Integrals Class 12 Exercise 7.9 Solutions | Mathematics Part II CBSE

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• Ex. 7.9


• Ex. 7.10


• Ex. 7.11


• Miscellaneous Exercise

Exercise 7.9 Solutions

Topic: Definite Integrals
Evaluate the following definite integrals:

Question 1. Evaluate \(\int_{-1}^{1} (x+1) dx\)

Let \(I = \int_{-1}^{1} (x+1) dx\)

Integration of \(x^n\) is \(\frac{x^{n+1}}{n+1}\).

$$I = \left[ \frac{x^2}{2} + x \right]_{-1}^{1}$$

Applying limits (Upper Limit - Lower Limit):

$$I = \left( \frac{1^2}{2} + 1 \right) - \left( \frac{(-1)^2}{2} + (-1) \right)$$ $$I = \left( \frac{1}{2} + 1 \right) - \left( \frac{1}{2} - 1 \right)$$ $$I = \frac{3}{2} - \left( -\frac{1}{2} \right)$$ $$I = \frac{3}{2} + \frac{1}{2} = \frac{4}{2} = 2$$

Answer: 2

Question 2. Evaluate \(\int_{2}^{3} \frac{1}{x} dx\)

Let \(I = \int_{2}^{3} \frac{1}{x} dx\)

We know that \(\int \frac{1}{x} dx = \log |x|\).

$$I = \left[ \log |x| \right]_{2}^{3}$$

Applying limits:

$$I = \log 3 - \log 2$$ $$I = \log \left( \frac{3}{2} \right)$$

Answer: \(\log \frac{3}{2}\)

Question 3. Evaluate \(\int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) dx\)

Let \(I = \int_{1}^{2} (4x^3 - 5x^2 + 6x + 9) dx\)

Integrating term by term:

$$I = \left[ 4\frac{x^4}{4} - 5\frac{x^3}{3} + 6\frac{x^2}{2} + 9x \right]_{1}^{2}$$ $$I = \left[ x^4 - \frac{5}{3}x^3 + 3x^2 + 9x \right]_{1}^{2}$$

Applying limits:

$$I = \left( 2^4 - \frac{5(2)^3}{3} + 3(2)^2 + 9(2) \right) - \left( 1^4 - \frac{5(1)^3}{3} + 3(1)^2 + 9(1) \right)$$ $$I = \left( 16 - \frac{40}{3} + 12 + 18 \right) - \left( 1 - \frac{5}{3} + 3 + 9 \right)$$ $$I = \left( 46 - \frac{40}{3} \right) - \left( 13 - \frac{5}{3} \right)$$ $$I = 46 - 13 - \frac{40}{3} + \frac{5}{3}$$ $$I = 33 - \frac{35}{3} = \frac{99 - 35}{3} = \frac{64}{3}$$

Answer: \(\frac{64}{3}\)

Question 4. Evaluate \(\int_{0}^{\pi/4} \sin 2x dx\)

Let \(I = \int_{0}^{\pi/4} \sin 2x dx\)

Integration of \(\sin ax\) is \(-\frac{\cos ax}{a}\).

$$I = \left[ -\frac{\cos 2x}{2} \right]_{0}^{\pi/4}$$ $$I = -\frac{1}{2} \left[ \cos 2(\frac{\pi}{4}) - \cos 2(0) \right]$$ $$I = -\frac{1}{2} \left[ \cos \frac{\pi}{2} - \cos 0 \right]$$ $$I = -\frac{1}{2} (0 - 1) = \frac{1}{2}$$

Answer: \(\frac{1}{2}\)

Question 5. Evaluate \(\int_{0}^{\pi/2} \cos 2x dx\)

Let \(I = \int_{0}^{\pi/2} \cos 2x dx\)

Integration of \(\cos ax\) is \(\frac{\sin ax}{a}\).

$$I = \left[ \frac{\sin 2x}{2} \right]_{0}^{\pi/2}$$ $$I = \frac{1}{2} \left[ \sin 2(\frac{\pi}{2}) - \sin 2(0) \right]$$ $$I = \frac{1}{2} (\sin \pi - \sin 0)$$ $$I = \frac{1}{2} (0 - 0) = 0$$

Answer: 0

Question 6. Evaluate \(\int_{4}^{5} e^x dx\)

