12th NCERT INTEGRALS Exercise 7.11 Questions 22
: Exercise :-
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### By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19

Question (1)

$\int_0^{\frac{\pi }{2}} {{{\cos }^2}x\;dx}$

Solution

$I = \int_0^{\frac{\pi }{2}} {{{\cos }^2}x\;dx} \quad - - - (1)$ By the property $I = \int_0^{\frac{\pi }{2}} {{{\cos }^2}\left( {\frac{\pi }{2} - x} \right)\;dx} \quad$ $I = \int_0^{\frac{\pi }{2}} {{{\sin }^2}x\;dx} \quad - - - (2)\quad$ Add (1) anad (2) we get $2I = \int_0^{\frac{\pi }{2}} {\left( {{{\sin }^2}x\; + {{\cos }^2}x} \right)dx}$ $I = \frac{1}{2}\int_0^{\frac{\pi }{2}} {dx}$ $I = \frac{1}{2}\left[ x \right]_0^{\frac{\pi }{2}}$ $I = \frac{1}{2}\left( {\frac{\pi }{2} - 0} \right)$ $I = \frac{\pi }{4}$

Question (2)

$\int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}\;} d$

Solution

$I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}\;} dx\quad - - - (1)$ By property $I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}\;} dx$ $I = \int_0^{\frac{\pi }{2}} {\frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}\;} dx\quad - - - (2)$ Add (1) and (2) $2I = \int_0^{\frac{\pi }{2}} {\left( {\frac{{\sqrt {\sin x} + \sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} \right)\;} dx$ $I = \frac{1}{2}\int_0^{\frac{\pi }{2}} {dx}$ $I = \frac{1}{2}\left[ x \right]_0^{\frac{\pi }{2}}$ $I = \frac{1}{2}\left( {\frac{\pi }{2}} \right)$ $I = \frac{\pi }{4}$

Question (3)

$\int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{\frac{3}{2}}}x}}{{{{\sin }^{\frac{3}{2}}}x + {{\cos }^{\frac{3}{2}}}x}}{\mkern 1mu} } dx$

Solution

$I = \int_0^{\frac{\pi}{2}} {\frac{{{{\sin }^{\frac{3}{2}}}x}}{{{{\sin }^{\frac{3}{2}}}x + {{\cos }^{\frac{3}{2}}}x}}\,} dx\quad - - - (1)$ By the property $I = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^{\frac{3}{2}}}\left( {\frac{\pi }{2} - x} \right)}}{{{{\sin }^{\frac{3}{2}}}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^{\frac{3}{2}}}\left( {\frac{\pi }{2} - x} \right)}}\,} dx$ $I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^{\frac{3}{2}}}x}}{{{{\cos }^{\frac{3}{2}}}x + {{\sin }^{\frac{3}{2}}}x}}{\mkern 1mu} } dx\quad - - - (2)$ Add (1) and (2) $2I = \int_0^{\frac{\pi }{2}} {\left( {\frac{{{{\sin }^{\frac{3}{2}}}x + co{x^{\frac{3}{2}}}x}}{{{{\sin }^{\frac{3}{2}}}x + co{x^{\frac{3}{2}}}x}}} \right)} dx$ $2I = \int_0^{\frac{\pi }{2}} {dx}$ $I = \frac{1}{2}\left[ x \right]_0^{\frac{\pi }{2}}$ $I = \frac{1}{2}\left( {\frac{\pi }{2} - 0} \right)$ $I = \frac{\pi }{4}$

Question (4)

$\int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}\;} dx$

Solution

$I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}x}}{{{{\sin }^5}x + {{\cos }^5}x}}\;} dx\quad - - - (1)$ By property $I = \int_0^{\frac{\pi }{2}} {\frac{{{{\cos }^5}\left( {\frac{\pi }{2} - x} \right)}}{{{{\sin }^5}\left( {\frac{\pi }{2} - x} \right) + {{\cos }^5}\left( {\frac{\pi }{2} - x} \right)}}\;} dx\quad$ $I = \int_0^{\frac{\pi }{2}} {\frac{{{{\sin }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}\;} dx\quad - - - (2)$ Add (1) and (2) $2I = \int_0^{\frac{\pi }{2}} {\left( {\frac{{{{\sin }^5}x + {{\cos }^5}x}}{{{{\cos }^5}x + {{\sin }^5}x}}} \right)} dx$ $2I = \int_0^{\frac{\pi }{2}} {dx}$ $2I = \left[ x \right]_0^{\frac{\pi }{2}}$ $2I = \frac{\pi }{2}$ $I = \frac{\pi }{4}$

