GSEB Class 12 Maths Solutions Chapter 7 સંકલન Exercise 7.11

નિયત સંકલનના ગુણધર્મોનો ઉપયોગ કરી પ્રશ્નો 1 થી 19 માં દર્શાવેલા સંકલિતોની કિંમત શોધો :

Question 1. \( \int_0^{\frac{\pi}{2}} \cos2x \, dx \)
Answer:
અહીં,
\( I = \int_0^{\frac{\pi}{2}} \cos^2 x \, dx \) ....(i)
સંકલનના ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરતાં,
\( I = \int_0^{\frac{\pi}{2}} \cos^2 \left( \frac{\pi}{2} - x \right) \, dx \)
\( I = \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \) ....(ii)
સમીકરણ (i) અને (ii) નો સરવાળો કરતાં,
\( 2I = \int_0^{\frac{\pi}{2}} \cos^2 x \, dx + \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \)
\( 2I = \int_0^{\frac{\pi}{2}} (\cos^2 x + \sin^2 x) \, dx \)
\( 2I = \int_0^{\frac{\pi}{2}} 1 \, dx \)
\( 2I = [x]_0^{\frac{\pi}{2}} \)
\( 2I = \frac{\pi}{2} - 0 \)
\( 2I = \frac{\pi}{2} \)
\( I = \frac{\pi}{4} \)
In simple words: પહેલા આપેલ સમીકરણને I માનો. પછી, સંકલનના નિયમનો ઉપયોગ કરીને \( \cos^2 x \) ને \( \sin^2 x \) માં બદલો. બંને સમીકરણોનો સરવાળો કરવાથી, \( \cos^2 x + \sin^2 x \) 1 બની જશે. 1 નું સંકલન \( x \) થાય છે, જેમાં કિંમતો મૂકીને જવાબ મેળવી શકાય છે.

Exam Tip: જ્યારે સંકલનમાં \( \cos^2 x \) અથવા \( \sin^2 x \) હોય અને મર્યાદા \( 0 \) થી \( \frac{\pi}{2} \) હોય, ત્યારે ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરીને સમસ્યાને સરળ બનાવવાનો પ્રયાસ કરો.

Question 2. \( \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x+\sqrt{\cos x}}}dx \)
Answer:
અહીં,
\( I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x+\sqrt{\cos x}}} \, dx \) ....(i)
સંકલનના ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરતાં,
\( I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin \left(\frac{\pi}{2} - x\right)}}{\sqrt{\sin \left(\frac{\pi}{2} - x\right)+\sqrt{\cos \left(\frac{\pi}{2} - x\right)}}} \, dx \)
\( I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x+\sqrt{\sin x}}} \, dx \) ....(ii)
સમીકરણ (i) અને (ii) નો સરવાળો કરતાં,
\( 2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x+\sqrt{\cos x}}} \, dx + \int_0^{\frac{\pi}{2}} \frac{\sqrt{\cos x}}{\sqrt{\cos x+\sqrt{\sin x}}} \, dx \)
\( 2I = \int_0^{\frac{\pi}{2}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x+\sqrt{\cos x}}} \, dx \)
\( 2I = \int_0^{\frac{\pi}{2}} 1 \, dx \)
\( 2I = [x]_0^{\frac{\pi}{2}} \)
\( 2I = \frac{\pi}{2} - 0 \)
\( 2I = \frac{\pi}{2} \)
\( I = \frac{\pi}{4} \)
In simple words: સૌપ્રથમ, આપેલ સંકલનને I ધારો. પછી, ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરીને \( \sin x \) ને \( \cos x \) માં અને \( \cos x \) ને \( \sin x \) માં બદલો. બંને સમીકરણોનો સરવાળો કરવાથી, અંશ અને છેદ સમાન બની જશે, જેથી સંકલન 1 નું થશે. આ રીતે ગણતરી કરતા, આપણને પરિણામ મળે છે.

Exam Tip: જ્યારે સંકલનમાં \( \sin x \) અને \( \cos x \) બંને હોય અને મર્યાદા \( 0 \) થી \( \frac{\pi}{2} \) હોય, ત્યારે \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મ લાગુ પાડવાથી સામાન્ય રીતે સમસ્યાનો ઉકેલ સરળ બને છે.

Question 3. \( \int_0^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x \, dx}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x} \)
Answer:
અહીં,
\( I = \int_0^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x+\cos^{\frac{3}{2}} x} \, dx \) ....(i)
સંકલનના ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરતાં,
\( I = \int_0^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} \left(\frac{\pi}{2} - x\right)}{\sin^{\frac{3}{2}} \left(\frac{\pi}{2} - x\right)+\cos^{\frac{3}{2}} \left(\frac{\pi}{2} - x\right)} \, dx \)
\( I = \int_0^{\frac{\pi}{2}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x+\sin^{\frac{3}{2}} x} \, dx \) ....(ii)
સમીકરણ (i) અને (ii) નો સરવાળો કરતાં,
\( 2I = \int_0^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x+\cos^{\frac{3}{2}} x} \, dx + \int_0^{\frac{\pi}{2}} \frac{\cos^{\frac{3}{2}} x}{\cos^{\frac{3}{2}} x+\sin^{\frac{3}{2}} x} \, dx \)
\( 2I = \int_0^{\frac{\pi}{2}} \frac{\sin^{\frac{3}{2}} x+\cos^{\frac{3}{2}} x}{\sin^{\frac{3}{2}} x+\cos^{\frac{3}{2}} x} \, dx \)
\( 2I = \int_0^{\frac{\pi}{2}} 1 \, dx \)
\( 2I = [x]_0^{\frac{\pi}{2}} \)
\( 2I = \frac{\pi}{2} - 0 \)
\( 2I = \frac{\pi}{2} \)
\( I = \frac{\pi}{4} \)
In simple words: પહેલા સંકલનને I માનો. પછી, \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) આ ગુણધર્મ વાપરો, જેનાથી \( \sin x \) \( \cos x \) માં અને \( \cos x \) \( \sin x \) માં બદલાઈ જશે. પછી બંને સમીકરણો ઉમેરવાથી, અંશ અને છેદ સરખા બની જશે, અને સંકલન 1 નું થઈ જશે.

