CBSE Class 12 Mathematics Indefinite And Definite Integrals Worksheet Set 07

CBSE Class 12 Mathematics Indefinite and Definite Integrals Worksheet (7). 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.

Question. Evaluate:
\[ I = \int_{0}^{2} \frac{1}{4+x-x^2} \, dx \]
Answer:
Using perfect square method:
\[ I = - \int_{0}^{2} \frac{1}{x^2-x-4} \, dx \]
\[ I = - \int_{0}^{2} \frac{1}{\left(x-\frac{1}{2}\right)^2 - \frac{1}{4} - 4} \, dx \]
\[ I = - \int_{0}^{2} \frac{1}{\left(x-\frac{1}{2}\right)^2 - \left(\frac{\sqrt{17}}{2}\right)^2} \, dx \]
\[ I = \int_{0}^{2} \frac{1}{\left(\frac{\sqrt{17}}{2}\right)^2 - \left(x-\frac{1}{2}\right)^2} \, dx {} \]
\[ I = \left[ \frac{1}{2 \times \frac{\sqrt{17}}{2}} \log \left| \frac{\frac{\sqrt{17}}{2} + x - \frac{1}{2}}{\frac{\sqrt{17}}{2} - x + \frac{1}{2}} \right| \right]_{0}^{2} \]
\[ I = \frac{1}{\sqrt{17}} \left[ \log \left| \frac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1} \right| \right]_{0}^{2} \]
\[ I = \frac{1}{\sqrt{17}} \left[ \log \left| \frac{\sqrt{17}+3}{\sqrt{17}-3} \right| - \log \left| \frac{\sqrt{17}-1}{\sqrt{17}+1} \right| \right] \]
\[ I = \frac{1}{\sqrt{17}} \log \left| \frac{\sqrt{17}+3}{\sqrt{17}-3} \times \frac{\sqrt{17}+1}{\sqrt{17}-1} \right| \quad \dots \{\log A - \log B = \log (A/B)\} \]
\[ I = \frac{1}{\sqrt{17}} \log \left| \frac{17+\sqrt{17}+3\sqrt{17}+3}{17-\sqrt{17}-3\sqrt{17}+3} \right| \]
\[ I = \frac{1}{\sqrt{17}} \log \left| \frac{20+4\sqrt{17}}{20-4\sqrt{17}} \right| \]
\[ I = \frac{1}{\sqrt{17}} \log \left| \frac{5+\sqrt{17}}{5-\sqrt{17}} \right| \]
\[ I = \frac{1}{\sqrt{17}} \log \left| \frac{(\sqrt{17}+5)(5+\sqrt{17})}{25-17} \right| \quad \dots (\text{rationalize}) \]
\[ I = \frac{1}{\sqrt{17}} \log \left( \frac{17+25+10\sqrt{17}}{8} \right) \]
\[ I = \frac{1}{\sqrt{17}} \log \left( \frac{42+10\sqrt{17}}{8} \right) \]
\[ \therefore I = \frac{1}{\sqrt{17}} \log \left( \frac{21+5\sqrt{17}}{4} \right) \quad \text{Ans.} \]

Question. Evaluate:
\[ I = \int_{1}^{2} \frac{5(x^2-x-1)}{x^2+3x+2} \, dx \]
Answer:
\[ I = 5 \int_{1}^{2} \frac{x^2-x-1}{x^2+3x+2} \, dx \]
Here, degree of numerator = degree of denominator.
Express as Quotient + \( \frac{\text{Remainder}}{\text{Divisor}} \):
\[ I = 5 \int_{1}^{2} \left( 1 - \frac{4x+3}{x^2+3x+2} \right) \, dx \]
\[ = 5 \int_{1}^{2} 1 \, dx - 5 \int_{1}^{2} \frac{4x+3}{x^2+3x+2} \, dx \]
\[ = 5 [x]_{1}^{2} - 5 \int_{1}^{2} \frac{4x+3}{(x+1)(x+2)} \, dx \]
\[ I = 5 - 5 \int_{1}^{2} \frac{4x+3}{(x+1)(x+2)} \, dx \]

Let \( \frac{4x+3}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2} \)
\[ \Rightarrow 4x + 3 = A(x+2) + B(x+1) \]
Comparing the coefficients of \( x \) and constant:
\( 4 = A + B \)
\( 3 = 2A + B \)

