alpha-1)}{\cos x-\cos \alpha} \, dx \)
\( = 2 \int \frac{\cos^2 x-\cos^2 \alpha}{\cos x-\cos \alpha} \, dx \)
\( = 2 \int \frac{(\cos x+\cos \alpha)(\cos x-\cos \alpha)}{\cos x-\cos \alpha} \, dx \)
\( I = 2[\sin x + x \cos \alpha] + c \)

Question. \( I = \int \frac{1+\cos(4x)}{\cot x-\tan x} \, dx \)
Answer: \( I = \int \frac{1+\cos(4x)}{\cot x-\tan x} \, dx \)
\( = \int \frac{2\cos^2(2x)}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}} \, dx \)
\( = \int \frac{2\cos^2(2x)}{\frac{\cos^2 x-\sin^2 x}{\sin x \cos x}} \, dx \)
\( = \int \frac{2\sin x \cos x \cdot \cos^2(2x)}{\cos^2 x-\sin^2 x} \, dx \)
\( = \int \frac{\sin(2x) \cdot \cos^2(2x)}{\cos(2x)} \, dx \)
\( = \int \sin(2x) \cdot \cos(2x) \, dx \)
\( = \frac{1}{2} \int 2\sin(2x) \cdot \cos(2x) \, dx \)
\( = \frac{1}{2} \int \sin(4x) \, dx \)
\( = \frac{1}{2} \left(-\frac{\cos(4x)}{4}\right) + c \)
\( I = -\frac{1}{8}\cos(4x) + c \)

Type : Rationalize : \( \int \frac{1}{1 \pm \sin x} \, dx, \int \frac{1}{1 \pm \cos x} \, dx \)

Question. \( I = \int \frac{1}{1+\sin x} \, dx \)
Answer: \( I = \int \frac{1}{1+\sin x} \, dx \)
Rationalize
\( = \int \frac{1}{1+\sin x} \times \frac{1-\sin x}{1-\sin x} \, dx \)
\( = \int \frac{1-\sin x}{\cos^2 x} \, dx \)
Separate
\( I = \int \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \, dx \)
\( = \int \sec^2 x - \tan x \sec x \, dx \)
\( I = \tan x - \sec x + c \)

Question. \( I = \int \frac{\sin x}{1-\sin x} \, dx \)
Answer: \( I = \int \frac{\sin x}{1-\sin x} \, dx \)
Rationalize
\( = \int \frac{\sin x(1+\sin x)}{(1-\sin x)(1+\sin x)} \, dx \)
\( = \int \frac{\sin x+\sin^2 x}{\cos^2 x} \, dx \)
Separate
\( = \int \tan x \sec x + \tan^2 x \, dx \)
\( = \int \tan x \sec x + \sec^2 x - 1 \, dx \)
\( I = \sec x + \tan x - x + c \)

Question. \( I = \int \frac{\cos x-\cos(2x)}{1-\cos x} \, dx \)
Answer: \( I = \int \frac{\cos x-\cos(2x)}{1-\cos x} \, dx \)
\( = \int \frac{\cos x-(2\cos^2 x-1)}{1-\cos x} \, dx \)
\( = -\int \frac{2\cos^2 x-\cos x-1}{1-\cos x} \, dx \)
\( = -\int \frac{(2\cos x+1)(\cos x-1)}{1-\cos x} \, dx \)
\( = -\int \frac{(2\cos x+1)(\cos x-1)}{-(\cos x-1)} \, dx \)
\( = \int 2\cos x + 1 \, dx \)
\( I = 2\sin x + x + c \)

Question. \( I = \int \tan^{-1}\sqrt{\frac{1-\cos(2x)}{1+\cos(2x)}} \, dx \)
Answer: \( I = \int \tan^{-1}\sqrt{\frac{1-\cos(2x)}{1+\cos(2x)}} \, dx \)
\( = \int \tan^{-1}\sqrt{\frac{2\sin^2 x}{2\cos^2 x}} \, dx \)
\( = \int \tan^{-1}(\tan x) \, dx \)
\( = \int x \, dx \)
\( I = \frac{x^2}{2} + c \)