Question (5)

$\int_{ - 5}^5 {\left| {x + 2} \right|dx}$

Solution

$I = \int_{ - 5}^5 {\left| {x + 2} \right|dx}$ |x+2| = x+2, x > -2
|x+2| = -(x+2), x < -2 $I = \int_{ - 5}^{ - 2} { - \left( {x + 2} \right)dx + \int_{ - 2}^5 {\left( {x + 2} \right)dx} }$ $I = - \int_{ - 5}^{ - 2} {xdx - 2\int_{ - 5}^{ - 2} {dx + \int_{ - 2}^5 {xdx + 2\int_{ - 2}^5 {dx} } } }$ $I = - \left[ {\frac{{{x^2}}}{2}} \right]_{ - 5}^{ - 2} - 2\left[ x \right]_{ - 5}^{ - 2} + \left[ {\frac{{{x^2}}}{2}} \right]_{ - 2}^5 + 2\left[ x \right]_{ - 2}^{ - 5}$
$= - \frac{1}{2}\left[ {{{\left( { - 2} \right)}^2} - {{\left( { - 5} \right)}^2}} \right] - 2\left[ {\left( { - 2} \right) - \left( { - 5} \right)} \right] + \frac{1}{2}\left[ {{{\left( 5 \right)}^2} - {{\left( { - 2} \right)}^2}} \right] + 2\left[ {5 - \left( { - 2} \right)} \right]$
$\begin{array}{l} = - \frac{1}{2}\left[ {{{\left( { - 2} \right)}^2} - {{\left( { - 5} \right)}^2}} \right] - 2\left[ {\left( { - 2} \right) - \left( { - 5} \right)} \right]\\\quad \quad + \frac{1}{2}\left[ {{{\left( 5 \right)}^2} - {{\left( { - 2} \right)}^2}} \right] + 2\left[ {5 - \left( { - 2} \right)} \right]\end{array}$
$I = - \frac{1}{2}\left( {4 - 25} \right) - 2\left( 3 \right) + \frac{1}{2}\left( {25 - 4} \right) + 2\left( 7 \right)$ $I = \frac{{21}}{2} - 6 + \frac{{21}}{2} + 14$ $I = 21 + 8$ $I = 29$

Question (6)

$\int_2^8 {\left| {x - 5} \right|dx}$

Solution

$I = \int_2^8 {\left| {x - 5} \right|dx}$ |x-5| = x - 5, x > 5
|x-5| = -(x-5), x < 5 $I = \int_2^5 { - \left( {x - 5} \right)dx + \int_5^8 {\left( {x - 5} \right)dx} }$ $I = - \int_2^5 {xdx} + 5\int_2^5 {dx} + \int_5^8 {xdx} - 5\int_5^8 {dx}$$I = \frac{{ - 1}}{2}\left[ {{x^2}} \right]_2^5 + 5\left[ x \right]_2^5 + \frac{1}{2}\left[ {{x^2}} \right]_5^8 - 5\left[ x \right]_5^8$ $I = \frac{{ - 1}}{2}\left[ {{x^2}} \right]_2^5 + 5\left[ x \right]_2^5 + \frac{1}{2}\left[ {{x^2}} \right]_5^8 - 5\left[ x \right]_5^8$ $= \frac{{ - 1}}{2}\left( {25 - 4} \right) + 5\left( {5 - 2} \right) + \frac{1}{2}\left( {64 - 25} \right) - 5\left( {8 - 5} \right)$ $I = \frac{{ - 21}}{2} + 15 + \frac{{39}}{2} - 15$ $I = \frac{{18}}{2}$ $I = 9$

Question (7)