Let \(I = \int_{4}^{5} e^x dx\)

$$I = \left[ e^x \right]_{4}^{5}$$ $$I = e^5 - e^4$$ $$I = e^4(e - 1)$$

Answer: \(e^4(e - 1)\)

Question 7. Evaluate \(\int_{0}^{\pi/4} \tan x dx\)

Let \(I = \int_{0}^{\pi/4} \tan x dx\)

We know \(\int \tan x dx = -\log|\cos x|\) or \(\log|\sec x|\).

$$I = \left[ -\log|\cos x| \right]_{0}^{\pi/4}$$ $$I = - [\log(\cos \frac{\pi}{4}) - \log(\cos 0)]$$ $$I = - [\log(\frac{1}{\sqrt{2}}) - \log(1)]$$ $$I = - [\log(2^{-1/2}) - 0]$$ $$I = - [-\frac{1}{2}\log 2] = \frac{1}{2}\log 2$$

Answer: \(\frac{1}{2}\log 2\)

Question 8. Evaluate \(\int_{\pi/6}^{\pi/4} \text{cosec } x dx\)

Let \(I = \int_{\pi/6}^{\pi/4} \text{cosec } x dx\)

Formula: \(\int \text{cosec } x dx = \log|\text{cosec } x - \cot x|\)

$$I = \left[ \log|\text{cosec } x - \cot x| \right]_{\pi/6}^{\pi/4}$$ $$I = \log|\text{cosec } \frac{\pi}{4} - \cot \frac{\pi}{4}| - \log|\text{cosec } \frac{\pi}{6} - \cot \frac{\pi}{6}|$$ $$I = \log|\sqrt{2} - 1| - \log|2 - \sqrt{3}|$$ $$I = \log \left( \frac{\sqrt{2} - 1}{2 - \sqrt{3}} \right)$$

Question 9. Evaluate \(\int_{0}^{1} \frac{dx}{\sqrt{1-x^2}}\)

Formula: \(\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1} x\)

$$I = \left[ \sin^{-1} x \right]_{0}^{1}$$ $$I = \sin^{-1}(1) - \sin^{-1}(0)$$ $$I = \frac{\pi}{2} - 0 = \frac{\pi}{2}$$

Answer: \(\frac{\pi}{2}\)

Question 10. Evaluate \(\int_{0}^{1} \frac{dx}{1+x^2}\)

Formula: \(\int \frac{dx}{1+x^2} = \tan^{-1} x\)

$$I = \left[ \tan^{-1} x \right]_{0}^{1}$$ $$I = \tan^{-1}(1) - \tan^{-1}(0)$$ $$I = \frac{\pi}{4} - 0 = \frac{\pi}{4}$$

Answer: \(\frac{\pi}{4}\)

Question 11. Evaluate \(\int_{2}^{3} \frac{dx}{x^2 - 1}\)

Formula: \(\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log \left| \frac{x-a}{x+a} \right|\). Here \(a=1\).

$$I = \frac{1}{2} \left[ \log \left| \frac{x-1}{x+1} \right| \right]_{2}^{3}$$ $$I = \frac{1}{2} \left( \log \left| \frac{3-1}{3+1} \right| - \log \left| \frac{2-1}{2+1} \right| \right)$$ $$I = \frac{1}{2} \left( \log \frac{2}{4} - \log \frac{1}{3} \right)$$ $$I = \frac{1}{2} \left( \log \frac{1}{2} - \log \frac{1}{3} \right)$$ $$I = \frac{1}{2} \log \left( \frac{1/2}{1/3} \right) = \frac{1}{2} \log \left( \frac{3}{2} \right)$$

Answer: \(\frac{1}{2} \log \frac{3}{2}\)

Question 12. Evaluate \(\int_{0}^{\pi/2} \cos^2 x dx\)

Use identity: \(\cos^2 x = \frac{1 + \cos 2x}{2}\)

$$I = \int_{0}^{\pi/2} \frac{1 + \cos 2x}{2} dx$$ $$I = \frac{1}{2} \left[ x + \frac{\sin 2x}{2} \right]_{0}^{\pi/2}$$ $$I = \frac{1}{2} \left[ (\frac{\pi}{2} + \frac{\sin \pi}{2}) - (0 + \frac{\sin 0}{2}) \right]$$ $$I = \frac{1}{2} \left( \frac{\pi}{2} + 0 - 0 \right) = \frac{\pi}{4}$$

Answer: \(\frac{\pi}{4}\)

Question 13. Evaluate \(\int_{2}^{3} \frac{x dx}{x^2+1}\)