Exam Tip: જ્યારે સંકલનમાં \( \sin x \) અને \( \cos x \) ની ઘાત સમાન હોય અને મર્યાદા \( 0 \) થી \( \frac{\pi}{2} \) હોય, ત્યારે \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મનો ઉપયોગ કરવાથી દાખલો ખૂબ સરળ બની જાય છે.

Question 4. \( \int_0^{\frac{\pi}{2}} \frac{\cos ^5 x}{\sin ^5 x+\cos ^5 x}dx \)
Answer:
અહીં,
\( I = \int_0^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^5 x+\cos^5 x} \, dx \) ....(i)
સંકલનના ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરતાં,
\( I = \int_0^{\frac{\pi}{2}} \frac{\cos^5 \left(\frac{\pi}{2} - x\right)}{\sin^5 \left(\frac{\pi}{2} - x\right)+\cos^5 \left(\frac{\pi}{2} - x\right)} \, dx \)
\( I = \int_0^{\frac{\pi}{2}} \frac{\sin^5 x}{\cos^5 x+\sin^5 x} \, dx \) ....(ii)
સમીકરણ (i) અને (ii) નો સરવાળો કરતાં,
\( 2I = \int_0^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^5 x+\cos^5 x} \, dx + \int_0^{\frac{\pi}{2}} \frac{\sin^5 x}{\cos^5 x+\sin^5 x} \, dx \)
\( 2I = \int_0^{\frac{\pi}{2}} \frac{\cos^5 x+\sin^5 x}{\sin^5 x+\cos^5 x} \, dx \)
\( 2I = \int_0^{\frac{\pi}{2}} 1 \, dx \)
\( 2I = [x]_0^{\frac{\pi}{2}} \)
\( 2I = \frac{\pi}{2} - 0 \)
\( 2I = \frac{\pi}{2} \)
\( I = \frac{\pi}{4} \)
In simple words: આપેલ સંકલનને I તરીકે લો. પછી, સંકલનનો ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) લાગુ પાડો, જેનાથી \( \cos x \) \( \sin x \) માં અને \( \sin x \) \( \cos x \) માં બદલાઈ જાય છે. બંને સમીકરણોનો સરવાળો કરવાથી, અંશ અને છેદ સમાન બનશે, અને સંકલન 1 નું થઈ જાય છે, જે પછી સરળતાથી ગણતરી કરી શકાય છે.

Exam Tip: આ પ્રકારના સંકલનોમાં, જો અંશ અને છેદમાં \( \sin x \) અને \( \cos x \) ની સમાન ઘાત હોય અને સીમાઓ \( 0 \) થી \( \frac{\pi}{2} \) હોય, તો ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરવાથી જવાબ \( \frac{a}{2} \) મળે છે.

Question 5. \( \int_{-5}^5|x + 2|dx \)
Answer:
સ્પષ્ટ છે કે \( |x + 2| = x + 2, \) જો \( x \ge -2 \)
અને \( |x + 2| = -(x + 2) = -x - 2, \) જો \( x < -2 \)
આથી,
\( I = \int_{-5}^5|x + 2| \, dx \)
\( I = \int_{-5}^{-2}|x + 2| \, dx + \int_{-2}^5|x + 2| \, dx \)
\( I = \int_{-5}^{-2}(-x - 2) \, dx + \int_{-2}^5(x + 2) \, dx \)
\( I = \left[-\frac{x^2}{2} - 2x\right]_{-5}^{-2} + \left[\frac{x^2}{2} + 2x\right]_{-2}^5 \)
\( I = \left[ \left(-\frac{(-2)^2}{2} - 2(-2)\right) - \left(-\frac{(-5)^2}{2} - 2(-5)\right) \right] + \left[ \left(\frac{5^2}{2} + 2(5)\right) - \left(\frac{(-2)^2}{2} + 2(-2)\right) \right] \)
\( I = \left[ \left(-\frac{4}{2} + 4\right) - \left(-\frac{25}{2} + 10\right) \right] + \left[ \left(\frac{25}{2} + 10\right) - \left(\frac{4}{2} - 4\right) \right] \)
\( I = \left[ (-2 + 4) - \left(-\frac{25}{2} + \frac{20}{2}\right) \right] + \left[ \left(\frac{25}{2} + \frac{20}{2}\right) - (2 - 4) \right] \)
\( I = \left[ 2 - \left(-\frac{5}{2}\right) \right] + \left[ \frac{45}{2} - (-2) \right] \)
\( I = \left[ 2 + \frac{5}{2} \right] + \left[ \frac{45}{2} + 2 \right] \)
\( I = \frac{4+5}{2} + \frac{45+4}{2} \)
\( I = \frac{9}{2} + \frac{49}{2} \)
\( I = \frac{58}{2} \)
\( I = 29 \)
In simple words: સૌપ્રથમ, \( |x+2| \) ને તેની વ્યાખ્યા પ્રમાણે બે ભાગમાં વહેંચી દો. પછી, સંકલનની સીમાઓને જ્યાં \( x+2 \) તેનું ચિહ્ન બદલે છે તે બિંદુ (-2) પર વિભાજીત કરો. દરેક ભાગનું અલગ-અલગ સંકલન કરો અને પછી સીમાઓ મૂકીને પરિણામોનો સરવાળો કરો.

Exam Tip: માનાંક કાર્ય (absolute value function) ના સંકલન કરતી વખતે, હંમેશા તે બિંદુ શોધો જ્યાં માનાંક કાર્ય શૂન્ય બને છે. તે બિંદુએ સંકલનની સીમાઓને વિભાજીત કરો અને પછી દરેક ભાગનું અલગથી સંકલન કરો.