$\int_0^1 {x{{\left( {1 - x} \right)}^n}dx}$

Solution

$I = \int_0^1 {x{{\left( {1 - x} \right)}^n}dx}$ By property $I = \int_0^1 {\left( {1 - x} \right)\left[ {1 - {{\left( {1 - x} \right)}}} \right]}^n \;dx$ $I = \int_0^1 {\left( {1 - x} \right){x^n}dx}$ $I = \int_0^1 {{x^n}} dx - \int_0^1 {{x^{n + 1}}} dx$ $I = \frac{1}{{n + 1}}\left[ {{x^{n + 1}}} \right]_0^1 - \frac{1}{{n + 2}}\left[ {{x^{n + 2}}} \right]_0^1$ $I = \frac{1}{{n + 1}}\left[ {1 - 0} \right] - \frac{1}{{n + 2}}\left[ {1 - 0} \right]$ $I = \frac{1}{{n + 1}} - \frac{1}{{n + 2}}$ $I = \frac{{n + 2 - n - 1}}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$ $I = \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}$

Question (8)

$\int_0^{\frac{\pi }{4}} {\log \left( {1 + \tan x} \right)dx}$

Solution

$I = \int_0^{\frac{\pi }{4}} {\log \left( {1 + \tan x} \right)dx} \quad - - - (1)$ By property $I = \int_0^{\frac{\pi }{4}} {\log \left[ {1 + \tan \left( {\frac{\pi }{4} - x} \right)} \right]dx}$ $I = \int_0^{\frac{\pi }{4}} {\log \left[ {1 + \frac{{\tan \frac{\pi }{4} - \tan x}}{{1 + \tan \frac{\pi }{4}\tan x}}} \right]dx}$ $I = \int_0^{\frac{\pi }{4}} {\log \left[ {1 + \frac{{1 - \tan x}}{{1 + \tan x}}} \right]\;} dx$ $I = \int_0^{\frac{\pi }{4}} {\log \left[ {\frac{{1 +\require{cancel} \cancel{\tan x} + 1 - \cancel{\tan x}}}{{1 + \tan x}}} \right]} \;dx$ $I = \int_0^{\frac{\pi }{4}} {\log \left[ {\frac{2}{{1 + \tan x}}} \right]} \;dx\quad - - - (2)$ Add (1) and (2) $2I = \int_0^{\frac{\pi }{4}} {\left[ {\log \left( {1 + \tan x} \right) + \log \left( {\frac{2}{{1 + \tan x}}} \right)} \right]} \;dx$ $2I = \int_0^{\frac{\pi }{4}} {\log \left[ {\left( \cancel{{1 + \tan x}} \right) \times \frac{2}{{\left(\cancel{ {1 + \tan x}} \right)}}} \right]} \;dx$ $2I = \int_0^{\frac{\pi }{4}} {\log 2\;dx}$ $2I = \log 2\int_0^{\frac{\pi }{4}} {dx}$ $2I = \log 2\left[ x \right]_0^{\frac{\pi }{4}}$ $2I = \log 2\left[ {\frac{\pi }{4} - 0} \right]$ $2I = \frac{\pi }{4}\log 2$ $I = \frac{\pi }{8}\log 2$

Question (9)

$\int_0^2 {x\sqrt {2 - x} dx}$

Solution

$I = \int_0^2 {x\sqrt {2 - x} dx}$ By property $I = \int_0^2 {\left( {2 - x} \right)\sqrt {2 - \left( {2 - x} \right)} dx}$ $I = \int_0^2 {\left( {2 - x} \right)\sqrt x } dx$ $I = 2\int_0^2 {{x^{\frac{1}{2}}}} dx - \int_0^2 {{x^{\frac{3}{2}}}} dx$ $I = 2\frac{{\left[ {{x^{\frac{3}{2}}}} \right]_0^2}}{{\frac{3}{2}}} - \frac{{\left[ {{x^{\frac{5}{2}}}} \right]_0^2}}{{\frac{5}{2}}}$ $I = \frac{4}{3}\left[ {{2^{\frac{3}{2}}} - 0} \right] - \frac{2}{5}\left[ {{2^{\frac{5}{2}}} - 0} \right]$ $I = \frac{4}{3}\left( {2\sqrt 2 } \right) - \frac{2}{5}\left( {4\sqrt 2 } \right)$ $I = \frac{{8\sqrt 2 }}{3} - \frac{{8\sqrt 2 }}{5}$ $I = 8\sqrt 2 \left( {\frac{1}{3} - \frac{1}{5}} \right)$ $I = 8\sqrt 2 \left( {\frac{{5 - 3}}{{15}}} \right)$ $I = \frac{{16\sqrt 2 }}{{15}}$