Let \(x^2 + 1 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2}\)

Change limits: When \(x=2, t=5\); When \(x=3, t=10\).

$$I = \int_{5}^{10} \frac{1}{t} \frac{dt}{2} = \frac{1}{2} [\log t]_{5}^{10}$$ $$I = \frac{1}{2} (\log 10 - \log 5) = \frac{1}{2} \log(\frac{10}{5})$$ $$I = \frac{1}{2} \log 2$$

Answer: \(\frac{1}{2} \log 2\)

Question 14. Evaluate \(\int_{0}^{1} \frac{2x+3}{5x^2+1} dx\)

Split the integral: \(I = \int_{0}^{1} \frac{2x}{5x^2+1} dx + \int_{0}^{1} \frac{3}{5x^2+1} dx\)

Part 1: Multiply numerator and denominator by 5 to get derivative of denominator.

$$I_1 = \frac{1}{5} \int_{0}^{1} \frac{10x}{5x^2+1} dx = \frac{1}{5} [\log(5x^2+1)]_{0}^{1}$$ $$I_1 = \frac{1}{5} (\log 6 - \log 1) = \frac{1}{5} \log 6$$

Part 2: \(I_2 = 3 \int_{0}^{1} \frac{1}{5(x^2 + 1/5)} dx = \frac{3}{5} \int_{0}^{1} \frac{dx}{x^2 + (1/\sqrt{5})^2}\)

$$I_2 = \frac{3}{5} \cdot \frac{1}{1/\sqrt{5}} [\tan^{-1}(\frac{x}{1/\sqrt{5}})]_{0}^{1}$$ $$I_2 = \frac{3\sqrt{5}}{5} [\tan^{-1}(\sqrt{5}x)]_{0}^{1} = \frac{3}{\sqrt{5}} \tan^{-1}(\sqrt{5})$$

Total Answer: \(\frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1} \sqrt{5}\)

Question 15. Evaluate \(\int_{0}^{1} x e^{x^2} dx\)

Let \(x^2 = t \Rightarrow 2x dx = dt \Rightarrow x dx = \frac{dt}{2}\)

Limits: When \(x=0, t=0\); When \(x=1, t=1\).

$$I = \int_{0}^{1} e^t \frac{dt}{2} = \frac{1}{2} [e^t]_{0}^{1}$$ $$I = \frac{1}{2} (e^1 - e^0) = \frac{1}{2} (e - 1)$$

Answer: \(\frac{1}{2} (e - 1)\)

Question 16. Evaluate \(\int_{1}^{2} \frac{5x^2}{x^2+4x+3} dx\)

Divide \(5x^2\) by \(x^2+4x+3\). Quotient is 5, Remainder is \(-(20x+15)\).

$$I = \int_{1}^{2} \left( 5 - \frac{20x+15}{x^2+4x+3} \right) dx$$ $$I = [5x]_{1}^{2} - \int_{1}^{2} \frac{20x+15}{x^2+4x+3} dx$$ $$I = 5(2-1) - I_1 = 5 - I_1$$

Solving \(I_1\): Let \(20x+15 = A(2x+4) + B\). \(20x+15 = 20x + 40 - 25\).

$$I_1 = \int \frac{10(2x+4) - 25}{x^2+4x+3} dx$$ $$I_1 = 10 \int \frac{2x+4}{x^2+4x+3} dx - 25 \int \frac{dx}{x^2+4x+3}$$ $$I_1 = 10 \log|x^2+4x+3| - 25 \int \frac{dx}{(x+2)^2 - 1^2}$$ $$I_1 = 10 \log|x^2+4x+3| - 25 \cdot \frac{1}{2} \log|\frac{x+2-1}{x+2+1}|$$

Applying limits 1 to 2 for \(I_1\):

$$I_1 = \left[ 10 \log|x^2+4x+3| - \frac{25}{2} \log|\frac{x+1}{x+3}| \right]_{1}^{2}$$ $$I_1 = 10(\log 15 - \log 8) - \frac{25}{2}(\log \frac{3}{5} - \log \frac{2}{4})$$ $$I_1 = 10 \log 15 - 10 \log 8 - \frac{25}{2} \log 3 + \frac{25}{2} \log 5 + \frac{25}{2} \log \frac{1}{2}$$ Simplifying this expression results in \(I_1 = 10 \log 5 + 10 \log 3 - 30 \log 2 - \frac{25}{2} \log 3 + \frac{25}{2} \log 5 - \frac{25}{2} \log 2\) Alternatively, substituting back into \(I = 5 - I_1\).