Question 6. \( \int_2^8|x - 5|dx \)
Answer:
સ્પષ્ટ છે કે \( |x - 5| = x - 5, \) જો \( x \ge 5 \)
અને \( |x - 5| = -(x - 5) = -x + 5, \) જો \( x < 5 \)
આથી,
\( I = \int_2^8|x - 5| \, dx \)
\( I = \int_2^5|x - 5| \, dx + \int_5^8|x - 5| \, dx \)
\( I = \int_2^5(-x + 5) \, dx + \int_5^8(x - 5) \, dx \)
\( I = \left[-\frac{x^2}{2} + 5x\right]_2^5 + \left[\frac{x^2}{2} - 5x\right]_5^8 \)
\( I = \left[ \left(-\frac{5^2}{2} + 5(5)\right) - \left(-\frac{2^2}{2} + 5(2)\right) \right] + \left[ \left(\frac{8^2}{2} - 5(8)\right) - \left(\frac{5^2}{2} - 5(5)\right) \right] \)
\( I = \left[ \left(-\frac{25}{2} + 25\right) - \left(-\frac{4}{2} + 10\right) \right] + \left[ \left(\frac{64}{2} - 40\right) - \left(\frac{25}{2} - 25\right) \right] \)
\( I = \left[ \frac{-25+50}{2} - (-2 + 10) \right] + \left[ (32 - 40) - \left(\frac{25-50}{2}\right) \right] \)
\( I = \left[ \frac{25}{2} - 8 \right] + \left[ -8 - \left(-\frac{25}{2}\right) \right] \)
\( I = \frac{25}{2} - \frac{16}{2} + \frac{-16+25}{2} \)
\( I = \frac{9}{2} + \frac{9}{2} \)
\( I = \frac{18}{2} \)
\( I = 9 \)
In simple words: સૌપ્રથમ, માનાંક કાર્ય \( |x-5| \) ને તેની વ્યાખ્યા મુજબ બે ભાગમાં વિભાજીત કરો: એક જ્યારે \( x \ge 5 \) હોય અને બીજું જ્યારે \( x < 5 \) હોય. પછી, સંકલનની મૂળભૂત સીમાઓને તે બિંદુ (5) પર વિભાજીત કરો જ્યાં કાર્ય તેનું ચિહ્ન બદલે છે. દરેક નવા ભાગનું અલગ-અલગ સંકલન કરો અને પરિણામોનો સરવાળો કરો.

Exam Tip: માનાંક સંકલન માટે, હંમેશા તે બિંદુ શોધીને સંકલન સીમાઓને વિભાજીત કરો જ્યાં માનાંક કાર્ય શૂન્ય બને છે. આનાથી કાર્યને યોગ્ય રીતે વ્યાખ્યાયિત કરીને સંકલન કરવામાં મદદ મળે છે.

Question 7. \( \int_0^1x(1-x)^n dx \)
Answer:
અહીં,
\( I = \int_0^1 x (1 - x)^n \, dx \)
સંકલનના ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરતાં,
\( I = \int_0^1 (1 - x) (1 - (1 - x))^n \, dx \)
\( I = \int_0^1 (1 - x) (x)^n \, dx \)
\( I = \int_0^1 (x^n - x^{n+1}) \, dx \)
\( I = \left[\frac{x^{n+1}}{n+1} - \frac{x^{n+2}}{n+2}\right]_0^1 \)
\( I = \left(\frac{1^{n+1}}{n+1} - \frac{1^{n+2}}{n+2}\right) - \left(\frac{0^{n+1}}{n+1} - \frac{0^{n+2}}{n+2}\right) \)
\( I = \frac{1}{n+1} - \frac{1}{n+2} \)
\( I = \frac{(n+2) - (n+1)}{(n+1)(n+2)} \)
\( I = \frac{1}{(n+1)(n+2)} \)
In simple words: પહેલાં આપેલ સંકલનને I ધારો. પછી, ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરો, જેનાથી \( x \) ને \( 1-x \) થી બદલી શકાય. આ બદલાવ કરવાથી, કૌંસ સરળ બનશે અને તમે પદાવલિને \( x^n - x^{n+1} \) તરીકે લખી શકશો. પછી, દરેક પદનું અલગથી સંકલન કરો અને સીમાઓ મૂકીને અંતિમ જવાબ મેળવો.

Exam Tip: જ્યારે સંકલનની સીમાઓ \( 0 \) થી \( 1 \) હોય અને \( x(1-x)^n \) જેવું પદ હોય, ત્યારે \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મનો ઉપયોગ કરીને \( x \) ને \( a-x \) (અહીં \( 1-x \)) થી બદલવાથી દાખલો સરળ બની શકે છે.

Question 8. \( \int_0^{\frac{\pi}{4}}log(1 +tan x)dx \)
Answer:
અહીં,
\( I = \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \) ....(i)
સંકલનના ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરતાં,
\( I = \int_0^{\frac{\pi}{4}} \log\left(1 + \tan\left(\frac{\pi}{4} - x\right)\right) \, dx \)
\( I = \int_0^{\frac{\pi}{4}} \log\left(1 + \frac{\tan \frac{\pi}{4} - \tan x}{1 + \tan \frac{\pi}{4} \tan x}\right) \, dx \)
\( I = \int_0^{\frac{\pi}{4}} \log\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) \, dx \)
\( I = \int_0^{\frac{\pi}{4}} \log\left(\frac{1 + \tan x + 1 - \tan x}{1 + \tan x}\right) \, dx \)
\( I = \int_0^{\frac{\pi}{4}} \log\left(\frac{2}{1 + \tan x}\right) \, dx \)
\( I = \int_0^{\frac{\pi}{4}} (\log 2 - \log(1 + \tan x)) \, dx \)
\( I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - \int_0^{\frac{\pi}{4}} \log(1 + \tan x) \, dx \)
\( I = \int_0^{\frac{\pi}{4}} \log 2 \, dx - I \)
\( 2I = \log 2 [x]_0^{\frac{\pi}{4}} \)
\( 2I = \log 2 \left(\frac{\pi}{4} - 0\right) \)
\( 2I = \frac{\pi}{4} \log 2 \)
\( I = \frac{\pi}{8} \log 2 \)
In simple words: પહેલા, સંકલનને I ધારો. પછી, \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મનો ઉપયોગ કરીને \( x \) ને \( \frac{\pi}{4} - x \) થી બદલો. \( \tan(A-B) \) સૂત્રનો ઉપયોગ કરીને પદાવલિને સરળ બનાવો, જેથી તમને \( \log\left(\frac{2}{1 + \tan x}\right) \) મળશે. પછી, \( \log\left(\frac{a}{b}\right) = \log a - \log b \) નિયમનો ઉપયોગ કરીને સંકલનને બે ભાગમાં વહેંચો. એક ભાગ I બરાબર હશે, અને બીજો ભાગ \( \log 2 \) નું સંકલન હશે. આ રીતે I ને ઉકેલી શકાય છે.

Exam Tip: જ્યારે સંકલનમાં \( \log(1 + \tan x) \) હોય અને સીમાઓ \( 0 \) થી \( \frac{\pi}{4} \) હોય, ત્યારે ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) વાપરવાથી સમસ્યા ઘણીવાર સરળ બની જાય છે. \( \tan(\frac{\pi}{4} - x) \) નું વિસ્તરણ યાદ રાખવું મહત્વનું છે.

