CBSE Class 12 Mathematics Indefinite And Definite Integrals Worksheet Set 01

CBSE Class 12 Mathematics Indefinite and Definite Integrals (1). 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.

Properties of Definite Integrals

• P-I: \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(t) \, dt \)
  e.g. \( \int_{0}^{\frac{\pi}{2}} \sin t \, dt = \int_{0}^{\frac{\pi}{2}} \sin x \, dx \)
• P-II: \( \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx \)
• P-III: \( \int_{a}^{b} f(x) \, dx = \int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx \) where \( a < c < b \)
  e.g. \( \int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{a}^{2a} f(x) \, dx \)
• P-IV: \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \)
  Proof: Taking RHS \( \int_{0}^{a} f(a - x) \, dx \)
  Put \( a - x = t \quad \text{when } x = 0 \Rightarrow t = a \)
  \( -dx = dt \Rightarrow dx = -dt \quad \text{when } x = a \Rightarrow t = 0 \)
  \( \therefore \text{RHS} = -\int_{a}^{0} f(t) \, dt \)
  \( = \int_{0}^{a} f(t) \, dt \quad \dots \text{(by P-II)} \)
  \( = \int_{0}^{a} f(x) \, dx \quad \dots \text{(by P-I)} \)
  \( = \text{LHS} \)
  \( \therefore \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \quad \text{proved} \)
• P-V: \( \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx \)
  Proof: Do yourself by put \( a + b - x = t \)
• P-VI: \( \int_{0}^{2a} f(x) \, dx = \begin{cases} 2\int_{0}^{a} f(x) \, dx & ; \text{ if } f(2a - x) = f(x) \\ 0 & ; \text{ if } f(2a - x) = -f(x) \end{cases} \)
  Mainly \( \int_{0}^{2a} f(x) \, dx = 2\int_{0}^{a} f(x) \, dx \)
• P-VII: Even - function property:
  \( \int_{-a}^{a} f(x) \, dx = \begin{cases} 2\int_{0}^{a} f(x) \, dx & ; \text{ if } f(x) \to \text{even} \\ 0 & ; \text{ if } f(x) \to \text{odd} \end{cases} \)
  If \( f(-x) = f(x) \) then \( f(x) \) is an even function
  If \( f(-x) = -f(x) \) then \( f(x) \) is an odd function

Question. Evaluate \[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \quad \dots (1) \]
\[ = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2} - x\right)}}{\sqrt{\sin \left(\frac{\pi}{2} - x\right)} + \sqrt{\cos \left(\frac{\pi}{2} - x\right)}} \, dx \quad \dots \left[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx \right] \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \quad \dots (2) \]
Adding (1) & (2):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \]
\[ = \int_{0}^{\frac{\pi}{2}} 1 \cdot dx \]
\[ = [x]_{0}^{\frac{\pi}{2}} \]
\[ 2I = \frac{\pi}{2} \]
\[ I = \frac{\pi}{4} \quad \text{ans.} \]

Question. Evaluate \[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x) \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\frac{\pi}{2}} \sin \left[2 \left(\frac{\pi}{2} - x\right)\right] \log \left[\tan \left(\frac{\pi}{2} - x\right)\right] \, dx \quad \dots \left[\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\right] \]
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(\pi - 2x) \log(\cot x) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\cot x) \, dx \quad \dots (2) \quad [\because \sin(\pi - 2x) = \sin(2x)] \]
Adding eq. (1) & (2):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \sin(2x) [\log(\tan x) + \log(\cot x)] \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(\tan x \cdot \cot x) \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \sin(2x) \log(1) \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} 0 \, dx \quad \dots \{\because \log 1 = 0\} \]
\[ \therefore I = 0 \quad \text{ans.} \]