Final simplified form: \(5 - [ 10 \log(\frac{15}{8}) - \frac{25}{2} \log(\frac{6}{5}) ]\)

Question 17. Evaluate \(\int_{0}^{\pi/4} (2\sec^2 x + x^3 + 2) dx\)

$$I = \left[ 2 \tan x + \frac{x^4}{4} + 2x \right]_{0}^{\pi/4}$$ $$I = \left( 2 \tan \frac{\pi}{4} + \frac{(\pi/4)^4}{4} + 2(\frac{\pi}{4}) \right) - (0)$$ $$I = 2(1) + \frac{\pi^4}{1024} + \frac{\pi}{2}$$ $$I = 2 + \frac{\pi}{2} + \frac{\pi^4}{1024}$$

Question 18. Evaluate \(\int_{0}^{\pi} (\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2}) dx\)

Use formula: \(\cos^2 A - \sin^2 A = \cos 2A\).

$$\sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} = -(\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}) = -\cos x$$

$$I = \int_{0}^{\pi} (-\cos x) dx = -[\sin x]_{0}^{\pi}$$ $$I = -(\sin \pi - \sin 0) = -(0 - 0) = 0$$

Answer: 0

Question 19. Evaluate \(\int_{0}^{2} \frac{6x+3}{x^2+4} dx\)

Split the integral:

$$I = \int_{0}^{2} \frac{6x}{x^2+4} dx + \int_{0}^{2} \frac{3}{x^2+4} dx$$ $$I = 3 \int_{0}^{2} \frac{2x}{x^2+4} dx + 3 \int_{0}^{2} \frac{dx}{x^2+2^2}$$

$$I = 3 [\log(x^2+4)]_{0}^{2} + 3 \cdot \frac{1}{2} [\tan^{-1} \frac{x}{2}]_{0}^{2}$$ $$I = 3 (\log 8 - \log 4) + \frac{3}{2} (\tan^{-1} 1 - \tan^{-1} 0)$$ $$I = 3 \log 2 + \frac{3}{2} (\frac{\pi}{4}) = 3 \log 2 + \frac{3\pi}{8}$$

Answer: \(3 \log 2 + \frac{3\pi}{8}\)

Question 20. Evaluate \(\int_{0}^{1} (x e^x + \sin \frac{\pi x}{4}) dx\)

Part 1: \(\int x e^x dx\). Using integration by parts: \(x e^x - \int e^x dx = x e^x - e^x\).

$$[x e^x - e^x]_{0}^{1} = (e - e) - (0 - 1) = 1$$

Part 2: \(\int \sin \frac{\pi x}{4} dx = \frac{-\cos(\pi x / 4)}{\pi / 4} = -\frac{4}{\pi} \cos \frac{\pi x}{4}\).

$$\left[ -\frac{4}{\pi} \cos \frac{\pi x}{4} \right]_{0}^{1} = -\frac{4}{\pi} (\cos \frac{\pi}{4} - \cos 0)$$ $$= -\frac{4}{\pi} (\frac{1}{\sqrt{2}} - 1) = \frac{4}{\pi} - \frac{2\sqrt{2}}{\pi}$$

Total \(I = 1 + \frac{4 - 2\sqrt{2}}{\pi}\)

Question 21. \(\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}\) equals

$$I = [\tan^{-1} x]_{1}^{\sqrt{3}}$$ $$I = \tan^{-1} \sqrt{3} - \tan^{-1} 1$$ $$I = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12}$$

A. \(\pi/3\)

B. \(2\pi/3\)

C. \(\pi/6\)

D. \(\pi/12\)

Correct Option: D

Question 22. \(\int_{0}^{2/3} \frac{dx}{4+9x^2}\) equals

$$I = \int_{0}^{2/3} \frac{dx}{(2)^2 + (3x)^2}$$ Using \(\int \frac{dx}{a^2 + (bx)^2} = \frac{1}{b} \cdot \frac{1}{a} \tan^{-1} \frac{bx}{a}\)

$$I = \frac{1}{3} \cdot \frac{1}{2} [\tan^{-1} \frac{3x}{2}]_{0}^{2/3}$$ $$I = \frac{1}{6} [\tan^{-1}(\frac{3}{2} \cdot \frac{2}{3}) - \tan^{-1} 0]$$ $$I = \frac{1}{6} (\tan^{-1} 1 - 0) = \frac{1}{6} \cdot \frac{\pi}{4} = \frac{\pi}{24}$$