Question (10)

$\int_0^{\frac{\pi }{2}} {\left[ {2\log \left( {\sin x} \right) - \log \left( {\sin 2x} \right)} \right]} \;dx$

Solution

$I = \int_0^{\frac{\pi }{2}} {\left[ {2\log \left( {\sin x} \right) - \log \left( {\sin 2x} \right)} \right]} \;dx$ $I = \int_0^{\frac{\pi }{2}} {\left[ {\log \left( {{{\sin }^2}x} \right) - \log \left( {\sin 2x} \right)} \right]} \;dx$ $I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{{{\sin }^2}x}}{{\sin 2x}}} \right)} \;dx$ $I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{{{\sin }^2}x}}{{2\sin x\cos x}}} \right)} \;dx$ $I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{\tan x}}{2}} \right)dx}$ $I = \int_0^{\frac{\pi }{2}} {\left[ {\log \left( {\tan x} \right) - \log 2} \right]} dx$ $I = \int_0^{\frac{\pi }{2}} {\log \left( {\tan x} \right)dx - \log 2\int_0^{\frac{\pi }{2}} {dx} }$ $I = {I_1} - \log 2\left[ x \right]_0^{\frac{\pi }{2}}$ $I = {I_1} - \frac{\pi }{2}\log 2$ ${I_1} = \int_0^{\frac{\pi }{2}} {\log \left( {\tan x} \right)dx\quad - - - (1)}$ By property ${I_1} = \int_0^{\frac{\pi }{2}} {\log \left[ {\tan \left( {\frac{\pi }{2} - x} \right)} \right]dx\quad }$ ${I_1} = \int_0^{\frac{\pi }{2}} {\log \left( {\cot x} \right)dx} \quad - - - (2)$ Add (1) and (2) $2{I_1} = \int_0^{\frac{\pi }{2}} {\left[ {\log \left( {\tan x} \right) + \log \left( {\cot x} \right)} \right]} \;dx$ $2{I_1} = \int\limits_0^{\frac{\pi }{2}} {\log \left( {\tan x \cdot \cot x} \right)dx}$ $2{I_1} = \int_0^{\frac{\pi }{2}} {\log 1 \cdot dx}$ Note log1=0
2I1 = 0
I1 = 0
$I = {I_1} - \frac{\pi }{2}\log 2$ $I = 0 - \frac{\pi }{2}\log 2$ $I = - \frac{\pi }{2}\log 2$

Question (11)

$\int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}xdx}$

Solution

$I = \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}xdx}$ Let f(x) = sin2x
f(-x) = sin2(-x)
f(-x) = (-sinx)2
f(-x)=sin2 x
f(-x) = f(x)
⇒ f is even function $I = \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}xdx}$ $I = 2\int_0^{\frac{\pi }{2}} {{{\sin }^2}xdx} \quad - - - (1)$ By property $I = 2\int_0^{\frac{\pi }{2}} {{{\sin }^2}\left( {\frac{\pi }{2} - x} \right)dx}$ $I = 2\int_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} \quad - - - (2)$ Add (1) and (2) $2I = 2\left[ {\int_0^{\frac{\pi }{2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)dx} } \right]$ $I = \int_0^{\frac{\pi }{2}} {dx}$ $I = \left[ x \right]_0^{\frac{\pi }{2}}$ $I = \frac{\pi }{2}$

Question (12)