Question 9. \( \int_0^2x\sqrt{2-x}dx \)
Answer:
અહીં,
\( I = \int_0^2 x\sqrt{2-x} \, dx \)
સંકલનના ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરતાં,
\( I = \int_0^2 (2 - x)\sqrt{2 - (2 - x)} \, dx \)
\( I = \int_0^2 (2 - x)\sqrt{x} \, dx \)
\( I = \int_0^2 (2\sqrt{x} - x\sqrt{x}) \, dx \)
\( I = \int_0^2 (2x^{\frac{1}{2}} - x^{\frac{3}{2}}) \, dx \)
\( I = \left[2\frac{x^{\frac{3}{2}}}{\frac{3}{2}} - \frac{x^{\frac{5}{2}}}{\frac{5}{2}}\right]_0^2 \)
\( I = \left[\frac{4}{3}x^{\frac{3}{2}} - \frac{2}{5}x^{\frac{5}{2}}\right]_0^2 \)
\( I = \left(\frac{4}{3}(2)^{\frac{3}{2}} - \frac{2}{5}(2)^{\frac{5}{2}}\right) - (0 - 0) \)
\( I = \left(\frac{4}{3} \cdot 2\sqrt{2} - \frac{2}{5} \cdot 4\sqrt{2}\right) \)
\( I = \frac{8\sqrt{2}}{3} - \frac{8\sqrt{2}}{5} \)
\( I = 8\sqrt{2} \left(\frac{1}{3} - \frac{1}{5}\right) \)
\( I = 8\sqrt{2} \left(\frac{5 - 3}{15}\right) \)
\( I = 8\sqrt{2} \left(\frac{2}{15}\right) \)
\( I = \frac{16\sqrt{2}}{15} \)
In simple words: પહેલા સંકલનને I માનો. પછી, \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મનો ઉપયોગ કરીને \( x \) ને \( 2-x \) થી બદલો. આનાથી વર્ગમૂળ પદ \( \sqrt{x} \) બનશે. પછી, પદાવલિને વિસ્તૃત કરો અને \( x \) ની ઘાતના નિયમોનો ઉપયોગ કરીને દરેક પદનું સંકલન કરો. અંતે, સીમાઓ મૂકીને જવાબ શોધો.

Exam Tip: જ્યારે સંકલનમાં \( x \) અને \( a-x \) બંને હોય, ત્યારે \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મનો ઉપયોગ કરવાથી પદાવલિ સરળ બને છે. ખાસ કરીને, \( \sqrt{a-x} \) ને \( \sqrt{x} \) માં બદલવામાં મદદ મળે છે.

Question 10. \( \int_0^{\frac{\pi}{2}}(2log sin x – log sin 2x) dx \)
Answer:
અહીં,
\( I = \int_0^{\frac{\pi}{2}} (2\log \sin x - \log \sin 2x) \, dx \)
\( I = \int_0^{\frac{\pi}{2}} (2\log \sin x - \log (2\sin x \cos x)) \, dx \)
\( I = \int_0^{\frac{\pi}{2}} (2\log \sin x - (\log 2 + \log \sin x + \log \cos x)) \, dx \)
\( I = \int_0^{\frac{\pi}{2}} (\log \sin x - \log \cos x - \log 2) \, dx \)
\( I = \int_0^{\frac{\pi}{2}} \log \sin x \, dx - \int_0^{\frac{\pi}{2}} \log \cos x \, dx - \int_0^{\frac{\pi}{2}} \log 2 \, dx \)
અહીં, \( \int_0^{\frac{\pi}{2}} \log \sin x \, dx = \int_0^{\frac{\pi}{2}} \log \cos x \, dx \) (સંકલનના ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરવાથી).
તેથી,
\( I = \int_0^{\frac{\pi}{2}} \log \sin x \, dx - \int_0^{\frac{\pi}{2}} \log \sin x \, dx - \log 2 \int_0^{\frac{\pi}{2}} 1 \, dx \)
\( I = 0 - \log 2 [x]_0^{\frac{\pi}{2}} \)
\( I = - \log 2 \left(\frac{\pi}{2} - 0\right) \)
\( I = - \frac{\pi}{2} \log 2 \)
અથવા
\( I = \frac{\pi}{2} \log \left(\frac{1}{2}\right) \)
In simple words: પહેલા, \( \sin 2x \) ને \( 2\sin x \cos x \) માં બદલો. પછી, \( \log \) ના ગુણધર્મોનો ઉપયોગ કરીને તેને અલગ-અલગ \( \log \) પદોમાં વહેંચો. ત્યારબાદ, \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મને કારણે \( \int_0^{\frac{\pi}{2}} \log \sin x \, dx \) અને \( \int_0^{\frac{\pi}{2}} \log \cos x \, dx \) સમાન બની જાય છે. આનાથી તે પદો રદ થાય છે અને ફક્ત \( \log 2 \) નું સંકલન કરવાનું રહે છે.

Exam Tip: જ્યારે સંકલનમાં \( \log \sin x \), \( \log \cos x \) અથવા \( \log \tan x \) જેવા પદો હોય અને સીમાઓ \( 0 \) થી \( \frac{\pi}{2} \) હોય, ત્યારે \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મનો ઉપયોગ કરવાથી ઘણીવાર પદો રદ થઈ જાય છે અથવા સંકલન સરળ બની જાય છે. \( \log(A \cdot B) = \log A + \log B \) અને \( \log(\frac{A}{B}) = \log A - \log B \) ગુણધર્મો યાદ રાખો.

