CBSE Class 12 Mathematics Indefinite And Definite Integrals Worksheet Set 05

CBSE Class 12 Mathematics Indefinite and Definite Integrals (5). 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.

DEFINITE INTEGRALS

Question. Evaluate:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) \, dx \]
Answer:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \log \left(\frac{2-\sin x}{2+\sin x}\right) \, dx \]
Here \( f(x) = \log \left[ \frac{2-\sin x}{2+\sin x} \right] \)
\[ f(-x) = \log \left[ \frac{2-\sin(-x)}{2+\sin(-x)} \right] \]
\[ = \log \left( \frac{2+\sin x}{2-\sin x} \right) \]
\[ = -\log \left( \frac{2-\sin x}{2+\sin x} \right) \quad \dots \left[\log \left(\frac{a}{b}\right) = -\log \left(\frac{b}{a}\right)\right] \]
\[ f(-x) = -f(x) \]
\( \therefore f(x) \to \) odd function
\( \therefore I = 0 \quad \text{Ans.} \)

Question. Evaluate:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx \]
Answer:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx \]
\( f(x) = \sin^2 x \)
\( f(-x) = \sin^2(-x) = (-\sin x)^2 = \sin^2 x = f(x) \)
\( \dots f(x) \to \) an even function
\[ \therefore I = 2 \int_{0}^{\frac{\pi}{2}} \sin^2 x \, dx \quad \dots \left\{\because \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx\right\} \]
\[ I = \frac{2}{2} \int_{0}^{\frac{\pi}{2}} [1 - \cos(2x)] \, dx \]
\[ I = \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{\frac{\pi}{2}} \]
\[ I = \left[ \frac{\pi}{2} - \frac{\sin(\pi)}{2} \right] - [0] \]
\[ I = \frac{\pi}{2} \quad \text{Ans.} \quad \{\because \sin(\pi) = 0\} \]

Question. Evaluate:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^4 x \, dx \]
Answer:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sin^4 x \, dx \]
Clearly \( f(x) \) is an even function
\[ \therefore I = 2 \int_{0}^{\frac{\pi}{4}} \sin^4 x \, dx \]
\[ = 2 \int_{0}^{\frac{\pi}{4}} (\sin^2 x)^2 \, dx \]
\[ = 2 \int_{0}^{\frac{\pi}{4}} \left( \frac{1-\cos(2x)}{2} \right)^2 \, dx \]
\[ = \frac{2}{4} \int_{0}^{\frac{\pi}{4}} [1 + \cos^2(2x) - 2\cos(2x)] \, dx \]
\[ = \frac{1}{2} \int_{0}^{\frac{\pi}{4}} \left[ 1 + \frac{1+\cos(4x)}{2} - 2\cos(2x) \right] \, dx \]
\[ = \frac{1}{4} \int_{0}^{\frac{\pi}{4}} [3 + \cos(4x) - 4\cos(2x)] \, dx \]
\[ = \frac{1}{4} \left[ 3x + \frac{\sin(4x)}{4} - 2\sin(2x) \right]_{0}^{\frac{\pi}{4}} \]
\[ = \frac{1}{4} \left[ \left( \frac{3\pi}{4} + \frac{\sin \pi}{4} - 2\sin \left(\frac{\pi}{2}\right) \right) - (0 + 0 - 0) \right] \]
\[ = \frac{1}{4} \left[ \frac{3\pi}{4} + 0 - 2 \right] \quad \dots \{\because \sin(\pi) = \sin(2\pi) = 0\} \]
\[ = \frac{3\pi}{16} - \frac{1}{2} \quad \text{Ans.} \]

Question. Evaluate:
\[ \int_{-a}^{a} \sqrt{\frac{a-x}{a+x}} \, dx \]
Answer:
\[ I = \int_{-a}^{a} \sqrt{\frac{a-x}{a+x}} \, dx \]
rationalize:
\[ I = \int_{-a}^{a} \frac{a-x}{\sqrt{a^2-x^2}} \, dx \]
\[ I = a \int_{-a}^{a} \frac{1}{\sqrt{a^2-x^2}} \, dx - \int_{-a}^{a} \frac{x}{\sqrt{a^2-x^2}} \, dx \]
Let \( f(x) = \frac{1}{\sqrt{a^2-x^2}} ; \quad g(x) = \frac{x}{\sqrt{a^2-x^2}} \)
\[ f(-x) = \frac{1}{\sqrt{a^2-(-x)^2}} ; \quad g(-x) = \frac{-x}{\sqrt{a^2-(-x)^2}} \]
\[ f(-x) = \frac{1}{\sqrt{a^2-x^2}} ; \quad g(-x) = \frac{-x}{\sqrt{a^2-x^2}} \]
\[ f(-x) = f(x) ; \quad g(-x) = -g(x) \]
\( f(x) \to \) even function \quad \( g(x) \to \) odd function
\[ \therefore I = 2a \int_{0}^{a} \frac{1}{\sqrt{a^2-x^2}} \, dx - 0 \quad \dots \left\{\because \int_{-a}^{a} f(x) \, dx = \begin{cases} 2\int_{0}^{a} f(x) \, dx & : f(x) \text{ even} \\ 0 & : f(x) \text{ odd} \end{cases}\right\} \]
\[ I = 2a \left( \sin^{-1} \frac{x}{a} \right)_{0}^{a} \]
\[ I = 2a (\sin^{-1}(1) - \sin^{-1}(0)) \]
\[ I = 2a \left(\frac{\pi}{2}\right) \]
\[ I = a\pi \quad \text{ans.} \]