Question. Evaluate \[ I = \int_{0}^{\frac{\pi}{4}} \log (1 + \tan x) \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{4}} \log (1 + \tan x) \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\frac{\pi}{4}} \log \left[1 + \tan \left(\frac{\pi}{4} - x\right)\right] \, dx \quad \dots \left[\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\right] \]
\[ I = \int_{0}^{\frac{\pi}{4}} \log \left[ 1 + \frac{1 - \tan x}{1 + \tan x} \right] \, dx \quad \dots \{\tan (A - B) \text{ formula}\} \]
\[ I = \int_{0}^{\frac{\pi}{4}} \log \left[ \frac{1 + \tan x + 1 - \tan x}{1 + \tan x} \right] \, dx \]
\[ I = \int_{0}^{\frac{\pi}{4}} \log \left( \frac{2}{1 + \tan x} \right) \, dx \quad \dots (2) \]
Eq. (1) + (2):
\[ 2I = \int_{0}^{\frac{\pi}{4}} \log \left( 1 + \tan x \cdot \frac{2}{1 + \tan x} \right) \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{4}} \log (2) \, dx \]
\[ 2I = \log 2 \, [x]_{0}^{\frac{\pi}{4}} \]
\[ 2I = \log 2 \left[\frac{\pi}{4} - 0\right] \]
\[ \therefore I = \frac{\pi}{8} \log 2 \quad \text{ans.} \]

Question. Evaluate \[ I = \int_{0}^{\frac{\pi}{2}} 2\log(\cos x) - \log(\sin(2x)) \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{2}} 2\log(\cos x) - \log(\sin(2x)) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log(\cos^2 x) - \log(\sin(2x)) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\cos^2 x}{\sin(2x)} \right) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\cos^2 x}{2\sin x \cos x} \right) \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left[ \frac{\cot x}{2} \right] \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left[ \frac{\cot(\frac{\pi}{2} - x)}{2} \right] \, dx \quad \dots \left[\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\right] \]
\[ I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\tan x}{2} \right) \, dx \quad \dots (2) \]
Eq. (1) + (2):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{\cot x}{2} \cdot \frac{\tan x}{2} \right) \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log \left( \frac{1}{4} \right) \, dx \quad \dots \{\tan x \cdot \cot x = 1\} \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \log(1) - \log(4) \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} -\log 4 \, dx \quad \dots \{\log 1 = 0\} \]
\[ 2I = -\log 4 \, [x]_{0}^{\frac{\pi}{2}} \]
\[ 2I = -\log 4 \left[\frac{\pi}{2}\right] \]
\[ I = -\frac{\pi}{4} \log 4 \quad \text{ans.} \]
(or) \( I = -\frac{\pi}{4} \log(2)^2 \)
\[ I = -\frac{\pi}{2} \log 2 \quad \text{ans.} \]

Question. Evaluate \[ I = \int_{0}^{1} \log \left(\frac{1}{x} - 1\right) \, dx \]
Answer:
\[ I = \int_{0}^{1} \log \left(\frac{1-x}{x}\right) \, dx \quad \dots (1) \]
\[ = \int_{0}^{1} \log \left[ \frac{1-(1-x)}{1-x} \right] \, dx \quad \dots \left[\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\right] \]
\[ = \int_{0}^{1} \log \left[ \frac{x}{1-x} \right] \, dx \quad \dots (2) \]
(1) + (2):
\[ 2I = \int_{0}^{1} \log \left( \frac{1-x}{x} \cdot \frac{x}{1-x} \right) \, dx \]
\[ = \int_{0}^{1} \log(1) \, dx \]
\[ 2I = 0 \quad \dots \{\because \log 1 = 0\} \]
\[ I = 0 \quad \text{ans.} \]