Question 11. \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2 x dx \)
Answer:
અહીં, \( f(x) = \sin^2 x \)
હવે, \( f(-x) = \sin^2 (-x) = (-\sin x)^2 = \sin^2 x = f(x) \)
આથી, \( f(x) \) યુગ્મ વિધેય છે. (એટલે કે, \( f(-x) = f(x) \))
જો \( f(x) \) યુગ્મ વિધેય હોય, તો \( \int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx \)
તેથી,
\( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \, dx \)
\( I = 2 \int_0^{\frac{\pi}{2}} \sin^2 x \, dx \)
આપણે જાણીએ છીએ કે \( \sin^2 x = \frac{1 - \cos 2x}{2} \)
\( I = 2 \int_0^{\frac{\pi}{2}} \frac{1 - \cos 2x}{2} \, dx \)
\( I = \int_0^{\frac{\pi}{2}} (1 - \cos 2x) \, dx \)
\( I = \left[x - \frac{\sin 2x}{2}\right]_0^{\frac{\pi}{2}} \)
\( I = \left(\frac{\pi}{2} - \frac{\sin 2(\frac{\pi}{2})}{2}\right) - \left(0 - \frac{\sin 2(0)}{2}\right) \)
\( I = \left(\frac{\pi}{2} - \frac{\sin \pi}{2}\right) - (0 - 0) \)
કારણ કે \( \sin \pi = 0 \),
\( I = \frac{\pi}{2} - 0 \)
\( I = \frac{\pi}{2} \)
In simple words: પહેલા, આપેલ વિધેય \( f(x) = \sin^2 x \) યુગ્મ છે કે વિયુગ્મ તે તપાસો. અહીં તે યુગ્મ વિધેય છે, તેથી સંકલન \( \int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx \) સૂત્રનો ઉપયોગ કરો. પછી, \( \sin^2 x \) ને \( \frac{1 - \cos 2x}{2} \) માં રૂપાંતરિત કરો. આનાથી સંકલન કરવાનું સરળ બનશે. છેલ્લે, સંકલન કરીને સીમાઓ મૂકીને જવાબ મેળવો.

Exam Tip: જ્યારે સંકલનની સીમાઓ \( -a \) થી \( a \) હોય, ત્યારે હંમેશા પહેલા તપાસો કે વિધેય યુગ્મ છે કે વિયુગ્મ. જો \( f(-x) = f(x) \) (યુગ્મ) હોય, તો \( \int_{-a}^a f(x) \, dx = 2 \int_0^a f(x) \, dx \). જો \( f(-x) = -f(x) \) (વિયુગ્મ) હોય, તો \( \int_{-a}^a f(x) \, dx = 0 \). આનાથી ગણતરી ઘણી સરળ બની શકે છે.

Question 12. \( \int_0^\pi \frac{x \, dx}{1+\sin x} \)
Answer:
અહીં,
\( I = \int_0^\pi \frac{x}{1 + \sin x} \, dx \) ....(i)
સંકલનના ગુણધર્મ \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) નો ઉપયોગ કરતાં,
\( I = \int_0^\pi \frac{(\pi - x)}{1 + \sin (\pi - x)} \, dx \)
\( I = \int_0^\pi \frac{\pi - x}{1 + \sin x} \, dx \) ....(ii) (કારણ કે \( \sin(\pi - x) = \sin x \))
સમીકરણ (i) અને (ii) નો સરવાળો કરતાં,
\( 2I = \int_0^\pi \frac{x}{1 + \sin x} \, dx + \int_0^\pi \frac{\pi - x}{1 + \sin x} \, dx \)
\( 2I = \int_0^\pi \frac{x + \pi - x}{1 + \sin x} \, dx \)
\( 2I = \int_0^\pi \frac{\pi}{1 + \sin x} \, dx \)
\( 2I = \pi \int_0^\pi \frac{1}{1 + \sin x} \, dx \)
હવે, \( \frac{1}{1 + \sin x} = \frac{1}{1 + \sin x} \cdot \frac{1 - \sin x}{1 - \sin x} = \frac{1 - \sin x}{1 - \sin^2 x} = \frac{1 - \sin x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} = \sec^2 x - \sec x \tan x \)
તેથી,
\( 2I = \pi \int_0^\pi (\sec^2 x - \sec x \tan x) \, dx \)
\( 2I = \pi [\tan x - \sec x]_0^\pi \)
\( 2I = \pi [(\tan \pi - \sec \pi) - (\tan 0 - \sec 0)] \)
\( 2I = \pi [(0 - (-1)) - (0 - 1)] \)
\( 2I = \pi [1 - (-1)] \)
\( 2I = \pi [1 + 1] \)
\( 2I = 2\pi \)
\( I = \pi \)
In simple words: પહેલાં આપેલ સંકલનને I ધારો. પછી, \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મનો ઉપયોગ કરો, જેનાથી \( x \) ને \( \pi - x \) થી બદલી શકાય. બંને સમીકરણોનો સરવાળો કરવાથી, \( x \) પદ રદ થઈ જશે અને \( \frac{\pi}{1 + \sin x} \) બાકી રહેશે. પછી, \( \frac{1}{1 + \sin x} \) ને \( \frac{1 - \sin x}{1 - \sin x} \) વડે ગુણીને સરળ બનાવો, જેથી \( \sec^2 x - \sec x \tan x \) મળશે. આ પદોનું સંકલન કરીને સીમાઓ મૂકીને અંતિમ જવાબ મેળવો.

Exam Tip: જ્યારે સંકલનમાં \( x \cdot f(\sin x) \) અથવા \( x \cdot f(\cos x) \) જેવા પદો હોય અને સીમાઓ \( 0 \) થી \( \pi \) હોય, ત્યારે \( \int_0^a f(x) \, dx = \int_0^a f(a - x) \, dx \) ગુણધર્મનો ઉપયોગ કરવો ખૂબ ફાયદાકારક છે. આનાથી ઘણીવાર \( x \) પદ રદ થઈ જાય છે અને સંકલન સરળ બની જાય છે.

Question 13. \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^7x dx \)
Answer: Let's consider the function \( f(x) = \sin^7 x \). If we find \( f(-x) \), it is \( \sin^7 (-x) \). This equals \( (-\sin x)^7 \), which simplifies to \( -\sin^7 x \). So, \( f(-x) = -f(x) \). This means \( f(x) \) is an odd function. For any odd function integrated over a symmetric interval from \( -a \) to \( a \), the value of the integral is always zero.
\( \implies \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^7x dx = 0 \)
In simple words: When you integrate an odd function over an interval that is the same on both positive and negative sides (like from \( -\pi/2 \) to \( \pi/2 \)), the answer is always zero.

Exam Tip: Always check the symmetry of the integrand for limits of the form \( \int_{-a}^a f(x) dx \). If f(x) is odd, the integral is 0. If f(x) is even, the integral is \( 2\int_0^a f(x) dx \).

