12th NCERT INTEGRALS Miscellaneous Exercise Questions 25 to 44
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Evalulate the definate integrals in Exercise 25 to 33

Question (25)

\[\int\limits_{\frac{\pi }{2}}^\pi {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx} \]

Solution

\[I = \int\limits_{\frac{\pi }{2}}^\pi {{e^x}\left( {\frac{{1 - \sin x}}{{1 - \cos x}}} \right)dx} \] \[I = \int\limits_{\frac{\pi }{2}}^\pi {{e^x}\left( {\frac{{1 - \sin x}}{{2{{\sin }^2}\frac{x}{2}}}} \right)dx} \] \[I = \int\limits_{\frac{\pi }{2}}^\pi {{e^x}\left( {\frac{1}{{2{{\sin }^2}\frac{x}{2}}} - \frac{{\require{cancel}\cancel{2\sin \frac{x}{2}}\cos \frac{x}{2}}}{{\cancel{2}{{\sin }^\cancel{2}}\frac{x}{2}}}} \right)dx} \] \[I = \int\limits_{\frac{\pi }{2}}^\pi {{e^x}\left( {\frac{{\cos e{c^2}\frac{x}{2}}}{2} - \cot \frac{x}{2}} \right)dt} \] \[Let \quad f\left( x \right) = - \cot \frac{x}{2}\] \[f'\left( x \right) = - \left( { - \cos e{c^2}\frac{x}{2}} \right) \cdot \frac{1}{2}\] \[f'\left( x \right) = \frac{1}{2}\cos e{c^2}\left( {\frac{x}{2}} \right)\] \[I = \int\limits_{\frac{\pi }{2}}^\pi {{e^x}\left[ {f\left( x \right) + f'\left( x \right)} \right]} dx\] \[I = \left[ {{e^x}f\left( x \right)} \right]_{\frac{\pi }{2}}^\pi \] \[I = \left[ { - {e^x}\cot \frac{x}{2}} \right]_{\frac{\pi }{2}}^\pi \] \[I = \left( { - {e^\pi }\cot \frac{\pi }{2}} \right) - \left( { - {e^{\frac{\pi }{2}}}\cot \frac{\pi }{4}} \right)\] \[I = \left( { - {e^\pi } \cdot 0} \right) - \left( { - {e^{\frac{\pi }{2}}} \cdot 1} \right)\] \[I = {e^{\frac{\pi }{2}}}\]

Question (26)

\[\int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx} \]

Solution

\[I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx} \] Divide by cos4x to numerator and denominator \[I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\tan x{{\sec }^2}x}}{{1 + {{\tan }^4}x}}dx} \] Let tan2x = t
2tanxsec2dx = dt
tanx sec2x dx = dt/2
When x = 0, t =tan20 = 0
x = π/4, t = tan2π/4 = 1 \[I = \int\limits_0^1 {\frac{{\frac{{dt}}{2}}}{{1 + {t^2}}}} \] \[I = \frac{1}{2}\left[ {{{\tan }^{ - 1}}t} \right]_0^1\] \[I = \frac{1}{2}\left[ {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right]\] \[I = \frac{1}{2}\left[ {\frac{\pi }{4} - 0} \right]\] \[I = \frac{\pi }{8}\]

Question (27)

\[\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^2}x}}{{{{\cos }^2}x + 4{{\sin }^2}x}}dx} \]

