CBSE Class 12 Mathematics Integration Worksheet Set 05

CBSE Class 12 Mathematics Integration Worksheet (3). 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 integral)

Question. \( I = \int \frac{x^2}{(a+bx)^2} \, dx \)
Answer: \( I = \int \frac{x^2}{(a+bx)^2} \, dx \)
put \( a+bx = t \)
\( b \, dx = dt \Rightarrow dx = \frac{dt}{b} \)
\( \therefore I = \frac{1}{b} \int \frac{x^2}{t^2} \, dt \)
\( = \frac{1}{b} \int \frac{(t-\frac{a}{b})^2}{t^2} \, dt \)
\( = \frac{1}{b} \cdot \frac{1}{b^2} \int \frac{t^2+a^2-2at}{t^2} \, dt \)
Separate
\( = \frac{1}{b^3} \int 1 + \frac{a^2}{t^2} - \frac{2a}{t} \, dt \)
\( = \frac{1}{b^3} \left[ t - \frac{a^2}{t} - 2a \log |t| \right] + c \)
\( = \frac{1}{b^3} \left[ (a+bx) - \frac{a^2}{a+bx} - 2a \log |a+bx| \right] + c \)

Type: When degree of Numerator \(\ge\) degree of Denominator then divide and write \( \int \frac{N}{D} \, dx = \int Q + \frac{R}{D} \, dx \)

Question. \( I = \int \frac{x^7}{x-1} \, dx \)
Answer: \( I = \int \frac{x^7}{x-1} \, dx \)
clearly degree of \( N' > \) degree of \( D' \) (then divide)
\( \therefore I = \int Q + \frac{R}{D} \, dt \)
\( = \int (x^6 - x^5 + x^4 - x^3 + x^2 - x + 1) - \frac{1}{x+1} \, dx \)
\( = \frac{x^7}{7} - \frac{x^6}{6} + \frac{x^5}{5} - \frac{x^4}{4} + \frac{x^3}{3} - \frac{x^2}{2} + x - \log |x+1| + c \)

Question. \( I = \int \frac{1}{x^{1/2}+x^{1/3}} \, dx \)
Answer: \( I = \int \frac{1}{x^{1/2}+x^{1/3}} \, dx \)
put \( x = t^6 \) .......{L.C.M of 2 & 3 = 6}
\( dx = 6t^5 \, dt \)
\( \therefore I = 6 \int \frac{t^5}{t^3+t^2} \, dt \)
\( = \int \frac{t^5}{t^2(t+1)} \, dt \)
\( = \int \frac{t^3}{t+1} \, dt \)
Degree of N > degree of D (then divide)
\( = \int (t^2 - t + 1) - \frac{1}{t+1} \, dt \)
\( I = 2t^3 - 3t^2 + 6t - 6 \log |t+1| + c \)
replacing \( t \) by \( x^{1/6} \)
\( \therefore I = 2x^{1/2} - 3x^{1/3} + 6x^{1/6} - 6 \log |x^{1/6}+1| + c \)

Question. \( I = \int \frac{e^{2x}-1}{e^{2x}+1} \, dx \)
Answer: \( I = \int \frac{e^{2x}-1}{e^{2x}+1} \, dx \)
take \( e^x \) common in N and D
\( = \int \frac{e^x(e^x-e^{-x})}{e^x(e^x+e^{-x})} \, dx \)
put \( e^x + e^{-x} = t \)
\( \therefore (e^x - e^{-x}) \, dx = dt \)
\( \therefore I = \int \frac{dt}{t} \)
\( = \log |t| + c \)
\( I = \log |e^x + e^{-x}| + c \)

Question. \( I = \int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} \, dx \)
Answer: \( I = \int \frac{\sqrt{\tan x}}{\sin x \cdot \cos x} \, dx \)
Divide N and D by \( \cos^2 x \)
\( I = \int \frac{\frac{\sqrt{\tan x}}{\cos^2 x}}{\frac{\sin x \cdot \cos x}{\cos^2 x}} \, dt \)
\( = \int \frac{\sqrt{\tan x} \cdot \sec^2 x}{\tan x} \)
put \( \tan x = t \)
\( \therefore \sec^2 x \, dx = dt \)
\( \therefore I = \int \frac{\sqrt{t}}{t} \, dt \)
\( = \int \frac{1}{\sqrt{t}} \, dt \)
\( I = 2\sqrt{t} + c \)
\( I = 2\sqrt{\tan x} + c \)