A. \(\pi/6\)

B. \(\pi/12\)

C. \(\pi/24\)

D. \(\pi/4\)

Correct Option: C

Integrals Class 12 Exercise 7.11 Solutions | Maths Part II CBSE

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• Miscellaneous Exercise

Integrals Class 12 Mathematics Part II CBSE Solution - Exercise 7.11

Detailed solutions for Definite Integrals using properties (Exercise 7.11).

Question 1. By using the properties of definite integrals, evaluate the integral: \(\int_{0}^{\pi/2} \cos^2 x \, dx\)

Let \( I = \int_{0}^{\pi/2} \cos^2 x \, dx \quad \dots(1) \)

Using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \):

\( I = \int_{0}^{\pi/2} \cos^2(\frac{\pi}{2} - x) \, dx \)
\( \Rightarrow I = \int_{0}^{\pi/2} \sin^2 x \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} (\cos^2 x + \sin^2 x) \, dx \)
\( 2I = \int_{0}^{\pi/2} 1 \, dx \)
\( 2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} - 0 \)
\( I = \frac{\pi}{4} \)

Question 2. Evaluate: \(\int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx\)

Let \( I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \quad \dots(1) \)

Using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \):

\( I = \int_{0}^{\pi/2} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} \, dx \)
\( I = \int_{0}^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \, dx \)
\( 2I = \int_{0}^{\pi/2} 1 \, dx \)
\( 2I = [x]_{0}^{\pi/2} = \frac{\pi}{2} \)
\( I = \frac{\pi}{4} \)

Question 3. Evaluate: \(\int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx\)

Let \( I = \int_{0}^{\pi/2} \frac{\sin^{3/2} x}{\sin^{3/2} x + \cos^{3/2} x} \, dx \quad \dots(1) \)

Using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \):

\( I = \int_{0}^{\pi/2} \frac{\sin^{3/2}(\frac{\pi}{2}-x)}{\sin^{3/2}(\frac{\pi}{2}-x) + \cos^{3/2}(\frac{\pi}{2}-x)} \, dx \)
\( I = \int_{0}^{\pi/2} \frac{\cos^{3/2} x}{\cos^{3/2} x + \sin^{3/2} x} \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} 1 \, dx \)
\( 2I = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4} \)

Question 4. Evaluate: \(\int_{0}^{\pi/2} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} \, dx\)

Let \( I = \int_{0}^{\pi/2} \frac{\cos^5 x}{\sin^5 x + \cos^5 x} \, dx \quad \dots(1) \)

Using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx \):

\( I = \int_{0}^{\pi/2} \frac{\cos^5(\frac{\pi}{2}-x)}{\sin^5(\frac{\pi}{2}-x) + \cos^5(\frac{\pi}{2}-x)} \, dx \)
\( I = \int_{0}^{\pi/2} \frac{\sin^5 x}{\cos^5 x + \sin^5 x} \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} 1 \, dx = \frac{\pi}{2} \)
\( I = \frac{\pi}{4} \)

Question 5. Evaluate: \(\int_{-5}^{5} |x+2| \, dx\)

We split the integral at \( x = -2 \) because \( |x+2| \) changes definition there.

\( |x+2| = -(x+2) \) for \( x < -2 \) and \( (x+2) \) for \( x \ge -2 \).

\( I = \int_{-5}^{-2} -(x+2) \, dx + \int_{-2}^{5} (x+2) \, dx \)
\( I = -[\frac{x^2}{2} + 2x]_{-5}^{-2} + [\frac{x^2}{2} + 2x]_{-2}^{5} \)

Calculating the first part:

\( -[(\frac{4}{2} - 4) - (\frac{25}{2} - 10)] = -[(2-4) - (12.5-10)] = -[-2 - 2.5] = 4.5 \)

Calculating the second part:

\( [(\frac{25}{2} + 10) - (\frac{4}{2} - 4)] = [(12.5+10) - (2-4)] = [22.5 - (-2)] = 24.5 \)

\( I = 4.5 + 24.5 = 29 \)

Question 6. Evaluate: \(\int_{2}^{8} |x-5| \, dx\)

Split the integral at \( x = 5 \).