Question. \( I = \int \tan^{-1}(\sec x + \tan x) \, dx \)
Answer: \( I = \int \tan^{-1}(\sec x + \tan x) \, dx \)
\( = \int \tan^{-1}\left( \frac{1}{\cos x} + \frac{\sin x}{\cos x} \right) \, dx \)
\( = \int \tan^{-1}\left( \frac{1+\sin x}{\cos x} \right) \, dx \)
\( = \int \tan^{-1}\left[ \frac{1+\cos(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x)} \right] \, dx \)
\( = \int \tan^{-1}\left( \frac{2\cos^2(\frac{\pi}{4}-\frac{x}{2})}{2\sin(\frac{\pi}{4}-\frac{x}{2})\cos(\frac{\pi}{4}-\frac{x}{2})} \right) \, dx \)
\( = \int \tan^{-1}\left( \cot\left(\frac{\pi}{4}-\frac{x}{2}\right) \right) \, dx \)
\( = \int \tan^{-1}\left[ \tan\left( \frac{\pi}{2} - \left(\frac{\pi}{4}-\frac{x}{2}\right) \right) \right] \, dx \)
\( = \int \frac{\pi}{2} - \frac{\pi}{4} + \frac{x}{2} \, dx \)
\( = \int \frac{\pi}{4} + \frac{x}{2} \, dx \)
\( I = \frac{\pi x}{4} + \frac{x^2}{4} + c \)

Question. \( I = \int \frac{\cos(2x)}{(\cos x+\sin x)^2} \, dx \)
Answer: \( I = \int \frac{\cos(2x)}{(\cos x+\sin x)^2} \, dx \)
\( = \int \frac{\cos^2 x-\sin^2 x}{(\cos x+\sin x)^2} \, dx \)
\( = \int \frac{(\cos x+\sin x)(\cos x-\sin x)}{(\cos x+\sin x)^2} \, dx \)
\( = \int \frac{\cos x-\sin x}{\cos x+\sin x} \, dx \)
put \( \cos x + \sin x = t \)
\( (-\sin x + \cos x) \, dx = dt \)
\( = \int \frac{dt}{t} \)
\( = \log |t| + c \)
\( I = \log |\cos x + \sin x| + c \)

Question. \( I = \int \frac{\cos x-\sin x}{1+\sin(2x)} \, dx \)
Answer: \( I = \int \frac{\cos x-\sin x}{1+\sin(2x)} \, dx \)
\( = \int \frac{\cos x-\sin x}{\sin^2 x+\cos^2 x+2\sin x \cos x} \, dx \)
\( = \int \frac{\cos x-\sin x}{(\sin x+\cos x)^2} \, dx \)
put \( \sin x + \cos x = t \)
\( (\cos x - \sin x) \, dx = dt \)
\( = \int \frac{dt}{t^2} \)
\( = -\frac{1}{t} + c \)
\( I = -\frac{1}{\sin x+\cos x} + c \)

Question. \( I = \int \frac{1-\tan x}{1+\tan x} \, dx \)
Answer: \( I = \int \frac{1-\tan x}{1+\tan x} \, dx \)
\( = \int \frac{1-\frac{\sin x}{\cos x}}{1+\frac{\sin x}{\cos x}} \, dx \)
\( = \int \frac{\cos x-\sin x}{\cos x+\sin x} \, dx \)
put \( \cos x + \sin x = t \)
\( (-\sin x + \cos x) \, dx = dt \)
\( I = \int \frac{dt}{t} \)
\( I = \log | \sin x + \cos x | + c \)