Question 14. \( \int_0^{2 \pi}\cos^5 x dx \)
Answer: Let's define the function \( f(x) = \cos^5 x \). First, we evaluate \( f(2\pi - x) \). This yields \( \cos^5 (2\pi - x) \), which is equal to \( \cos^5 x \). So, we can observe that \( f(2\pi - x) = f(x) \). According to a property of definite integrals, if \( f(2a - x) = f(x) \), then \( \int_0^{2a} f(x) dx = 2\int_0^a f(x) dx \). Applying this rule, we get:
\( \int_0^{2 \pi}\cos^5 x dx = 2\int_0^{\pi}\cos^5 x dx \)
Next, let's examine \( f(\pi - x) \). This equals \( \cos^5 (\pi - x) \). Since \( \cos(\pi - x) = -\cos x \), we have \( (-\cos x)^5 \), which simplifies to \( -\cos^5 x \). Thus, \( f(\pi - x) = -f(x) \). When this condition is met for an integral from \( 0 \) to \( a \), specifically \( \int_0^a f(x) dx = 0 \) if \( f(a-x) = -f(x) \), we find that:
\( \int_0^{\pi}\cos^5 x dx = 0 \)
Substituting this back into our earlier expression, we conclude that:
\( \int_0^{2 \pi}\cos^5 x dx = 2 \times 0 = 0 \)
In simple words: We check the function's behaviour over the given range using integral properties. First, we find that the integral from \( 0 \) to \( 2\pi \) is twice the integral from \( 0 \) to \( \pi \). Then, we see that the function is 'odd' over the \( 0 \) to \( \pi \) range, making that integral zero. So, the total integral also becomes zero.

Exam Tip: For definite integrals, always check for symmetry properties like \( f(2a-x)=f(x) \) or \( f(a-x)=-f(x) \). These properties can greatly simplify the integral's evaluation, often reducing it to zero or half its range.

Question 15. \( \int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x}dx \)
Answer: Let the given integral be \( I \). We have \( I = \int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x}dx \). We can use the property of definite integrals that says \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). In this case, \( a = \frac{\pi}{2} \). If we change \( x \) to \( \frac{\pi}{2} - x \) inside the integral, we find:
\( I = \int_0^{\frac{\pi}{2}} \frac{\sin(\frac{\pi}{2}-x)-\cos(\frac{\pi}{2}-x)}{1+\sin(\frac{\pi}{2}-x)\cos(\frac{\pi}{2}-x)}dx \)
We are aware that \( \sin(\frac{\pi}{2}-x) = \cos x \) and \( \cos(\frac{\pi}{2}-x) = \sin x \). So, the integral transforms into:
\( I = \int_0^{\frac{\pi}{2}} \frac{\cos x-\sin x}{1+\cos x \sin x}dx \)
We are able to rewrite the top part as \( -(\sin x - \cos x) \):
\( I = \int_0^{\frac{\pi}{2}} \frac{-(\sin x-\cos x)}{1+\sin x \cos x}dx \)
This shows \( I = -\int_0^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x}dx \).
Therefore, \( I = -I \). If we add \( I \) to both sides, we obtain \( 2I = 0 \).
This indicates that \( I = 0 \).
Hence, the integral's value is 0.
In simple words: By using a special property for integrals, we swapped the sine and cosine terms. This made the new integral the negative of the original one. When an integral is equal to its own negative, its value must be zero.

Exam Tip: For definite integrals with limits \( 0 \) to \( \frac{\pi}{2} \), always try the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). This often simplifies the integrand or leads to a solvable system of equations, especially when trigonometric functions are involved.

Question 16. \( \int_0^\pi \log(1 + \cos x)dx \)
Answer: Let the given integral be \( I \). We have:
\( I = \int_0^\pi \log(1 + \cos x) dx \) (i)
We apply the property of definite integrals, \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). Replacing \( x \) with \( \pi - x \):
\( I = \int_0^\pi \log(1 + \cos(\pi - x)) dx \)
Since \( \cos(\pi - x) = -\cos x \), this becomes:
\( I = \int_0^\pi \log(1 - \cos x) dx \) (ii)
Now, we add Equation (i) and Equation (ii):
\( 2I = \int_0^\pi [\log(1 + \cos x) + \log(1 - \cos x)] dx \)
Using the logarithm property \( \log A + \log B = \log(AB) \):
\( 2I = \int_0^\pi \log[(1 + \cos x)(1 - \cos x)] dx \)
We know that \( (1+\cos x)(1-\cos x) = 1-\cos^2 x = \sin^2 x \):
\( 2I = \int_0^\pi \log(\sin^2 x) dx \)
Using another logarithm property \( \log A^n = n \log A \):
\( 2I = \int_0^\pi 2 \log(\sin x) dx \)
Dividing by 2, we obtain:
\( I = \int_0^\pi \log(\sin x) dx \)
Next, we use the property \( \int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx \) if \( f(2a-x) = f(x) \). Here, \( a = \pi \), so \( 2a = 2\pi \). However, our limit is \( \pi \). For \( \int_0^\pi \log(\sin x) dx \), let \( f(x) = \log(\sin x) \). Then \( f(\pi - x) = \log(\sin(\pi - x)) = \log(\sin x) = f(x) \). This means we can write:
\( I = 2 \int_0^{\frac{\pi}{2}} \log(\sin x) dx \)
Let \( I_1 = \int_0^{\frac{\pi}{2}} \log(\sin x) dx \) (iii)
Applying the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \) to \( I_1 \):
\( I_1 = \int_0^{\frac{\pi}{2}} \log(\sin(\frac{\pi}{2} - x)) dx \)
\( I_1 = \int_0^{\frac{\pi}{2}} \log(\cos x) dx \) (iv)
Adding Equation (iii) and Equation (iv):
\( 2I_1 = \int_0^{\frac{\pi}{2}} [\log(\sin x) + \log(\cos x)] dx \)
\( 2I_1 = \int_0^{\frac{\pi}{2}} \log(\sin x \cos x) dx \)
Multiply and divide the term inside the logarithm by 2:
\( 2I_1 = \int_0^{\frac{\pi}{2}} \log\left(\frac{2 \sin x \cos x}{2}\right) dx \)
Using the identity \( 2 \sin x \cos x = \sin 2x \):
\( 2I_1 = \int_0^{\frac{\pi}{2}} \log\left(\frac{\sin 2x}{2}\right) dx \)
Using logarithm property \( \log(\frac{A}{B}) = \log A - \log B \):
\( 2I_1 = \int_0^{\frac{\pi}{2}} (\log(\sin 2x) - \log 2) dx \)
Separating the integrals:
\( 2I_1 = \int_0^{\frac{\pi}{2}} \log(\sin 2x) dx - \log 2 \int_0^{\frac{\pi}{2}} 1 dx \)
\( 2I_1 = \int_0^{\frac{\pi}{2}} \log(\sin 2x) dx - \log 2 [x]_0^{\frac{\pi}{2}} \)
\( 2I_1 = \int_0^{\frac{\pi}{2}} \log(\sin 2x) dx - \frac{\pi}{2} \log 2 \)
Let \( I_2 = \int_0^{\frac{\pi}{2}} \log(\sin 2x) dx \). We use substitution: Let \( t = 2x \). Then \( dt = 2 dx \implies dx = \frac{1}{2} dt \). When \( x=0, t=0 \). When \( x=\frac{\pi}{2}, t=\pi \).
\( I_2 = \int_0^{\pi} \log(\sin t) \frac{1}{2} dt \)
\( I_2 = \frac{1}{2} \int_0^{\pi} \log(\sin t) dt \)
From earlier steps, we established that \( I = \int_0^\pi \log(\sin x) dx \), and \( I = 2I_1 \). So, \( \int_0^\pi \log(\sin t) dt = 2I_1 \).
Therefore, \( I_2 = \frac{1}{2} (2I_1) = I_1 \).
Substituting \( I_2 = I_1 \) back into the equation for \( 2I_1 \):
\( 2I_1 = I_1 - \frac{\pi}{2} \log 2 \)
Subtracting \( I_1 \) from both sides:
\( I_1 = -\frac{\pi}{2} \log 2 \)
Finally, recall that \( I = 2I_1 \). Substituting the value of \( I_1 \):
\( I = 2 \times (-\frac{\pi}{2} \log 2) \)
\( I = -\pi \log 2 \)
In simple words: We used several integral properties and logarithm rules. First, we transformed the integral by adding it to itself after a change of variable, simplifying it to an integral of \( \log(\sin x) \). Then, we split this new integral and used substitution again to simplify it further. By solving for the components and putting them back together, we found the final answer.