Solution

\[I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^2}x}}{{{{\cos }^2}x + 4{{\sin }^2}x}}dx} \] \[I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^2}x}}{{{{\cos }^2}x + 4\left( {1 - {{\cos }^2}x} \right)}}dx} \] \[I = \int\limits_0^{\frac{\pi }{2}} {\frac{{{{\cos }^2}x}}{{4 - 3{{\cos }^2}x}}dx} \]
$$ \require{enclose} \begin{array}{r1} - \frac{1}{3} \quad\; \\ -3cos^2x + 4 \enclose{longdiv}{\; {cos^2x}+{0} +\;0}\ \\ \underline {-cos^2x \mp \ \frac{4}{3}}\ \\ {0 + \frac{4}{3}\phantom{00} }\ \\ \end{array} $$
\[I = \int\limits_0^{\frac{\pi }{2}} {\left[ {\frac{{ - 1}}{3} + \frac{{\frac{4}{3}}}{{4 - 3{{\cos }^2}x}}} \right]} dx\] \[I = \frac{{ - 1}}{3}\int\limits_0^{\frac{\pi }{2}} {dx} + \frac{4}{3}\int\limits_0^{\frac{\pi }{2}} {\frac{1}{{4 - 3{{\cos }^2}x}}} dx\] divide numerator and denominator by cos2x \[I = \frac{{ - 1}}{3}\left[ x \right]_0^{\frac{\pi }{2}} + \frac{4}{3}\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}x}}{{4{{\sec }^2}x - 3}}} dx\] \[I = \frac{{ - \pi }}{6} + \frac{4}{3}\int\limits_0^{\frac{\pi }{2}} {\frac{{{{\sec }^2}x}}{{1 + 4{{\tan }^2}x}}} dx\] Let tanx = t
sec2x dx = dt
when x = 0, t=tan0 = 0
x = π/2, t = tanπ/2 = ∞ \[I = - \frac{\pi }{6} + \frac{4}{3}\int\limits_0^\infty {\frac{{dt}}{{1 + 4{t^2}}}} \] \[I = - \frac{\pi }{6} + \frac{\cancel {4}}{3} \cdot \frac{1}{\cancel{4}}\int\limits_0^\infty {\frac{{dt}}{{\frac{1}{4} + {t^2}}}} \] \[I = - \frac{\pi }{6} + \frac{1}{3} \cdot \frac{1}{{\frac{1}{2}}}\left[ {{{\tan }^{ - 1}}\left( {\frac{t}{{\frac{1}{2}}}} \right)} \right]_0^\infty \] \[I = \frac{{ - \pi }}{6} + \frac{2}{3}\left[ {{{\tan }^{ - 1}}2t} \right]_0^\infty \] \[I = \frac{{ - \pi }}{6} + \frac{2}{3}\left( {{{\tan }^{ - 1}}\infty - {{\tan }^{ - 1}}0} \right)\] \[I = \frac{{ - \pi }}{6} + \frac{2}{3}\left( {\frac{\pi }{2} - 0} \right)\] \[I = \frac{{ - \pi }}{6} + \frac{\pi }{3}\] \[I = \frac{\pi }{6}\]

Question (28)

\[\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x + \cos x}}{{\sqrt {sin2x} }}dx} \]

Solution

\[I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x + \cos x}}{{\sqrt {sin2x} }}dx} \] \[I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x + \cos x}}{{\sqrt {1 - \left( {1 - \sin 2x} \right)} }}dx} \]
1 -sin2x = (sinx - cosx)2
\[I = \int\limits_{\frac{\pi }{6}}^{\frac{\pi }{3}} {\frac{{\sin x + \cos x}}{{\sqrt {1 - {{\left( {\sin x - \cos x} \right)}^2}} }}dx} \] Let sinx - cosx = t
∴ [cosx - (-sinx)]dx = dt
(sinx + cosx )dx = dt
When \[x = \frac{\pi }{6};\] \[t = \sin \frac{\pi }{6} - \cos \frac{\pi }{6} = \frac{1}{2} - \frac{{\sqrt 3 }}{2} = \frac{{1 - \sqrt 3 }}{2}\] \[x = \frac{\pi }{3};\] \[t = \sin \frac{\pi }{3} - \cos \frac{\pi }{3} = \frac{{\sqrt 3 }}{2} - \frac{1}{2} = \frac{{\sqrt {3 - 1} }}{2}\] \[I = \int\limits_{\frac{{1 - \sqrt 2 }}{2}}^{\frac{{\sqrt 3 - 1}}{2}} {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} \] \[I = \left[ {{{\sin }^{ - 1}}t} \right]_{\frac{{1 - \sqrt 2 }}{2}}^{\frac{{\sqrt 3 - 1}}{2}}\] \[I = {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right) - {\sin ^{ - 1}}\left( {\frac{{1 - \sqrt 3 }}{2}} \right)\] \[I = {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right) - {\sin ^{ - 1}}\left[ { - \left( {\frac{{\sqrt 3 - 1}}{2}} \right)} \right]\] \[I = {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right) + {\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right)\] \[I = 2{\sin ^{ - 1}}\left( {\frac{{\sqrt 3 - 1}}{2}} \right)\]