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

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

Question. \( I = \int 5^{x+\tan^{-1} x} \cdot \left( \frac{x^2+2}{x^2+1} \right) \, dx \)
Answer: \( I = \int 5^{x+\tan^{-1} x} \cdot \left( \frac{x^2+2}{x^2+1} \right) \, dx \)
Hint: put \( x + \tan^{-1} x = t \)
\( \left( 1 + \frac{1}{1+x^2} \right) \, dx = dt \)
\( \frac{5^{x+\tan^{-1} x}}{\log 5} + c \)

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

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

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

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

Question. \( I = \int \frac{e^{5\log x} - e^{4\log x}}{e^{3\log x} - e^{2\log x}} \, dx \)
Answer: \( I = \int \frac{e^{5\log x} - e^{4\log x}}{e^{3\log x} - e^{2\log x}} \, dx \)
\( = \int \frac{e^{\log x^5} - e^{\log x^4}}{e^{\log x^3} - e^{\log x^2}} \, dx \)
\( = \int \frac{x^5-x^4}{x^3-x^2} \, dx \) ......{\( \because e^{\log x} = x \)}
\( = \int \frac{x^4(x-1)}{x^2(x-1)} \, dx \)
\( = \int x^2 \, dx \)
\( I = \frac{x^3}{3} + c \)

Question. \( I = \int (x^4 + 1)^{-1} \cdot e^{3\log x} \, dx \)
Answer: \( I = \int (x^4 + 1)^{-1} \cdot e^{3\log x} \, dx \)
\( = \int \frac{e^{\log x^3}}{x^4+1} \, dx \)
\( = \int \frac{x^3}{x^4+1} \, dx \)
put \( x^4 + 1 = t \)
\( 4x^3 \, dx = dt \Rightarrow x^3 \, dx = \frac{dt}{4} \)
\( \therefore I = \frac{1}{4} \int \frac{dt}{t} \)
\( = \frac{1}{4} \log |x^4 + 1| + c \)

Question. \( I = \int e^{\log \sqrt{x}} \, dx \)
Answer: \( I = \int \sqrt{x} \, dx \)
\( = \frac{2}{3}x^{3/2} + c \)

Question. \( I = \int \frac{(a^x+b^x)^2}{a^x b^x} \, dx \)
Answer: \( I = \int \frac{(a^x+b^x)^2}{a^x b^x} \, dx \)
\( = \int \frac{a^{2x}+b^{2x}+2a^x b^x}{a^x b^x} \, dx \)
Separate
\( = \int \frac{a^{2x}}{a^x b^x} + \frac{b^{2x}}{a^x b^x} + \frac{2a^x b^x}{a^x b^x} \, dx \)
\( = \int \frac{a^x}{b^x} + \frac{b^x}{a^x} + 2 \, dx \)
\( = \int \left(\frac{a}{b}\right)^x + \left(\frac{b}{a}\right)^x + 2 \, dx \)
\( I = \frac{(\frac{a}{b})^x}{\log(a/b)} + \frac{(\frac{b}{a})^x}{\log(b/a)} + 2x + c \)

Question. \( I = \int (2\tan x - 3\cot x)^2 \, dx \)
Answer: \( I = \int (2\tan x - 3\cot x)^2 \, dx \)
\( = \int 4\tan^2 x + 9\cot^2 x - 12\tan x \cdot \cot x \, dx \)
\( = \int 4(\sec^2 x - 1) + 9(\csc^2 x - 1) - 12 \, dx \)
\( = 4(\tan x - x) + 9(-\cot x - x) - 12x + c \)
\( = 4\tan x - 9\cot x - 25x + c \)

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

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

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

Question. \( I = \int \frac{\log(\tan \frac{x}{2})}{\sin x} \, dx \)
Answer: \( I = \int \frac{\log(\tan \frac{x}{2})}{\sin x} \, dx \)
put \( \log(\tan \frac{x}{2}) = t \)
\( \frac{1}{\tan \frac{x}{2}} \cdot \sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} \, dx = dt \)
\( \Rightarrow \frac{\cos(x/2)}{\sin(x/2)} \cdot \frac{1}{\cos^2(x/2)} \cdot \frac{1}{2} \, dx = dt \)
\( \Rightarrow \frac{1}{2\sin(x/2)\cos(x/2)} \, dx = dt \)
\( \Rightarrow \frac{1}{\sin x} \, dx = dt \)
\( \therefore I = \int t \, dt \)
\( = \frac{t^2}{2} + c \)
\( = \frac{(\log(\tan \frac{x}{2}))^2}{2} + c \)