Question. \( I = \int \frac{\sin^8 x-\cos^8 x}{1-2\sin^2 x \cdot \cos^2 x} \, dx \)
Answer: \( I = \int \frac{\sin^8 x-\cos^8 x}{1-2\sin^2 x \cdot \cos^2 x} \, dx \)
\( = \int \frac{(\sin^4 x)^2-(\cos^4 x)^2}{(\sin^2 x+\cos^2 x)^2-2\sin^2 x \cdot \cos^2 x} \, dx \) ......{\( 1 = \sin^2 x + \cos^2 x \)}
\( = \int \frac{(\sin^4 x+\cos^4 x)(\sin^4 x-\cos^4 x)}{\sin^4 x+\cos^4 x+2\sin^2 x \cdot \cos^2 x-2\sin^2 x \cdot \cos^2 x} \, dx \)
\( = \int \frac{(\sin^4 x+\cos^4 x)(\sin^4 x-\cos^4 x)}{\sin^4 x+\cos^4 x} \, dx \)
\( = \int (\sin^2 x + \cos x)(\sin^2 x - \cos^2 x) \, dx \)
\( = \int (1)[-\cos(2x)] \, dx \) ......{\( \cos(2x) = \cos^2 x - \sin^2 x \)}
\( = -\int \cos(2x) \, dx \)
\( = -\frac{\sin(2x)}{2} + c \)

Question. \( I = \int \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} \, dx \)
Answer: \( I = \int \frac{1}{\sqrt{\sin^3 x \sin(x+\alpha)}} \, dx \)
\( = \int \frac{1}{\sqrt{\sin^3 x \cdot (\sin x\cos\alpha+\cos x\sin\alpha)}} \, dx \)
take \( \sin x \) common
\( = \int \frac{1}{\sqrt{\sin^4 x \cdot (\cos\alpha+\cot x\sin\alpha)}} \, dx \)
\( = \int \frac{1}{\sin^2 x \cdot (\cos\alpha+\cot x\sin\alpha)^{1/2}} \, dx \)
\( = \int \frac{\csc^2 x}{\sqrt{\cos\alpha+\cot x\sin\alpha}} \, dx \)
put \( \cos\alpha + \cot x\sin\alpha = t \)
\( -\csc^2 x \cdot \sin\alpha \, dx = dt \)
\( \csc^2 x \, dx = \frac{-dt}{\sin\alpha} \)
\( I = -\frac{1}{\sin\alpha} \int \frac{dt}{\sqrt{t}} \)
\( = -\frac{1}{\sin\alpha} \times 2\sqrt{t} + c \)
\( I = -\frac{1}{\sin\alpha} \cdot 2\sqrt{\cos\alpha + \cot x\sin\alpha} + c \)

Question. \( I = \int \frac{\sin(2x)}{(a+b\cos x)^2} \, dx \)
Answer: \( I = \int \frac{\sin(2x)}{(a+b\cos x)^2} \, dx \)
\( = 2\int \frac{\sin x \cdot \cos x}{(a+b\cos x)^2} \, dx \)
put \( a + b\cos x = t \)
\( -b\sin x \, dx = dt \)
\( \sin x \, dx = \frac{-dt}{b} \)
\( I = \frac{-2}{b} \int \frac{\cos x}{t^2} \, dt \)
\( = \frac{-2}{b^2} \int \frac{1}{t^2} \left( \frac{t-a}{b} \right) \, dt \)
\( = \frac{-2}{b^2} \int \frac{t-a}{t^2} \, dt \)
\( = \frac{-2}{b^2} \int \frac{1}{t} - \frac{a}{t^2} \, dt \)
\( = \frac{-2}{b^2} \left[ \log | t | + \frac{a}{t} \right] + c \)
\( I = \frac{-2}{b^2} \left[ \log | a + b\cos x | + \frac{a}{a+b\cos x} \right] + c \)