$\int_0^\pi {\frac{x}{{1 + \sin x}}} dx$

Solution

$I = \int_0^\pi {\frac{x}{{1 + \sin x}}} dx\quad - - - (1)$ By property $I = \int_0^\pi {\frac{{\left( {\pi - x} \right)}}{{1 + \sin \left( {\pi - x} \right)}}} dx$ $I = \int_0^\pi {\frac{{\pi - x}}{{1 + \sin x}}dx} \quad - - - (2)$ Add (1) and (2) $2I = \int_0^\pi {\frac{{\cancel{x} + \pi - \cancel{x}}}{{1 + \sin x}}} dx$ $I = \frac{\pi }{2}\int_0^\pi {\frac{1}{{1 + \sin x}}dx}$ $I = \frac{\pi }{2}\int_0^\pi {\frac{{1 - \sin x}}{{\left( {1 + \sin x} \right)\left( {1 - \sin x} \right)}}} dx$ $I = \frac{\pi }{2}\int_0^\pi {\frac{{1 - \sin x}}{{1 - {{\sin }^2}x}}} dx$ $I = \frac{\pi }{2}\int_0^\pi {\frac{{1 - \sin x}}{{{{\cos }^2}x}}} dx$ $I = \frac{\pi }{2}\left[ {\int_0^\pi {\frac{1}{{{{\cos }^2}x}}dx - \int_0^\pi {\frac{{\sin x}}{{{{\cos }^2}x}}dx} } } \right]$ $I = \frac{\pi }{2}\left[ {\int_0^\pi {{{\sec }^2}xdx - \int_0^\pi {\tan x\sec xdx} } } \right]$ $I = \frac{\pi }{2}\left( {\left[ {\tan x} \right]_0^\pi - \left[ {\sec x} \right]_0^\pi } \right)$ $I = \frac{\pi }{2}\left[ {\left( {\tan \pi - \tan 0} \right) - \left( {\sec \pi - \sec \0 } \right)} \right]$ $I = \frac{\pi }{2}\left[ {\left( {0 - 0} \right) - \left( { - 1 - 1} \right)} \right]$ $I = \frac{\pi }{\cancel{2}}\left(\cancel{2} \right)$ $I = \pi$

Question (13)

$\int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^7}x\;dx}$

Solution

$I = \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}x\;dx}$ f(x) = sin7 x
f(-x)=sin7(-x)
f(-x)= (-sinx)7
f(-x)= -sin7x
f(-x)=-f(x)
⇒ f is odd function $I = \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^7}xdx = 0}$

Question (14)

$\int_0^{2\pi } {{{\cos }^5}xdx}$

Solution

$I = \int_0^{2\pi } {{{\cos }^5}xdx}$ $I = \int\limits_0^{2\pi } {{{\cos }^5}xdx}$ $I = \int\limits_0^{2\pi } {{{\cos }^4}x\cos xdx}$ $I = \int\limits_0^{2\pi } {{{\left( {1 - {{\sin }^2}x} \right)}^2}\cos xdx}$ Let sinx = t
∴ cosx dx= dt
when x = 0 ; t = sin0 =0
x = 2π ; t = sin2π = 0 $\therefore I = \int_0^0 {{{\left( {1 - {t^2}} \right)}^2}dt} = 0$

Question (15)

$\int_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx}$$I = \int_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx} \quad - - - (1)$

Solution

$I = \int_0^{\frac{\pi }{2}} {\frac{{\sin x - \cos x}}{{1 + \sin x\cos x}}dx} \quad - - - (1)$ By property $I = \int_0^{\frac{\pi }{2}} {\frac{{\sin \left( {\frac{\pi }{2} - x} \right) - \cos \left( {\frac{\pi }{2} - x} \right)}}{{1 + \sin \left( {\frac{\pi }{2} - x} \right)\cos \left( {\frac{\pi }{2} - x} \right)}}dx}$ $I = \int_0^{\frac{\pi }{2}} {\frac{{\cos x - \sin x}}{{1 + \cos x\sin x}}dx} \quad - - - (2)$ Add (1) and (2) $2I = \int_0^{\frac{\pi }{2}} {\left( {\frac{{\sin x - \cos x + \cos x - \sin x}}{{1 + \sin x\cos x}}} \right)dx}$ $2I = \int_0^{\frac{\pi }{2}} {0dx} = 0$

Question (16)