\( |x-5| = -(x-5) \) for \( x < 5 \) and \( (x-5) \) for \( x \ge 5 \).

\( I = \int_{2}^{5} -(x-5) \, dx + \int_{5}^{8} (x-5) \, dx \)
\( I = -[\frac{x^2}{2} - 5x]_{2}^{5} + [\frac{x^2}{2} - 5x]_{5}^{8} \)

\( I = -[(\frac{25}{2}-25) - (\frac{4}{2}-10)] + [(\frac{64}{2}-40) - (\frac{25}{2}-25)] \)
\( I = -[-12.5 - (-8)] + [-8 - (-12.5)] \)
\( I = -[-4.5] + [4.5] = 4.5 + 4.5 = 9 \)

Answer: \( I = 9 \)

Question 7. Evaluate: \(\int_{0}^{1} x(1-x)^n \, dx\)

Using property \( \int_{0}^{a} f(x)dx = \int_{0}^{a} f(a-x)dx \):

\( I = \int_{0}^{1} (1-x)(1-(1-x))^n \, dx \)
\( I = \int_{0}^{1} (1-x)x^n \, dx \)
\( I = \int_{0}^{1} (x^n - x^{n+1}) \, dx \)

\( I = [\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2}]_{0}^{1} \)
\( I = (\frac{1}{n+1} - \frac{1}{n+2}) - 0 \)
\( I = \frac{(n+2)-(n+1)}{(n+1)(n+2)} = \frac{1}{(n+1)(n+2)} \)

Question 8. Evaluate: \(\int_{0}^{\pi/4} \log(1+\tan x) \, dx\)

Let \( I = \int_{0}^{\pi/4} \log(1+\tan x) \, dx \quad \dots(1) \)

Using property \( x \to \frac{\pi}{4}-x \):

\( I = \int_{0}^{\pi/4} \log(1+\tan(\frac{\pi}{4}-x)) \, dx \)
Since \( \tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B} \), we have \( \tan(\frac{\pi}{4}-x) = \frac{1-\tan x}{1+\tan x} \).

\( I = \int_{0}^{\pi/4} \log(1 + \frac{1-\tan x}{1+\tan x}) \, dx \)
\( I = \int_{0}^{\pi/4} \log(\frac{1+\tan x + 1 - \tan x}{1+\tan x}) \, dx \)
\( I = \int_{0}^{\pi/4} \log(\frac{2}{1+\tan x}) \, dx \)
\( I = \int_{0}^{\pi/4} (\log 2 - \log(1+\tan x)) \, dx \)

\( I = \int_{0}^{\pi/4} \log 2 \, dx - I \)
\( 2I = \log 2 [x]_{0}^{\pi/4} = \frac{\pi}{4} \log 2 \)
\( I = \frac{\pi}{8} \log 2 \)

Question 9. Evaluate: \(\int_{0}^{2} x\sqrt{2-x} \, dx\)

Using property \( x \to 2-x \):

\( I = \int_{0}^{2} (2-x)\sqrt{2-(2-x)} \, dx \)
\( I = \int_{0}^{2} (2-x)\sqrt{x} \, dx \)
\( I = \int_{0}^{2} (2x^{1/2} - x^{3/2}) \, dx \)

\( I = [2\frac{x^{3/2}}{3/2} - \frac{x^{5/2}}{5/2}]_{0}^{2} \)
\( I = [\frac{4}{3}x^{3/2} - \frac{2}{5}x^{5/2}]_{0}^{2} \)
\( I = \frac{4}{3}(2\sqrt{2}) - \frac{2}{5}(4\sqrt{2}) \)
\( I = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} = 8\sqrt{2}(\frac{5-3}{15}) = \frac{16\sqrt{2}}{15} \)

Question 10. Evaluate: \(\int_{0}^{\pi/2} (2\log \sin x - \log \sin 2x) \, dx\)

\( I = \int_{0}^{\pi/2} (2\log \sin x - \log(2\sin x \cos x)) \, dx \)
\( I = \int_{0}^{\pi/2} (2\log \sin x - \log 2 - \log \sin x - \log \cos x) \, dx \)
\( I = \int_{0}^{\pi/2} (\log \sin x - \log \cos x - \log 2) \, dx \quad \dots(1) \)

Using property \( P_4 \):

\( I = \int_{0}^{\pi/2} (\log \cos x - \log \sin x - \log 2) \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{\pi/2} (-2\log 2) \, dx \)
\( 2I = -2\log 2 [x]_{0}^{\pi/2} = -\pi \log 2 \)
\( I = -\frac{\pi}{2} \log 2 \) or \( \frac{\pi}{2} \log(\frac{1}{2}) \)

Question 11. Evaluate: \(\int_{-\pi/2}^{\pi/2} \sin^2 x \, dx\)

Let \( f(x) = \sin^2 x \). Since \( f(-x) = \sin^2(-x) = (-\sin x)^2 = \sin^2 x = f(x) \), \( f(x) \) is an even function.