Solving these equations, we get:
\( A = -1 \) & \( B = 5 \)

\[ \therefore I = 5 - 5 \int_{1}^{2} \left( \frac{-1}{x+1} + \frac{5}{x+2} \right) \, dx \]
\[ I = 5 - 5 \left[ -\log|x+1| + 5\log|x+2| \right]_{1}^{2} \]
\[ = 5 - 5 \left[ (-\log 3 + 5\log 4) - (-\log 2 + 5\log 3) \right] \]
\[ = 5 - 5 [ -\log 3 + 10\log 2 + \log 2 - 5\log 3 ] \]
\[ = 5 - 5 [ 11\log 2 - 6\log 3 ] \]
\[ I = 5 - 55\log 2 + 30\log 3 \quad \text{Ans.} \]

Question. If \( \int_{a}^{b} x^3 \, dx = 0 \) and \( \int_{a}^{b} x^2 \, dx = \frac{2}{3} \), find the values of \( a \) & \( b \).
Answer:
Consider \( \int_{a}^{b} x^3 \, dx = 0 \)
\[ \Rightarrow \left[ \frac{x^4}{4} \right]_{a}^{b} = 0 \]
\[ \Rightarrow \frac{1}{4} [b^4 - a^4] = 0 \]
\[ \Rightarrow a^4 = b^4 \]
\[ \Rightarrow a = -b \]

Consider \( \int_{a}^{b} x^2 \, dx = \frac{2}{3} \)
\[ \Rightarrow \left[ \frac{x^3}{3} \right]_{a}^{b} = \frac{2}{3} \]
\[ \Rightarrow \frac{1}{3} [b^3 - a^3] = \frac{2}{3} \]
\[ \Rightarrow b^3 - a^3 = 2 \]
\[ \Rightarrow (-a)^3 - a^3 = 2 \]
\[ \Rightarrow -a^3 - a^3 = 2 \]
\[ \Rightarrow -2a^3 = 2 \]
\[ \Rightarrow a^3 = -1 \]
\[ \Rightarrow a = -1 \]

Since \( b = -a \Rightarrow b = 1 \).
\( \therefore a = -1 \) & \( b = 1 \) \quad \text{Ans.}

Question. Evaluate:
\[ I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos(2x) \cdot \log(\sin x) \, dx \]
Answer:
Integrating by parts:
\[ I = \left[ \log(\sin x) \cdot \frac{\sin(2x)}{2} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\sin x} \cdot \cos x \cdot \frac{\sin(2x)}{2} \, dx \]
\[ = \left[ \left(\log 1 \cdot \frac{\sin \pi}{2}\right) - \left(\log \left(\frac{1}{\sqrt{2}}\right) \cdot \frac{\sin \frac{\pi}{2}}{2}\right) \right] - \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\sin x} \cdot \cos x \cdot 2\sin x \cos x \, dx \]
\[ = \left[ 0 + \log(\sqrt{2}) \cdot \frac{1}{2} \right] - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos^2 x \, dx \quad \dots \{\log(a/b) = -\log(b/a)\} \]
\[ = \frac{1}{2} \log 2^{\frac{1}{2}} - \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} [1 + \cos(2x)] \, dx \]
\[ = \frac{1}{4} \log 2 - \frac{1}{2} \left[ x + \frac{\sin(2x)}{2} \right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} \]
\[ = \frac{1}{4} \log 2 - \frac{1}{2} \left[ \left( \frac{\pi}{2} + 0 \right) - \left( \frac{\pi}{4} + \frac{1}{2} \right) \right] \]
\[ = \frac{1}{4} \log 2 - \frac{\pi}{4} + \frac{\pi}{8} + \frac{1}{4} \]
\[ I = \frac{1}{4} \log 2 - \frac{\pi}{8} + \frac{1}{4} \quad \text{Ans.} \]

Question. Evaluate:
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{\sin(2x)}} \, dx \]
Answer:
\[ I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{1-1+\sin(2x)}} \, dx \]
\[ = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{1-(1-\sin 2x)}} \, dx \]
\[ = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{1-[\sin^2 x + \cos^2 x - 2\sin x \cos x]}} \, dx \]
\[ = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x + \cos x}{\sqrt{1-(\sin x - \cos x)^2}} \, dx \]