Question. Evaluate \[ I = \int_{0}^{5} \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{5 - x}} \, dx \]
Answer:
\[ I = \int_{0}^{5} \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{5 - x}} \, dx \quad \dots (1) \]
\[ I = \int_{0}^{5} \frac{\sqrt[3]{5-x}}{\sqrt[3]{5-x} + \sqrt[3]{5-(5-x)}} \, dx \quad \dots \left[\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\right] \]
\[ I = \int_{0}^{5} \frac{\sqrt[3]{5-x}}{\sqrt[3]{5-x} + \sqrt[3]{x}} \, dx \quad \dots (2) \]
(1) + (2):
\[ 2I = \int_{0}^{5} \frac{\sqrt[3]{x} + \sqrt[3]{5-x}}{\sqrt[3]{x} + \sqrt[3]{5-x}} \, dx \]
\[ = \int_{0}^{5} 1 \cdot dx \]
\[ = [x]_{0}^{5} \]
\[ 2I = 5 \]
\[ I = \frac{5}{2} \quad \text{ans.} \]

Question. Show that \[ \int_{0}^{2a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(2a - x) \, dx \]
Answer:
\[ \text{R.H.S} = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} f(2a - x) \, dx \]
In the second integral, put \( 2a - x = t \quad \text{when } x = 0 \Rightarrow t = 2a \)
\( -dx = dt \Rightarrow dx = -dt \quad \text{when } x = a \Rightarrow t = a \)
\[ \therefore \text{R.H.S} = \int_{0}^{a} f(x) \, dx - \int_{2a}^{a} f(t) \, dt \]
\[ = \int_{0}^{a} f(x) \, dx + \int_{a}^{2a} f(t) \, dt \quad \dots \left[\int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx\right] \]
\[ = \int_{0}^{a} f(x) \, dx + \int_{a}^{2a} f(x) \, dx \quad \dots \left[\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(t) \, dt\right] \]
\[ = \int_{0}^{2a} f(x) \, dx \quad \dots \left[\int_{a}^{c} f(x) \, dx + \int_{c}^{b} f(x) \, dx = \int_{a}^{b} f(x) \, dx\right] \]
\[ = \text{LHS} \quad \text{Proved} \]

Question. Show that \[ I = \int_{0}^{1} x(1 - x)^n \, dx = \frac{1}{(n+1)(n+2)} \]
Answer:
\[ I = \int_{0}^{1} (1 - x)[1 - (1 - x)]^n \, dx \quad \dots \left[\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx\right] \]
\[ I = \int_{0}^{1} (1 - x)(x)^n \, dx \]
\[ I = \int_{0}^{1} (x^n - x^{n+1}) \, dx \]
\[ I = \left[ \frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2} \right]_{0}^{1} \]
\[ I = \left[ \frac{1}{n+1} - \frac{1}{n+2} \right] - [0 - 0] \]
\[ I = \frac{n+2-n-1}{(n+1)(n+2)} \]
\[ I = \frac{1}{(n+1)(n+2)} \quad \text{ans.} \]