Question (29)

\[\int\limits_0^1 {\frac{1}{{\sqrt {1 + x} - \sqrt x }}} dx\]

Solution

\[I = \int\limits_0^1 {\frac{1}{{\sqrt {1 + x} - \sqrt x }}} dx\] Rationalised denominator we get \[I = \int\limits_0^1 {\frac{{\sqrt {1 + x} + \sqrt x }}{{\left( {1 + x} \right) - x}}} dx\] \[I = \int\limits_0^1 {{{\left( {1 + x} \right)}^{\frac{1}{2}}}dx + \int\limits_0^1 {{x^{\frac{1}{2}}}dx} } \] \[I = \frac{{\left[ {{{\left( {1 + x} \right)}^{\frac{3}{2}}}} \right]_0^1}}{{\frac{3}{2}}} + \frac{{\left[ {{{\left( x \right)}^{\frac{3}{2}}}} \right]_0^1}}{{\frac{3}{2}}}\] \[I = \frac{2}{3}\left[ {{{\left( 2 \right)}^{\frac{3}{2}}} - {{\left( 1 \right)}^{\frac{3}{2}}}} \right] + \frac{2}{3}\left[ {{{\left( 1 \right)}^{\frac{3}{2}}} - 0} \right]\] \[I = \frac{2}{3}\left( {2\sqrt 2 - 1} \right) + \frac{2}{3}\left( {1 - 0} \right)\] \[I = \frac{{4\sqrt 2 }}{3} - \frac{2}{3} + \frac{2}{3}\] \[I = \frac{{4\sqrt 2 }}{3}\]

Question (30)

\[\int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}dx} \]

Solution

\[I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}dx} \] \[I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x + \cos x}}{{9 + 16\left( {2\sin x\cos x} \right)}}dx} \] \[I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x + \cos x}}{{9 + 32\sin x\cos x}}dx} \] \[I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x + \cos x}}{{25 - 16 + 32\sin x\cos x}}dx} \] \[I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x + \cos x}}{{25 - 16\left( {1 - 2\sin x\cos x} \right)}}dx} \] \[I = \int\limits_0^{\frac{\pi }{4}} {\frac{{\sin x + \cos x}}{{25 - 16{{\left( {\sin x - \cos x} \right)}^2}}}dx} \] Let sinx-cosx = t
[cosx - (-sinx)] dx = dt
(cosx+sinx)dx = dt
When \[{\rm{x = 0 ; t = sin0 - cos0 = 0 - 1 = - 1}}\] \[{\rm{x = }}\frac{\pi }{4}{\rm{ ; t = sin }}\frac{\pi }{4}{\rm{ - cos}}\frac{\pi }{4}{\rm{ = }}\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }} = 0\] \[I = \int\limits_{ - 1}^0 {\frac{{dt}}{{25 - 16{t^2}}}} \] \[I = \frac{1}{{16}}\int\limits_{ - 1}^0 {\frac{{dt}}{{\frac{{25}}{{16}} - {t^2}}}} \] \[I = \frac{1}{{16}}\int\limits_{ - 1}^0 {\frac{{dt}}{{{{\left( {\frac{5}{4}} \right)}^2} - {t^2}}}} \]\[I = \frac{1}{{16}} \times \frac{{ - 1}}{{2\left( {\frac{5}{4}} \right)}}\left[ {\log \left| {\frac{{\frac{5}{4} - t}}{{\frac{5}{4} + t}}} \right|} \right]_{ - 1}^0\] \[I = \frac{1}{{\cancel{16}_4}} \times \frac{{ - 1}}{{2\left( {\frac{5}{\cancel{4}}} \right)}}\left[ {\log \left| {\frac{{\frac{5}{4} - t}}{{\frac{5}{4} + t}}} \right|} \right]_{ - 1}^0\] \[I = - \frac{1}{{40}}\left[ {\log \left| {\frac{{\frac{5}{4} - t}}{{\frac{5}{4} + t}}} \right|} \right]_{ - 1}^0\] \[I = - \frac{1}{{40}}\left[ {\log \left| {\frac{\cancel{5}}{\cancel{5}}} \right| - \log \left| {\frac{{5 + 4}}{{5 - 4}}} \right|} \right]\] \[I = - \frac{1}{{40}}\left[ {\log 1 - \log \left| {\frac{9}{1}} \right|} \right]\] \[I = - \frac{1}{{40}}\left( {0 - \log 9} \right)\] \[I = \frac{1}{{40}}\log 9\] \[I = - \frac{1}{{40}} \times \left( { - 2\log 3} \right)\] \[I = \frac{{\log 3}}{{20}}\]