$\int_0^\pi {\log \left( {1 + \cos x} \right)dx}$

Solution

$I = \int_0^\pi {\log \left( {1 + \cos x} \right)dx} \quad - - - (1)$ By property $I = \int_0^\pi {\log \left( {1 + \cos \left( {\pi - x} \right)} \right)dx}$ Now cos(π - x ) = - cosx $I = \int_0^\pi {\log \left( {1 - \cos x} \right)} dx\quad - - - (2)$ Add (1) and (2) $2I = \int_0^\pi {\left[ {\log \left( {1 + \cos x} \right) + \log \left( {1 - \cos x} \right)} \right]} dx$ $2I = \int_0^\pi {\log \left[ {\left( {1 + \cos x} \right)\left( {1 - \cos x} \right)} \right]} dx$ $2I = \int_0^\pi {\log \left( {{{\sin }^2}x} \right)} dx$ $2I = \int_0^\pi {2\log \left( {\sin x} \right)} dx$ $\cancel{2}I = \cancel{2}\int_0^\pi {\log \left( {\sin x} \right)} dx$ $I = \int_0^\pi {\log \left( {\sin x} \right)} dx$ Now sin is positive in in First and Second quadrant $I = 2\int_0^{\frac{\pi }{2}} {\log \left( {\sin x} \right)} dx\quad - - - (3)$ By property $I = 2\int_0^{\frac{\pi }{2}} {\log } \left( {\sin \left( {\frac{\pi }{2} - x} \right)} \right)dx$ $I = 2\int_0^{\frac{\pi }{2}} {\log \left( {\cos x} \right)} dx\quad - - - (4)$ Add (3) and (4) $2I = 2\left[ {\int_0^{\frac{\pi }{2}} {\left( {\log \sin x + \log \cos x} \right)dx} } \right]$ $I = \int_0^{\frac{\pi }{2}} {\log \left( {\sin x\cos x} \right)dx}$ $I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{2\sin x\cos x}}{2}} \right)dx}$ $I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{\sin 2x}}{2}} \right)dx}$ $I = \int_0^{\frac{\pi }{2}} {\left[ {\log \sin 2x - \log 2} \right]dx}$ $I = \int_0^{\frac{\pi }{2}} {\log \sin 2xdx - \log 2\int_0^{\frac{\pi }{2}} {dx} }$ Let 2x = t
∴ 2dx = dt
dx = dt/2
when x = 0, t = 0
When x = π/2 , t = π
$I = \int_0^\pi {\log \sin t\frac{{dt}}{2} - \frac{\pi }{2}\log 2}$ Now sin is positive in in First and Second quadrant $I = \cancel{2} \cdot \frac{1}{\cancel{2}}\int_0^{\frac{\pi }{2}} {\log \sin tdt - \frac{\pi }{2}\log 2}$
By the property $\int_0^a {f\left( x \right)dx = \int_0^a {f\left( t \right)dt} }$
$I = \int_0^{\frac{\pi }{2}} {\log \sin xdx - \frac{\pi }{2}\log 2}$ $I = \frac{I}{2} - \frac{\pi }{2}\log 2$ $I - \frac{I}{2} = - \frac{\pi }{2}\log 2$ $\frac{I}{2} = - \frac{\pi }{2}\log 2$ $I = - \pi \log 2$

Question (17)

$\int_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a + x} }}dx}$

Solution

$I = \int_0^a {\frac{{\sqrt x }}{{\sqrt x + \sqrt {a + x} }}dx} \quad - - - (1)$ By property $I = \int_0^a {\frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt {a - \left( {a - x} \right)} }}dx}$ $I = \int_0^a {\frac{{\sqrt {a - x} }}{{\sqrt {a - x} + \sqrt x }}} dx\quad - - - (2)$ Add (1) and (2) $2I = \int_0^a {\left( {\frac{{\sqrt {a - x} + \sqrt x }}{{\sqrt x + \sqrt {a - x} }}} \right)} dx$ $2I = \int_0^a {dx}$ $2I = \left[ x \right]_0^a$ $2I = a$ $I = \frac{a}{2}$

Question (18)

$\int_0^4 {\left| {x - 1} \right|dx}$

Solution

$I = \int_0^4 {\left| {x - 1} \right|dx}$ |x-1| = x - 1 ; x > 1
|x-1| = -(x - 1) ; x < 1 $I = \int_0^1 { - \left( {x - 1} \right)dx + \int_1^4 {\left( {x - 1} \right)dx} }$ $I = - \int_0^1 {xdx} + \int_0^1 {dx} + \int_1^4 {xdx} - \int_1^4 {dx}$ $I = - \left[ {\frac{{{x^2}}}{2}} \right]_0^1 + \left[ x \right]_0^1 + \left[ {\frac{{{x^2}}}{2}} \right]_1^4 - \left[ x \right]_1^4$ $I = \frac{{ - 1}}{2}\left( {1 - 0} \right) + \left( {1 - 0} \right) + \frac{1}{2}\left( {16 - 1} \right) - \left( {4 - 1} \right)$ $I = \frac{{ - 1}}{2} + 1 + \frac{{15}}{2} - 3$ $I = 7 + 1 - 3$ $I = 5$