Put \( \sin x - \cos x = t \), when \( x = \frac{\pi}{6} \Rightarrow t = \frac{1}{2} - \frac{\sqrt{3}}{2} = \frac{1-\sqrt{3}}{2} \)
\( (\cos x + \sin x) \, dx = dt \), when \( x = \frac{\pi}{3} \Rightarrow t = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3}-1}{2} \)

\[ \therefore I = \int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{dt}{\sqrt{1-t^2}} \]
\[ = \left[ \sin^{-1} t \right]_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \]
\[ = \sin^{-1} \left( \frac{\sqrt{3}-1}{2} \right) - \sin^{-1} \left( \frac{1-\sqrt{3}}{2} \right) \]
\[ = \sin^{-1} \left( \frac{\sqrt{3}-1}{2} \right) + \sin^{-1} \left( \frac{\sqrt{3}-1}{2} \right) \quad \dots \{\because \sin^{-1}(-x) = -\sin^{-1} x\} \]
\[ I = 2 \sin^{-1} \left( \frac{\sqrt{3}-1}{2} \right) \quad \text{Ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9+16\sin(2x)} \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9+16[1-1+\sin(2x)]} \, dx \]
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9+16[1-(1-\sin 2x)]} \, dx \]
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9+16[1-(\sin^2 x + \cos^2 x - 2\sin x \cos x)]} \, dx \]
\[ I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{9+16[1-(\sin x - \cos x)^2]} \, dx \]

Put \( \sin x - \cos x = t \), when \( x = 0 \Rightarrow t = 0 - 1 = -1 \)
\( (\cos x + \sin x) \, dx = dt \), when \( x = \frac{\pi}{4} \Rightarrow t = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0 \)

\[ \therefore I = \int_{-1}^{0} \frac{dt}{9+16(1-t^2)} \]
\[ = \int_{-1}^{0} \frac{dt}{25-16t^2} \]
\[ = \frac{1}{16} \int_{-1}^{0} \frac{dt}{\left(\frac{5}{4}\right)^2 - t^2} \]
\[ = \frac{1}{16} \times \frac{1}{2 \times \frac{5}{4}} \left[ \log \left| \frac{\frac{5}{4} + t}{\frac{5}{4} - t} \right| \right]_{-1}^{0} \]
\[ = \frac{1}{40} \left[ \log \left| \frac{5+4t}{5-4t} \right| \right]_{-1}^{0} \]
\[ = \frac{1}{40} \left[ \log |1| - \log \left| \frac{1}{9} \right| \right] \]
\[ = \frac{1}{40} [ 0 + \log 9 ] \]
\[ I = \frac{1}{40} \log 9 \]
\[ I = \frac{1}{20} \log 3 \quad \text{Ans.} \]

Question. Evaluate:
\[ I = \int_{1}^{2} e^{2x} \left( \frac{1}{x} - \frac{1}{2x^2} \right) \, dx \]
Answer:
\[ I = \int_{1}^{2} e^{2x} \cdot \frac{1}{x} \, dx - \frac{1}{2} \int_{1}^{2} e^{2x} \cdot \frac{1}{x^2} \, dx \]
Integrating the first term by parts:
\[ = \left[ \frac{1}{x} \cdot \frac{e^{2x}}{2} \right]_{1}^{2} - \int_{1}^{2} \left( -\frac{1}{x^2} \right) \cdot \frac{e^{2x}}{2} \, dx - \frac{1}{2} \int_{1}^{2} e^{2x} \cdot \frac{1}{x^2} \, dx \]
\[ = \left[ \frac{1}{2} \cdot \frac{e^4}{2} \right] - \left[ \frac{1}{1} \cdot \frac{e^2}{2} \right] \]
\[ = \frac{e^4}{4} - \frac{e^2}{2} \]
\[ = \frac{e^2}{2} \left( \frac{e^2}{2} - 1 \right) \quad \text{Ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos^2 x + 4\sin^2 x} \, dx \]
Answer:
Divide numerator and denominator by \( \cos^4 x \):
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 x}{\sec^2 x + 4\tan^2 x \cdot \sec^2 x} \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 x}{\sec^2 x(1 + 4\tan^2 x)} \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 x}{(1+\tan^2 x)(1+4\tan^2 x)} \, dx \]

Put \( \tan x = t \Rightarrow \sec^2 x \, dx = dt \)
When \( x = 0 \Rightarrow t = 0 \)
When \( x = \frac{\pi}{2} \Rightarrow t = \infty \)

\[ I = \int_{0}^{\infty} \frac{dt}{(1+t^2)(1+4t^2)} \]

Type: partial fraction type 4
Let \( t^2 = y \) (temporary variable)
\[ \therefore \frac{1}{(1+t^2)(1+4t^2)} = \frac{1}{(1+y)(1+4y)} \]
Let \( \frac{1}{(1+y)(1+4y)} = \frac{A}{1+y} + \frac{B}{1+4y} \)
\[ 1 = A(1+4y) + B(1+y) \]
Comparing coefficients:
\( 0 = 4A + B \)
\( 1 = A + B \)