Exam Tip: This type of integral (King's property applications) is very common. Memorize the standard result \( \int_0^{\frac{\pi}{2}} \log(\sin x) dx = -\frac{\pi}{2} \log 2 \). This saves significant time in multi-step problems where it appears as a sub-problem.

Question 17. \( \int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx \)
Answer: Let's call the given integral \( I \). We have:
\( I = \int_0^a \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}dx \) (i)
We will use a standard property of definite integrals, which states that \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). Applying this rule to our integral, we replace \( x \) with \( a-x \).
So, \( I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{a-(a-x)}}dx \).
Simplifying the term \( a-(a-x) \) to \( x \), we get:
\( I = \int_0^a \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}dx \) (ii)
Now, we add Equation (i) and Equation (ii) together:
\( 2I = \int_0^a \left(\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} + \frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\right)dx \)
Since the denominators are the same, we can combine the fractions:
\( 2I = \int_0^a \frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}dx \)
The numerator and denominator are identical, so the fraction simplifies to 1:
\( 2I = \int_0^a 1 dx \)
Integrating 1 with respect to \( x \) gives \( x \):
\( 2I = [x]_0^a \)
Evaluating from 0 to \( a \):
\( 2I = a - 0 \)
So, \( 2I = a \).
Finally, dividing by 2, we obtain \( I = \frac{a}{2} \).
In simple words: We used a key property of integrals that helps when the limits are from 0 to 'a'. By applying this, the integral changed its form but remained the same value. Adding the original and changed forms made the top and bottom of the fraction identical, which simplified the whole thing to 1. Integrating 1 gave us the simple answer of \( a/2 \).

Exam Tip: Integrals of this type, where the integrand is of the form \( \frac{f(x)}{f(x)+f(a-x)} \) and the limits are from \( 0 \) to \( a \), often simplify to \( \frac{a}{2} \). Recognizing this pattern can save time during exams.

Question 18. \( \int_0^4|x-1|dx \)
Answer: To calculate the integral \( \int_0^4|x-1|dx \), we must look at how the absolute value function is defined. The expression \( |x-1| \) acts differently based on if \( x-1 \) is positive or negative. If \( x-1 \) is greater than or equal to 0, meaning \( x \ge 1 \), then \( |x-1| \) is simply \( x-1 \). If \( x-1 \) is less than 0, meaning \( x < 1 \), then \( |x-1| \) becomes \( -(x-1) \), which is \( 1-x \).
The range of integration is from 0 to 4. We see that the point where the function's definition changes is \( x=1 \). So, we separate the integral into two parts: from 0 to 1 and from 1 to 4.
\( \int_0^4|x-1|dx = \int_0^1 (1-x)dx + \int_1^4 (x-1)dx \)
Now, we perform the integration for each part:
For the first integral: \( \int_0^1 (1-x)dx = [x - \frac{x^2}{2}]_0^1 = (1 - \frac{1^2}{2}) - (0 - \frac{0^2}{2}) = (1 - \frac{1}{2}) - 0 = \frac{1}{2} \).
For the second integral: \( \int_1^4 (x-1)dx = [\frac{x^2}{2} - x]_1^4 = (\frac{4^2}{2} - 4) - (\frac{1^2}{2} - 1) = (\frac{16}{2} - 4) - (\frac{1}{2} - 1) = (8 - 4) - (-\frac{1}{2}) = 4 - (-\frac{1}{2}) = 4 + \frac{1}{2} = \frac{9}{2} \).
Finally, we combine the outcomes from both parts:
Total integral \( = \frac{1}{2} + \frac{9}{2} = \frac{10}{2} = 5 \).
Therefore, the integral's value \( \int_0^4|x-1|dx \) is 5.
In simple words: Since the absolute value changes its rule at \( x=1 \), we split the integral into two sections. We solved each section separately and then added the results to get the total value.

Exam Tip: When integrating functions with absolute values, always identify the points where the expression inside the absolute value changes sign. Split the integral at these points and define the absolute value expression appropriately in each sub-interval before integrating.