Question. Evaluate:
\[ I = \int_{-\pi}^{\pi} \frac{2x(1+\sin x)}{1+\cos^2 x} \, dx \]
Answer:
\[ I = \int_{-\pi}^{\pi} \frac{2x(1+\sin x)}{1+\cos^2 x} \, dx {} \]
\[ I = \int_{-\pi}^{\pi} \frac{2x + 2x \sin x}{1+\cos^2 x} \, dx \]
\[ I = \int_{-\pi}^{\pi} \frac{2x}{1+\cos^2 x} \, dx + 2 \int_{-\pi}^{\pi} \frac{x \sin x}{1+\cos^2 x} \, dx \]
\[ f(x) = \frac{2x}{1+\cos^2 x} ; \quad g(x) = \frac{x \sin x}{1+\cos^2 x} \]
\[ f(-x) = \frac{-2x}{1+\cos^2 x} ; \quad g(-x) = \frac{(-x)\sin(-x)}{1+\cos^2 x} \]
\[ f(-x) = -f(x) ; \quad g(-x) = \frac{x \sin x}{1+\cos^2 x} = g(x) \]
\( f(x) \to \) odd function ; \( g(x) \to \) even function
\[ \therefore I = 0 + 4 \int_{0}^{\pi} \frac{x \sin x}{1+\cos^2 x} \, dx \]

Removal of \( x \) (already done)
Do yourself
\( I = \pi^2 \quad \text{ans.} \)

Question. Evaluate:
\[ I = \int_{0}^{2\pi} \cos^5 x \, dx \]
Answer:
\[ I = 2 \int_{0}^{\pi} \cos^5 x \, dx \quad \dots \left\{ \int_{0}^{2a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \text{ if } f(2a-x) = f(x) \right\} \]
\[ I = 2 \int_{0}^{\pi} \cos^5 x \, dx \quad \dots (1) \]
\[ I = 2 \int_{0}^{\pi} \cos^5(\pi - x) \, dx \quad \dots \left[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \right] \]
\[ I = -2 \int_{0}^{\pi} \cos^5 x \, dx \quad \dots (2) \quad \dots [\because \cos(\pi - x) = -\cos x] \]
\[ (1) + (2) \]
\[ 2I = 0 \]
\( I = 0 \quad \text{Ans.} \)

Question. Evaluate:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^3 + x^2\tan x + \sin x - x^2) \, dx \]
Answer:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^3 + x^2\tan x + \sin x) \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \, dx \]
Let \( f(x) = x^3 + x^2\tan x + \sin x \); \( \quad g(x) = x^2 \)
\[ f(-x) = -x^3 - x^2\tan x - \sin x ; \quad g(-x) = (-x)^2 = x^2 \]
\[ f(-x) = -(x^3 + x^2\tan x + \sin x) ; \quad g(-x) = g(x) \]
\[ f(-x) = -f(x) ; \quad g(x) \to \text{even function} \]
\( \therefore f(x) \to \) odd function
\[ \therefore I = 0 - 2 \int_{0}^{\frac{\pi}{2}} x^2 \, dx \quad \dots \{\text{property-VII}\} \]
\[ = -2 \left( \frac{x^3}{3} \right)_{0}^{\frac{\pi}{2}} \]
\[ = -2 \left( \frac{\pi^3}{24} \right) \]
\[ = \frac{-\pi^3}{12} \quad \text{ans.} \]

Question. Evaluation using limit as sum:
\[ I = \int_{2}^{4} 3x - 2 \, dx \]
Answer:
Here \( a = 2, b = 4 \text{ and } nh = 4 - 2 = 2 \)
\[ I = \lim_{h \to 0} h[f(2) + f(2+h) + f(2+2h) + \dots + f(2 + (n-1)h)] \]
\[ = \lim_{h \to 0} h[(6-2) + (6+3h-2) + (6+6h-2) + \dots + (6+3(n-1)h-2)] \]
\[ = \lim_{h \to 0} h[4 + (4+3h) + (4+6h) + \dots + (4+3(n-1)h)] \]
\[ = \lim_{h \to 0} h[(4+4+4 \dots n \text{ term}) + (3h+6h+\dots 3(n-1)h)] \]
\[ = \lim_{h \to 0} h[4n + 3h(1+2+\dots(n-1))] \]
\[ = \lim_{h \to 0} h\left[4n + 3h \cdot \frac{n(n-1)}{2}\right] \]
\[ = \lim_{h \to 0} \left[4nh + 3\frac{(nh)(nh-h)}{2}\right] \]
Put \( nh = 2 \):
\[ \therefore I = \lim_{h \to 0} \left[8 + \frac{3(2)(2-h)}{2}\right] \]
\[ = 8 + \frac{3(2)(2)}{2} \]
\[ I = 14 \quad \text{Ans.} \]
Check: \( I = \int_{2}^{4} 3x-2 \, dx = \left[ \frac{3x^2}{2} - 2x \right]_{2}^{4} = [(24-8) - (6-4)] = 16-2 = 14 \)