Using the property for even functions:

\( I = 2 \int_{0}^{\pi/2} \sin^2 x \, dx \)

From Question 1, we know \( \int_{0}^{\pi/2} \sin^2 x \, dx = \frac{\pi}{4} \).

\( I = 2(\frac{\pi}{4}) = \frac{\pi}{2} \)

Question 12. Evaluate: \(\int_{0}^{\pi} \frac{x}{1+\sin x} \, dx\)

Using property \( x \to \pi - x \):

\( I = \int_{0}^{\pi} \frac{\pi - x}{1+\sin(\pi-x)} \, dx = \int_{0}^{\pi} \frac{\pi - x}{1+\sin x} \, dx \)

\( 2I = \int_{0}^{\pi} \frac{x + \pi - x}{1+\sin x} \, dx = \pi \int_{0}^{\pi} \frac{1}{1+\sin x} \, dx \)

Multiply numerator and denominator by \( 1-\sin x \):

\( 2I = \pi \int_{0}^{\pi} \frac{1-\sin x}{\cos^2 x} \, dx = \pi \int_{0}^{\pi} (\sec^2 x - \tan x \sec x) \, dx \)
\( 2I = \pi [\tan x - \sec x]_{0}^{\pi} \)
\( 2I = \pi [(\tan \pi - \sec \pi) - (\tan 0 - \sec 0)] \)
\( 2I = \pi [(0 - (-1)) - (0 - 1)] = \pi [1+1] = 2\pi \)
\( I = \pi \)

Question 13. Evaluate: \(\int_{-\pi/2}^{\pi/2} \sin^7 x \, dx\)

Let \( f(x) = \sin^7 x \).

\( f(-x) = \sin^7(-x) = (-\sin x)^7 = -\sin^7 x = -f(x) \).

Since \( f(x) \) is an odd function, the integral over \( [-a, a] \) is zero.

\( I = 0 \)

Question 14. Evaluate: \(\int_{0}^{2\pi} \cos^5 x \, dx\)

Let \( f(x) = \cos^5 x \).

\( f(2\pi - x) = \cos^5(2\pi - x) = \cos^5 x = f(x) \).
So, \( I = 2 \int_{0}^{\pi} \cos^5 x \, dx \).

Now consider the integral over \( [0, \pi] \). Let \( g(x) = \cos^5 x \).

\( g(\pi - x) = \cos^5(\pi - x) = (-\cos x)^5 = -\cos^5 x = -g(x) \).

Thus, \( \int_{0}^{\pi} \cos^5 x \, dx = 0 \).

Answer: \( I = 0 \)

Question 15. Evaluate: \(\int_{0}^{\pi/2} \frac{\sin x - \cos x}{1+\sin x \cos x} \, dx\)

Using property \( x \to \frac{\pi}{2} - x \):

\( I = \int_{0}^{\pi/2} \frac{\sin(\frac{\pi}{2}-x) - \cos(\frac{\pi}{2}-x)}{1 + \sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)} \, dx \)
\( I = \int_{0}^{\pi/2} \frac{\cos x - \sin x}{1 + \cos x \sin x} \, dx \)
\( I = - \int_{0}^{\pi/2} \frac{\sin x - \cos x}{1 + \sin x \cos x} \, dx \)

\( I = -I \Rightarrow 2I = 0 \Rightarrow I = 0 \)

Question 16. Evaluate: \(\int_{0}^{\pi} \log(1+\cos x) \, dx\)

\( I = \int_{0}^{\pi} \log(1+\cos(\pi-x)) \, dx = \int_{0}^{\pi} \log(1-\cos x) \, dx \)
\( 2I = \int_{0}^{\pi} \log((1+\cos x)(1-\cos x)) \, dx = \int_{0}^{\pi} \log(1-\cos^2 x) \, dx \)
\( 2I = \int_{0}^{\pi} \log(\sin^2 x) \, dx = 2\int_{0}^{\pi} \log \sin x \, dx \)
\( I = \int_{0}^{\pi} \log \sin x \, dx \)

Using property \( f(2a-x)=f(x) \) where \( 2a=\pi \), we have \( \sin(\pi-x)=\sin x \).