Question. \( I = \int \frac{1}{\sqrt{1-\sin x}} \, dx \)
Answer: \( I = \int \frac{1}{\sqrt{1-\sin x}} \, dx \)
\( = \int \frac{1}{\sqrt{1-\cos(\frac{\pi}{2}-x)}} \, dx \)
\( = \int \frac{1}{\sqrt{2\sin^2(\frac{\pi}{4}-\frac{x}{2})}} \, dx \)
\( = \frac{1}{\sqrt{2}} \int \csc\left(\frac{\pi}{4}-\frac{x}{2}\right) \, dx \)
\( = \frac{1}{\sqrt{2}} \cdot \log \left| \csc\left(\frac{\pi}{4}-\frac{x}{2}\right) - \cot\left(\frac{\pi}{4}-\frac{x}{2}\right) \right| \times (-2) + c \)
\( = -\sqrt{2} \cdot \log \left| \csc\left(\frac{\pi}{4}-\frac{x}{2}\right) - \cot\left(\frac{\pi}{4}-\frac{x}{2}\right) \right| + c \)

Type : \( \int \text{linear}\sqrt{\text{Linear}} \, dx, \int \frac{\text{linear}}{\sqrt{\text{Linear}}} \, dx, \int \frac{\text{linear}}{\sqrt{(\text{Linear})^n}} \, dx \)

Put Linear = \( t \) or \( t^2 \) (Or) make adjustments

Question. \( I = \int x\sqrt{x+2} \, dx \)
Answer: \( I = \int x\sqrt{x+2} \, dx \)
put \( x + 2 = t^2 \)
\( dx = 2t \, dt \)
\( I = 2\int x \cdot \sqrt{t^2} \cdot t \, dt \)
\( = 2\int (t^2-2) \cdot t \cdot t \, dt \)
\( = 2\int t^4 - 2t^2 \, dt \)
\( = 2 \left[ \frac{t^5}{5} - \frac{2t^3}{3} \right] + c \)
replacing \( t \) by \( (x+2)^{1/2} \)
\( = 2 \left[ \frac{(x+2)^{5/2}}{5} - 2\frac{(x+2)^{3/2}}{3} \right] + c \)

Alternate Method: (adjustment)
\( I = \int x\sqrt{x+2} \, dx \)
\( = \int (x+2-2)\sqrt{x+2} \, dx \)
\( = \int (x+2)^{3/2} - 2\sqrt{x+2} \, dx \)
\( = \frac{2}{5}(x+2)^{5/2} - 2 \times \frac{2}{3}(x+2)^{3/2} + c \)

Question. \( I = \int (7x-2)\sqrt{3x+2} \, dx \)
Answer: \( I = \int (7x-2)\sqrt{3x+2} \, dx \)
put \( 3x + 2 = t^2 \)
\( 3 \, dx = 2t \, dt \)
\( dx = \frac{2}{3}t \, dt \)
\( I = \frac{2}{3} \int (7x-2)\sqrt{t^2} \cdot t \, dt \)
\( = \frac{2}{3} \int \left[ 7\left(\frac{t^2-2}{3}\right) - 2 \right] t \cdot t \, dt \)
\( = \frac{2}{3} \int \frac{(7t^2-14-6)}{3} t^2 \, dt \)
\( = \frac{2}{9} \int 7t^4 - 20t^2 \, dt \)
\( = \frac{2}{9} \left[ \frac{7t^5}{5} - \frac{20t^3}{3} \right] + c \)
replacing \( t \) by \( (3x+2)^{1/2} \)
\( I = \frac{2}{9} \left[ \frac{7}{5}(3x+2)^{5/2} - \frac{20}{3}(3x+2)^{3/2} \right] + c \)

Question. \( I = \int \frac{2x+3}{(x-1)^2} \, dx \)
Answer: \( I = \int \frac{2x+3}{(x-1)^2} \, dx \)
put \( x - 1 = t \)
\( dx = dt \)
\( I = \int \frac{2x+3}{t^2} \, dt \)
\( = \int \frac{2(t+1)+3}{t^2} \, dt \)
\( = \int \frac{2t+5}{t^2} \, dt \)
\( = \int \frac{2}{t} + \frac{5}{t^2} \, dt \)
\( = 2\log | t | - \frac{5}{t} + c \)
\( I = 2\log | x-1 | - \frac{5}{x-1} + c \)