Question. Evaluate:
\[ I = \int_{-1}^{2} 3x^2 + 2x - 5 \, dx \]
Answer:
Here \( a = -1, b = 2 \text{ and } nh = 3 \)
\[ I = \lim_{h \to 0} h[f(-1) + f(-1+h) + f(-1+2h) + \dots + f(-1+(n-1)h)] \]
\[ = \lim_{h \to 0} h[(3-2-5) + [3(1+h^2-2h) - 2 + 2h - 5] + [3(1+4h^2-4h) - 2 + 4h - 5] + \dots + [3(1+(n-1)^2h^2-2(n-1)h) - 2 + 2(n-1)h - 5]] \]
\[ = \lim_{h \to 0} h[(-4) + (3h^2-4h-5) + (12h^2-8h-4) + \dots + (3(n-1)^2h^2 - 4(n-1)h - 4)] \]
\[ = \lim_{h \to 0} h[(-4-4-4 \dots n \text{ terms}) + (3h^2+12h^2+\dots 3(n-1)^2h^2) + (-4h-8h \dots -4(n-1)h)] \]
\[ = \lim_{h \to 0} h[-4n + 3h^2(1^2+2^2\dots(n-1)^2) - 4h(1+2+3\dots(n-1))] \]
\[ = \lim_{h \to 0} h\left[-4n + 3h^2 \cdot \frac{n(n-1)(2n-1)}{6} - 4h \cdot \frac{n(n-1)}{2}\right] \]
\[ = \lim_{h \to 0} \left[-4nh + \frac{(nh)(nh-h)(2nh-h)}{2} - 2(nh)(nh-h)\right] \]
Put \( nh = 3 \):
\[ \therefore I = \lim_{h \to 0} \left[-12 + \frac{(3)(3-h)(6-h)}{2} - 2(3)(3-h)\right] \]
\[ = \left[-12 + \frac{(3)(3)(6)}{2} - 2(3)(3)\right] \]
\[ = -12 + 27 - 18 \]
\[ = -30 + 27 \]
\[ I = -3 \quad \text{Ans.} \]
Check: \( I = \int_{-1}^{2} 3x^2+2x-5 \, dx = (x^3+x^2-5x)_{-1}^{2} = (8+4-10) - (-1+1+5) = 2-5 = -3 \quad \text{Ans.} \)

Question. Evaluate:
\[ I = \int_{2}^{5} e^{2x+1} \, dx \]
Answer:
Here \( a = 2, b = 5 \text{ \& } nh = 3 \)
\[ I = \lim_{h \to 0} h[f(2) + f(2+h) + f(2+2h) + \dots + f(2+(n-1)h)] \]
\[ = \lim_{h \to 0} h[e^{4+1} + e^{4+2h+1} + e^{4+4h+1} + \dots + e^{4+2(n-1)h+1}] \]
\[ = \lim_{h \to 0} h[e^5 + e^{2h+5} + e^{4h+5} + \dots + e^{2(n-1)h+5}] \]
\[ = e^5 \lim_{h \to 0} h[1 + e^{2h} + e^{4h} + \dots + e^{2(n-1)h}] \]
G.P. \( a = 1, r = e^{2h} \):
\[ = e^5 \lim_{h \to 0} h \left[1 \cdot \left(\frac{(e^{2h})^n-1}{e^{2h}-1}\right)\right] \]
\[ = e^5 \lim_{h \to 0} \left[ \frac{e^{2nh}-1}{e^{2h}-1} \right] \]
\[ = e^5 \lim_{h \to 0} h \left[ \frac{e^{2nh}-1}{\frac{e^{2h}-1}{2h}\times 2h} \right] \quad \dots \{\text{adjustment}\} \]
\[ = \frac{e^5[e^6-1]}{\lim_{h \to 0}\left(\frac{e^{2h}-1}{2h}\right)\times 2} \quad \dots \{\because nh = 3\} \]
\[ I = \frac{e^5(e^6-1)}{2} \quad \text{Ans} \quad \dots \left\{ \because \lim_{x \to 0}\left(\frac{e^x-1}{x}\right) = 1 \right\} \]
Check: \( I = \int_{2}^{5} e^{2x+1} \, dx = \left( \frac{e^{2x+1}}{2} \right)_{2}^{5} = \frac{1}{2}[e^{11} - e^5] = \frac{e^5(e^6-1)}{2} \quad \text{ans.} \)

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