Question. If \( f'(x) = x+b \), \( f(1) = 5 \) and \( f(2) = 13 \). Find \( f(x) \).
Answer: We have, \( f(x) = \int f'(x) \, dx \)
\( f(x) = \int (x+b) \, dx \)
\( f(x) = \frac{x^2}{2} + bx + c \)
\( f(1) = 5 \) and \( f(2) = 13 \) .....(given)
\( 5 = \frac{1}{2} + b + c \)
\( \frac{9}{2} = b + c \) .....(1)
and \( 13 = 2 + 2b + c \)
\( 11 = 2b + c \) .....(2)
solving (1) & (2)
\( b = \frac{13}{2} \) and \( c = -2 \)
\( \therefore f(x) = \frac{x^2}{2} + \frac{13}{2}x - 2 \)

Question. If \( f'(x) = 3x^2 - \frac{2}{x^3} \) and \( f(1) = 0 \). Find \( f(x) \).
Answer: \( f(x) = x^3 + \frac{1}{x^2} - 2 \)

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

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

Question. \( I = \int \frac{(x+1)e^x}{\cos^2(xe^x)} \, dx \)
Answer: \( I = \int \frac{(x+1)e^x}{\cos^2(xe^x)} \, dx \)
put \( xe^x = t \)
\( (xe^x + e^x) \, dx = dt \)
\( e^x(x+1) \, dx = dt \)
\( \therefore I = \int \frac{dt}{\cos^2 t} \)
\( = \int \sec^2 t \, dt \)
\( = \tan t + c \)
\( = \tan(xe^x) + c \)

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

Question. \( I = \int \frac{1}{16-9x^2} \, dx \)
Answer: \( I = \int \frac{1}{16-9x^2} \, dx \)
\( = \frac{1}{9} \int \frac{1}{(\frac{4}{3})^2-x^2} \, dx \)
\( = \frac{1}{9} \times \frac{1}{2 \times \frac{4}{3}} \log \left| \frac{\frac{4}{3}+x}{\frac{4}{3}-x} \right| + c \) .....\( \int \frac{1}{a^2-x^2} \, dx = \frac{1}{2a}\log \left| \frac{a+x}{a-x} \right| \)
\( = \frac{1}{24} \log \left| \frac{4+3x}{4-3x} \right| + c \)

Question. \( I = \int \frac{1}{\sqrt{16-9x^2}} \, dx \)
Answer: \( I = \int \frac{1}{\sqrt{16-9x^2}} \, dx \)
\( = \frac{1}{3} \int \frac{1}{\sqrt{(\frac{4}{3})^2-x^2}} \, dx \)
\( = \frac{1}{3} \sin^{-1} \left( \frac{x}{\frac{4}{3}} \right) + c \) .....\( \int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1} \left( \frac{x}{a} \right) + c \)
\( = \frac{1}{3} \sin^{-1} \left( \frac{3x}{4} \right) + c \)

Question. \( I = \int \frac{1}{4+9x^2} \, dx \)
Answer: \( I = \int \frac{1}{4+9x^2} \, dx \)
\( = \frac{1}{9} \int \frac{1}{(\frac{2}{3})^2+x^2} \, dx \)
\( = \frac{1}{9} \times \frac{1}{\frac{2}{3}} \tan^{-1}\left( \frac{x}{\frac{2}{3}} \right) + c \)
\( = \frac{1}{6} \tan^{-1}\left( \frac{3x}{2} \right) + c \)

Question. \( I = \int \frac{1}{\sqrt{4+9x^2}} \, dx \)
Answer: \( I = \int \frac{1}{\sqrt{4+9x^2}} \, dx \)
\( = \frac{1}{3} \int \frac{1}{\sqrt{(\frac{2}{3})^2+x^2}} \, dx \)
\( = \frac{1}{3} \log \left| x + \sqrt{(\frac{2}{3})^2+x^2} \right| + c \)

Question. \( I = \int \frac{1}{9x^2-4} \, dx \)
Answer: \( I = \int \frac{1}{9x^2-4} \, dx \)
\( = \frac{1}{9} \int \frac{1}{x^2-(\frac{2}{3})^2} \, dx \)
\( = \frac{1}{9} \times \frac{1}{2\times\frac{2}{3}} \log \left| \frac{x-\frac{2}{3}}{x+\frac{2}{3}} \right| + c \)
\( = \frac{1}{12} \log \left| \frac{3x-2}{3x+2} \right| + c \)