\( I = 2\int_{0}^{\pi/2} \log \sin x \, dx \)

We know \( \int_{0}^{\pi/2} \log \sin x \, dx = -\frac{\pi}{2} \log 2 \).

\( I = 2(-\frac{\pi}{2} \log 2) = -\pi \log 2 \)

Question 17. Evaluate: \(\int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} \, dx\)

Let \( I = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a-x}} \, dx \quad \dots(1) \)

Using property \( x \to a-x \):

\( I = \int_{0}^{a} \frac{\sqrt{a-x}}{\sqrt{a-x} + \sqrt{x}} \, dx \quad \dots(2) \)

Adding (1) and (2):

\( 2I = \int_{0}^{a} 1 \, dx = [x]_{0}^{a} = a \)
\( I = \frac{a}{2} \)

Question 18. Evaluate: \(\int_{0}^{4} |x-1| \, dx\)

Split at \( x=1 \).

\( I = \int_{0}^{1} -(x-1) \, dx + \int_{1}^{4} (x-1) \, dx \)
\( I = [x - \frac{x^2}{2}]_{0}^{1} + [\frac{x^2}{2} - x]_{1}^{4} \)

\( I = (1 - 0.5) + [(8-4) - (0.5-1)] \)
\( I = 0.5 + [4 + 0.5] = 5 \)

Question 19. Show that \(\int_{0}^{a} f(x)g(x) \, dx = 2\int_{0}^{a} f(x) \, dx\), if \( f \) and \( g \) are defined as \( f(x)=f(a-x) \) and \( g(x)+g(a-x)=4 \).

Let \( I = \int_{0}^{a} f(x)g(x) \, dx \).

Using property \( x \to a-x \):

\( I = \int_{0}^{a} f(a-x)g(a-x) \, dx \)
Since \( f(a-x) = f(x) \) and \( g(a-x) = 4 - g(x) \):
\( I = \int_{0}^{a} f(x)(4 - g(x)) \, dx \)
\( I = 4\int_{0}^{a} f(x) \, dx - \int_{0}^{a} f(x)g(x) \, dx \)
\( I = 4\int_{0}^{a} f(x) \, dx - I \)

\( 2I = 4\int_{0}^{a} f(x) \, dx \)
\( I = 2\int_{0}^{a} f(x) \, dx \)

Hence proved.

Question 20. The value of \(\int_{-\pi/2}^{\pi/2} (x^3 + x\cos x + \tan^5 x + 1) \, dx\) is

Let \( f(x) = x^3 + x\cos x + \tan^5 x + 1 \).

\( \int_{-\pi/2}^{\pi/2} x^3 \, dx = 0 \) (Odd function)
\( \int_{-\pi/2}^{\pi/2} x\cos x \, dx = 0 \) (Odd function)
\( \int_{-\pi/2}^{\pi/2} \tan^5 x \, dx = 0 \) (Odd function)
\( \int_{-\pi/2}^{\pi/2} 1 \, dx = [x]_{-\pi/2}^{\pi/2} = \pi \)

Total value = \( \pi \).

C. \(\pi\)

Question 21. The value of \(\int_{0}^{\pi/2} \log(\frac{4+3\sin x}{4+3\cos x}) \, dx\) is

Let \( I = \int_{0}^{\pi/2} \log(\frac{4+3\sin x}{4+3\cos x}) \, dx \).

Using property \( x \to \frac{\pi}{2}-x \):

\( I = \int_{0}^{\pi/2} \log(\frac{4+3\sin(\pi/2-x)}{4+3\cos(\pi/2-x)}) \, dx \)
\( I = \int_{0}^{\pi/2} \log(\frac{4+3\cos x}{4+3\sin x}) \, dx \)
\( I = \int_{0}^{\pi/2} \log((\frac{4+3\sin x}{4+3\cos x})^{-1}) \, dx \)
\( I = -\int_{0}^{\pi/2} \log(\frac{4+3\sin x}{4+3\cos x}) \, dx \)
\( I = -I \)

\( 2I = 0 \Rightarrow I = 0 \)

C. 0