Solving these equations:
\( A = -\frac{1}{3} \) & \( B = \frac{4}{3} \)

\[ \therefore I = \int_{0}^{\infty} \left[ \frac{-1}{3(1+t^2)} + \frac{4}{3(1+4t^2)} \right] \, dt \]
\[ = -\frac{1}{3} \int_{0}^{\infty} \frac{1}{1+t^2} \, dt + \frac{4}{3 \times 4} \int_{0}^{\infty} \frac{1}{\frac{1}{4}+t^2} \, dt \]
\[ = -\frac{1}{3} \int_{0}^{\infty} \frac{1}{1+t^2} \, dt + \frac{1}{3} \int_{0}^{\infty} \frac{1}{\left(\frac{1}{2}\right)^2 + t^2} \, dt \]
\[ = -\frac{1}{3} [\tan^{-1} t]_{0}^{\infty} + \frac{1}{3} \times 2 [\tan^{-1}(2t)]_{0}^{\infty} \]
\[ = -\frac{1}{3} (\tan^{-1}\infty - \tan^{-1}0) + \frac{2}{3} [\tan^{-1}(\infty) - \tan^{-1}(0)] \]
\[ = -\frac{1}{3} \left[\frac{\pi}{2}\right] + \frac{2}{3} \left[\frac{\pi}{2} - 0\right] \]
\[ = -\frac{\pi}{6} + \frac{\pi}{3} \]
\[ I = \frac{\pi}{6} \quad \text{Ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{1} \sqrt{\frac{1-x}{1+x}} \, dx \]
Answer:
Rationalizing the numerator:
\[ I = \int_{0}^{1} \frac{1-x}{\sqrt{1-x^2}} \, dx \]
Separating the terms:
\[ I = \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \, dx - \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx \]

Put \( 1 - x^2 = t \), when \( x = 0 \Rightarrow t = 1 \)
\( x \, dx = -\frac{dt}{2} \), when \( x = 1 \Rightarrow t = 0 \)

\[ \therefore I = \int_{0}^{1} \frac{1}{\sqrt{1-x^2}} \, dx + \frac{1}{2} \int_{1}^{0} \frac{dt}{\sqrt{t}} \]
\[ I = [\sin^{-1} x]_{0}^{1} + \frac{1}{2} [2\sqrt{t}]_{1}^{0} \]
\[ I = \left(\frac{\pi}{2} - 0\right) + \frac{1}{2} (0 - 2) \]
\[ I = \frac{\pi}{2} - 1 \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{0}^{\frac{1}{\sqrt{2}}} \frac{\sin^{-1} x}{(1-x^2)^{3/2}} \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\boxplus}{\sqrt{2}}} \frac{\sin^{-1} x}{(1-x^2)\sqrt{1-x^2}} \, dx \]

Put \( \sin^{-1} x = t \), when \( x = 0 \Rightarrow t = 0 \)
\( \frac{1}{\sqrt{1-x^2}} \, dx = dt \), when \( x = \frac{1}{\sqrt{2}} \Rightarrow t = \frac{\pi}{4} \)

\[ I = \int_{0}^{\frac{\pi}{4}} \frac{t}{1-x^2} \, dt \]
\[ = \int_{0}^{\frac{\pi}{4}} \frac{t}{1-\sin^2 t} \, dt \quad \dots \{\because x = \sin t\} \]
\[ = \int_{0}^{\frac{\pi}{4}} t \sec^2 t \, dt \]

Integrating by parts:
\[ = [t \tan t]_{0}^{\frac{\pi}{4}} - \int_{0}^{\frac{\pi}{4}} 1 \cdot \tan t \, dt \]
\[ = \left( \frac{\pi}{4} \cdot \tan \frac{\pi}{4} - 0 \right) - \left[ \log|\sec t| \right]_{0}^{\frac{\pi}{4}} \]
\[ = \frac{\pi}{4} - [ \log(\sqrt{2}) - \log(1) ] \quad \dots \{\because \sec \frac{\pi}{4} = \sqrt{2}, \sec 0 = 1\} \]
\[ = \frac{\pi}{4} - \left[ \frac{1}{2} \log 2 \right] \]
\[ I = \frac{\pi}{4} - \frac{1}{2} \log 2 \quad \text{Ans.} \]

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