Question. \( I = \int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \, dx \)
Answer: \( I = \int \frac{1}{\sqrt{x+a}+\sqrt{x+b}} \, dx \)
Rationalize
\( I = \int \frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)} \, dx \)
\( = \int \frac{\sqrt{x+a}-\sqrt{x+b}}{a-b} \, dx \)
\( = \frac{1}{a-b} \int \sqrt{x+a}-\sqrt{x+b} \, dx \)
\( = \frac{1}{a-b} \left[ \frac{2}{3}(x+a)^{3/2} - \frac{2}{3}(x+b)^{3/2} \right] + c \)

Question. \( I = \int \frac{(x^4-x)^{1/4}}{x^5} \, dx \)
Answer: \( I = \int \frac{(x^4-x)^{1/4}}{x^5} \, dx \)
take \( x^4 \) common
\( = \int \frac{x\left(1-\frac{x}{x^4}\right)^{1/4}}{x^5} \, dx \)
\( = \int \frac{\left(1-\frac{1}{x^3}\right)^{1/4}}{x^4} \, dx \)
put \( 1 - \frac{1}{x^3} = t \)
\( \frac{3}{x^4} \, dx = dt \Rightarrow \frac{dx}{x^4} = \frac{dt}{3} \)
\( I = \frac{1}{3} \int t^{1/4} \, dt \)
\( = \frac{1}{3} \cdot \frac{4}{5}t^{5/4} + c \)
\( I = \frac{4}{15} \left( 1 - \frac{1}{x^3} \right)^{5/4} + c \)

Question. \( I = \int \frac{1}{x^2(x^4+1)^{3/4}} \, dx \)
Answer: \( I = \int \frac{1}{x^2(x^4+1)^{3/4}} \, dx \)
take \( x^4 \) common
\( = \int \frac{1}{x^2 \cdot x^3(1+\frac{1}{x^4})^{3/4}} \, dx \)
\( = \int \frac{1}{x^5(1+\frac{1}{x^4})^{3/4}} \, dx \)
Put \( 1 + \frac{1}{x^4} = t \)
\( \frac{-4}{x^5} \, dx = dt \Rightarrow \frac{dx}{x^5} = \frac{-dt}{4} \)
\( I = -\frac{1}{4} \int t^{-3/4} \, dt \)
\( = -\frac{1}{4} (t^{1/4} \times 4) + c \)
\( I = -\left(1 + \frac{1}{x^4}\right)^{1/4} + c \)

Question. \( I = \int \frac{1}{x\sqrt{ax-x^2}} \, dx \)
Answer: \( I = \int \frac{1}{x\sqrt{ax-x^2}} \, dx \)
take \( x^2 \) common
\( = \int \frac{1}{x \cdot x \sqrt{\frac{a}{x}-1}} \, dx \)
\( = \int \frac{1}{x^2 \sqrt{\frac{a}{x}-1}} \, dx \)
put \( \frac{a}{x} - 1 = t \)
\( \frac{-a}{x^2} \, dx = dt \Rightarrow \frac{1}{x^2} \, dx = -\frac{dt}{a} \)
\( I = -\frac{1}{a} \int \frac{dt}{\sqrt{t}} \)
\( = -\frac{1}{a} \times 2\sqrt{t} + c \)
\( I = -\frac{2}{a} \sqrt{\frac{a}{x}-1} + c \)

Question. \( I = \int \frac{1}{x(x^n+1)} \, dx \)
Answer: \( I = \int \frac{1}{x(x^n+1)} \, dx \)
take \( x^n \) common
\( = \int \frac{1}{x^{n+1}\left(1+\frac{1}{x^n}\right)} \, dx \)
put \( 1 + \frac{1}{x^n} = t \)
\( \frac{-n}{x^{n+1}} \, dx = dt \)
\( \Rightarrow \frac{1}{x^{n+1}} \, dx = -\frac{dt}{n} \)
\( I = -\frac{1}{n} \int \frac{1}{t} \, dt \)
\( = -\frac{1}{n} \log | t | + c \)
\( I = -\frac{1}{n} \log \left| 1 + \frac{1}{x^n} \right| + c \)