Question 19. If \( f \) and \( g \) are functions such that \( f(x) = f(a-x) \) and \( g(x) + g(a-x) = 4 \), then prove that \( \int_0^a f(x) g(x) dx = 2 \int_0^a f(x) dx \)
Answer: Let's assume the given integral is \( I \). So, we have \( I = \int_0^a f(x)g(x) dx \) (Equation i). We will use a key property of definite integrals, which states that \( \int_0^a h(x) dx = \int_0^a h(a-x) dx \). Applying this rule to our integral \( I \):
\( I = \int_0^a f(a-x)g(a-x) dx \)
From the problem statement, we are given two conditions: \( f(x) = f(a-x) \) and \( g(x) + g(a-x) = 4 \). From the second condition, we can determine that \( g(a-x) = 4 - g(x) \).
Now, we substitute these expressions back into our integral for \( I \):
\( I = \int_0^a f(x)(4 - g(x)) dx \)
We can expand the integrand:
\( I = \int_0^a [4f(x) - f(x)g(x)] dx \)
Next, we can separate the integral:
\( I = 4\int_0^a f(x) dx - \int_0^a f(x)g(x) dx \)
Notice that the second term on the right side is identical to our original integral \( I \):
\( I = 4\int_0^a f(x) dx - I \)
To solve for \( I \), we add \( I \) to both sides of the equation:
\( 2I = 4\int_0^a f(x) dx \)
Finally, we divide both sides by 2:
\( I = 2\int_0^a f(x) dx \)
This result matches the expression we were asked to prove, thus completing the proof.
In simple words: We started with the integral and applied a standard integral property. Using the given conditions about functions \( f \) and \( g \), we transformed the integral. This led to an equation where the original integral appeared on both sides, allowing us to solve for it and prove the required identity.

Exam Tip: Problems involving properties of definite integrals often require a substitution \( x \rightarrow a-x \) and then adding or subtracting the original and modified integrals. Keep a clear track of given conditions like \( f(x) = f(a-x) \) and \( g(x) + g(a-x) = C \) for effective substitution.

Question 20. \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3 + x \cos x + \tan^5 x + 1) dx \) नु मूल्य
Answer: Let's define the given integral as \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3 + x \cos x + \tan^5 x + 1) dx \). We can separate the integrand into parts to evaluate it more easily. Let's consider the function \( f(x) = x^3 + x \cos x + \tan^5 x \). To determine if this function is odd or even, we find \( f(-x) \):
\( f(-x) = (-x)^3 + (-x)\cos(-x) + \tan^5(-x) \)
Since \( (-x)^3 = -x^3 \), \( \cos(-x) = \cos x \), and \( \tan(-x) = -\tan x \), this simplifies to:
\( = -x^3 - x\cos x - \tan^5 x \)
We can factor out a negative sign:
\( = -(x^3 + x \cos x + \tan^5 x) \)
Thus, we see that \( f(-x) = -f(x) \). This tells us that \( f(x) \) is an odd function. A known property of definite integrals states that for any odd function \( f(x) \), the integral over a symmetric interval \( [-a, a] \) is always 0. So, \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^3 + x \cos x + \tan^5 x) dx = 0 \).
Now, we can split the original integral \( I \) into two parts:
\( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (x^3 + x \cos x + \tan^5 x) dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 dx \)
Substituting the value for the first part:
\( I = 0 + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 1 dx \)
The integral of 1 with respect to \( x \) is \( x \). Evaluating this from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \):
\( = [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \)
This gives us \( \frac{\pi}{2} - (-\frac{\pi}{2}) = \frac{\pi}{2} + \frac{\pi}{2} = \pi \).
Therefore, the value of the integral \( I \) is \( \pi \).
In simple words: We broke the integral into two parts: one with odd functions and one with a constant. Odd functions integrated over equal positive and negative limits always result in zero. The constant part just integrates to the length of the interval. Adding these gives the final value.

Exam Tip: When integrating over symmetric limits \( [-a, a] \), always separate the integrand into its odd and even components. The integral of an odd function is 0, and the integral of an even function is \( 2 \times \int_0^a f(x) dx \). This is a fundamental property for simplifying such integrals.

Question 21. \( \int_0^{\frac{\pi}{2}}\log\left(\frac{4+3 \sin x}{4+3 \cos x}\right)dx \) नु मूल्य
Answer: Let the given integral be \( I \). So, we have:
\( I = \int_0^{\frac{\pi}{2}}\log\left(\frac{4+3 \sin x}{4+3 \cos x}\right)dx \) (i)
We can use a property of definite integrals which states that \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). Applying this with \( a = \frac{\pi}{2} \):
\( I = \int_0^{\frac{\pi}{2}}\log\left(\frac{4+3 \sin(\frac{\pi}{2}-x)}{4+3 \cos(\frac{\pi}{2}-x)}\right)dx \)
Knowing that \( \sin(\frac{\pi}{2}-x) = \cos x \) and \( \cos(\frac{\pi}{2}-x) = \sin x \), we can change the expression to:
\( I = \int_0^{\frac{\pi}{2}}\log\left(\frac{4+3 \cos x}{4+3 \sin x}\right)dx \) (ii)
Next, we add Equation (i) and Equation (ii) together:
\( 2I = \int_0^{\frac{\pi}{2}}\left[\log\left(\frac{4+3 \sin x}{4+3 \cos x}\right) + \log\left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right]dx \)
Using the logarithm property \( \log A + \log B = \log(AB) \):
\( 2I = \int_0^{\frac{\pi}{2}}\log\left[\left(\frac{4+3 \sin x}{4+3 \cos x}\right) \times \left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right]dx \)
The terms inside the logarithm cancel each other out, leaving 1:
\( 2I = \int_0^{\frac{\pi}{2}}\log(1)dx \)
Since the logarithm of 1 is always 0, the integral simplifies to:
\( 2I = \int_0^{\frac{\pi}{2}} 0 dx \)
This integral evaluates to 0:
\( 2I = 0 \)
Therefore, \( I = 0 \).
In simple words: By using a standard integral property and adding the original integral to its transformed version, the terms inside the logarithm multiplied to 1. Since \( \log(1) \) is zero, the entire integral also became zero.

Exam Tip: Whenever you see an integral with limits \( 0 \) to \( \frac{\pi}{2} \) and a logarithmic function involving \( \sin x \) and \( \cos x \), try the property \( \int_0^a f(x) dx = \int_0^a f(a-x) dx \). This often leads to terms cancelling out, resulting in a zero integral or a simpler form.