12th NCERT INTEGRALS Exercise 7.11 Questions 22

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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 \[I = \int_{ - 5}^{ - 2} { - \left( {x + 2} \right)dx + \int_{ - 2}^5 {\left( {x + 2} \right)dx} } \] \[I = - \int_{ - 5}^{ - 2} {xdx - 2\int_{ - 5}^{ - 2} {dx + \int_{ - 2}^5 {xdx + 2\int_{ - 2}^5 {dx} } } } \] \[I = - \left[ {\frac{{{x^2}}}{2}} \right]_{ - 5}^{ - 2} - 2\left[ x \right]_{ - 5}^{ - 2} + \left[ {\frac{{{x^2}}}{2}} \right]_{ - 2}^5 + 2\left[ x \right]_{ - 2}^{ - 5}\] \[I = - \frac{1}{2}\left( {4 - 25} \right) - 2\left( 3 \right) + \frac{1}{2}\left( {25 - 4} \right) + 2\left( 7 \right)\] \[I = \frac{{21}}{2} - 6 + \frac{{21}}{2} + 14\] \[I = 21 + 8\] \[I = 29\]

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 \[I = \cancel{2} \cdot \frac{1}{\cancel{2}}\int_0^{\frac{\pi }{2}} {\log \sin tdt - \frac{\pi }{2}\log 2} \] \[I = \int_0^{\frac{\pi }{2}} {\log \sin xdx - \frac{\pi }{2}\log 2} \] \[I = \frac{I}{2} - \frac{\pi }{2}\log 2\] \[I - \frac{I}{2} = - \frac{\pi }{2}\log 2\] \[\frac{I}{2} = - \frac{\pi }{2}\log 2\] \[I = - \pi \log 2\]

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