Question (19)

Show that $\int_0^a {f\left( x \right)} g\left( x \right)dx = 2\int_0^a {f\left( x \right)} dx$ if f and g are defined as f(x) = f(a-x) and g(x) + g(a-x)=4

Solution

$I = \int_0^a {f\left( x \right)} g\left( x \right)dx\quad - - - (1)$ By property $I = \int_0^a {f\left( {a - x} \right)g\left( {a - x} \right)dx}$ Note f(a-x) = f(x) $I = \int_0^a {f\left( x \right)g\left( {a - x} \right)dx} \quad - - - (2)$ Add (1) and (2) $2I = \int_0^a {\left[ {f\left( x \right)g\left( x \right) + f\left( x \right)g\left( {a - x} \right)} \right]} dx$ $2I = \int_0^a {f\left( x \right)} \left[ {g\left( x \right) + g\left( {a - x} \right)} \right]dx$ Now g(x) + g(a-x) = 4 $2I = \int_0^a {f\left( x \right) \cdot 4dx}$ $2I = 4\int_0^a {f\left( x \right)dx}$ $I = 2\int_0^a {f\left( x \right)dx}$ $\therefore \quad\int_0^a {f\left( x \right)} g\left( x \right)dx = 2\int_0^a {f\left( x \right)} dx$

#### Choose the correct answer in Exercises 20 and 21

Question (20)

The value of $\int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\left( {{x^3} + x\cos x + {{\tan }^5}x + 1} \right)dx}$ (A) 0   (B) 2   (C) π   (D) 1

Solution

$I = \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\left( {{x^3} + x\cos x + {{\tan }^5}x + 1} \right)dx}$ $I = \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {\left( {{x^3} + x\cos x + {{\tan }^5}x} \right)dx + \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {dx} }$ $I = {I_1} + \left[ x \right]_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}}$ $I = {I_1} + \left[ {\frac{\pi }{2} - \left( { - \frac{\pi }{2}} \right)} \right]$ $I = {I_1} + \pi$ ${I_1} = \int_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{x^3} + x\cos x + {{\tan }^5}x}$ Let f(x) = x3 + xcosx + tan5x
f(-x) = (-x)3 + (-x) cos(-x) + tan5(-x)
f(-x) = -x3 - xcos(-x) - tan5x
f(-x) = -(x3 + xcosx +tan5x
f(-x) = - f(x)
⇒ f is odd function
∴ I1 = 0
∴ I = I1 + π
I= 0 + π = π
So option "C" is correct option

Question (21)

The value of $\int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)dx}$

Solution

$I = \int_0^{\frac{\pi }{2}} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)dx} \quad - - - (1)$ By property $I = \int_0^{\frac{\pi }{2}} {\log \left[ {\frac{{4 + 3\sin \left( {\frac{\pi }{2} - x} \right)}}{{4 + 3\cos \left( {\frac{\pi }{2} - x} \right)}}} \right]} dx$ $I = \int_0^{\frac{\pi }{2}} {\log \left[ {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right]} dx\quad - - - (2)$ Add (1) and (2) $2I = \int_0^{\frac{\pi }{2}} {\left[ {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) + \log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \right]} dx$ $2I = \int_0^{\frac{\pi }{2}} {\log \left[ {\frac{{\left( {4 + 3\sin x} \right)}}{{\left( {4 + 3\cos x} \right)}} \times \frac{{\left( {4 + 3\cos x} \right)}}{{\left( {4 + 3\sin x} \right)}}} \right]} dx$ $2I = \int_0^{\frac{\pi }{2}} {\log 1dx}$ $2I = \int_0^{\frac{\pi }{2}} {0dx}$ $2I = 0$ $I = 0$ ∴ Option "C" is correct