Question. Evaluate \[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)} \, dx \quad \dots \text{(P-IV)} \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos x + \sin x} \, dx \quad \dots (2) \]
(1) + (2):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} \, dx \]
(Type: - single \(\sin x\) & \(\cos x\))
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}} + \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}} \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{1 + \tan^2\left(\frac{x}{2}\right)}{2\tan\frac{x}{2} + 1 - \tan^2\frac{x}{2}} \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2\left(\frac{x}{2}\right)}{2\tan\frac{x}{2} + 1 - \tan^2\left(\frac{x}{2}\right)} \, dx \]
Put \(\tan\left(\frac{x}{2}\right) = t \quad \text{when } x = 0 \Rightarrow \tan(0) = t \implies t = 0\)
\(\sec^2\left(\frac{x}{2}\right) \cdot \frac{1}{2} \, dx = dt \implies \sec^2\left(\frac{x}{2}\right) \, dx = 2dt \quad \text{when } x = \frac{\pi}{2} \Rightarrow \tan\left(\frac{\pi}{4}\right) = t \implies t = 1\)
\[ \therefore 2I = 2 \int_{0}^{1} \frac{dt}{-t^2 + 2t + 1} \]
\[ I = -\int_{0}^{1} \frac{1}{t^2 - 2t - 1} \, dt \]
\[ = -\int_{0}^{1} \frac{1}{(t-1)^2 - 1 - 1} \, dt \]
\[ = -\int_{0}^{1} \frac{1}{(t-1)^2 - (\sqrt{2})^2} \, dt \]
\[ = \int_{0}^{1} \frac{1}{(\sqrt{2})^2 - (t-1)^2} \, dt \]
\[ = \frac{1}{2\sqrt{2}} \left[ \log \left| \frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1} \right| \right]_{0}^{1} \]
\[ = \frac{1}{2\sqrt{2}} \left[ \log \left| \frac{\sqrt{2} - 0}{\sqrt{2} + 0} \right| - \log \left| \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \right| \right] \]
\[ = \frac{1}{2\sqrt{2}} \left[ \log(1) - \log \left( \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \right) \right] \]
\[ I = -\frac{1}{2\sqrt{2}} \log \left( \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \right) \quad \text{ans.} \]
(Or)
\[ I = -\frac{1}{2\sqrt{2}} \log \left[ \frac{(\sqrt{2} - 1)(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)} \right] \quad \dots \{\text{Rationalize}\} \]
\[ = -\frac{1}{2\sqrt{2}} \log \left[ \frac{(\sqrt{2} - 1)^2}{2 - 1} \right] \]
\[ = -\frac{2}{2\sqrt{2}} \log(\sqrt{2} - 1) \]
\[ I = -\frac{1}{\sqrt{2}} \log(\sqrt{2} - 1) \quad \text{ans.} \]

Question. Evaluate \[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + \sin x \cdot \cos x} \, dx \]
Answer:
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2 x}{1 + \sin x \cdot \cos x} \, dx \quad \dots (1) \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos^2\left(\frac{\pi}{2} - x\right)}{1 + \sin\left(\frac{\pi}{2} - x\right) \cdot \cos\left(\frac{\pi}{2} - x\right)} \, dx \]
\[ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2 x}{1 + \cos x \cdot \sin x} \, dx \quad \dots (2) \]
(1) + (2):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \sin x \cdot \cos x} \, dx \quad \dots \{\sin^2 x + \cos^2 x = 1\} \]
Type: Divide by \(\cos^2 x\)
Divide N & D by \(\cos^2 x\):
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 x}{\sec^2 x + \tan x} \, dx \]
\[ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sec^2 x}{1 + \tan^2 x + \tan x} \, dx \]
Put \(\tan x = t \quad \text{when } x = 0 \Rightarrow t = 0\)
\(\sec^2 x \cdot dx = dt \quad \text{when } x = \frac{\pi}{2} \Rightarrow t = \infty\)
\[ \therefore 2I = \int_{0}^{\infty} \frac{dt}{t^2 + t + 1} \]
Perfect square:
\[ 2I = \int_{0}^{\infty} \frac{1}{\left(t + \frac{1}{2}\right)^2 - \frac{1}{4} + 1} \, dt \]
\[ 2I = \int_{0}^{\infty} \frac{dt}{\left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \]
\[ 2I = \frac{2}{\sqrt{3}} \left[ \tan^{-1} \left( \frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) \right]_{0}^{\infty} \]
\[ 2I = \frac{2}{\sqrt{3}} \left[ \tan^{-1} \left( \frac{2t + 1}{\sqrt{3}} \right) \right]_{0}^{\infty} \]
\[ 2I = \frac{2}{\sqrt{3}} \left[ \tan^{-1}(\infty) - \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \right] \]
\[ 2I = \frac{2}{\sqrt{3}} \left[ \frac{\pi}{2} - \frac{\pi}{6} \right] \]
\[ 2I = \frac{2}{\sqrt{3}} \left[ \frac{\pi}{3} \right] \]
\[ I = \frac{\pi}{3\sqrt{3}} \quad \text{ans.} \]