Question (31)

\[\int\limits_0^{\frac{\pi }{2}} {\sin 2x{{\tan }^{ - 1}}\left( {\sin x} \right)dx} \]

Solution

\[I = \int\limits_0^{\frac{\pi }{2}} {\sin 2x{{\tan }^{ - 1}}\left( {\sin x} \right)dx} \] \[I = \int\limits_0^{\frac{\pi }{2}} {2sinx\cos x{{\tan }^{ - 1}}\left( {\sin x} \right)dx} \] \[I = 2\int\limits_0^{\frac{\pi }{2}} {{{\tan }^{ - 1}}\left( {\sin x} \right) \cdot \sin x \cdot \cos xdx} \] Let sinx = t
cosx dx = dt
When x = 0; t = sin0 = 0
x = π/2 ; t = sin π/2 = 1 \[I = 2\int\limits_0^1 {{{\tan }^{ - 1}}t \cdot t \cdot dt} \]
By part u = tan-1t and v = t
\[I = 2\left[ {{{\tan }^{ - 1}}t\int {t \cdot dt} } \right]_0^1 - 2\int\limits_0^1 {\left( {\frac{d}{{dt}}{{\tan }^{ - 1}}t\int {t \cdot dt} } \right)} dt\] \[I = 2\left[ {{{\tan }^{ - 1}}t \cdot \frac{{{t^2}}}{2}} \right]_0^1 - 2\int\limits_0^1 {\frac{1}{{1 + {t^2}}} \cdot \frac{{{t^2}}}{2}dt} \] \[I = \cancel{2}\left[ {\frac{1}{\cancel{2}}{{\tan }^{ - 1}}1 - 0} \right] - \frac{2}{2}\int\limits_0^1 {\frac{{{t^2}}}{{1 + {t^2}}}dt} \] \[I = \frac{\pi }{4} - \int\limits_0^1 {\frac{{1 + {t^2} - 1}}{{1 + {t^2}}}dt} \] \[I = \frac{\pi }{4} - \int\limits_0^1 {dt} + \int\limits_0^1 {\frac{1}{{1 + {t^2}}}dt} \] \[I = \frac{\pi }{4} - \left[ t \right]_0^1 + \left[ {{{\tan }^{ - 1}}t} \right]_0^1\] \[I = \frac{\pi }{4} - \left( {1 - 0} \right) + \left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right)\] \[I = \frac{\pi }{4} - 1 + \frac{\pi }{4}\] \[I = \frac{\pi }{2} - 1\]

Question (32)

\[\int\limits_0^\pi {\frac{{x\tan x}}{{\sec x + \tan x}}dx} \]

Solution

\[I = \int\limits_0^\pi {\frac{{x\tan x}}{{\sec x + \tan x}}dx} \quad - - - (1)\] By property \[I = \int\limits_0^\pi {\frac{{\left( {\pi - x} \right)\tan \left( {\pi - x} \right)}}{{\sec \left( {\pi - x} \right) + \tan \left( {\pi - x} \right)}}dx} \] \[I = \int\limits_0^\pi {\frac{{ - \left( {\pi - x} \right)\tan x}}{{ - \sec x - \tan x}}dx} \] \[I = \int\limits_0^\pi {\left[ {\frac{{ \cancel{-} \left( {\pi - x} \right)\tan x}}{{\cancel{-} \left( {\sec x + \tan x} \right)}}} \right]dx} \quad - - - (2)\] Add (1) and (2) \[2I = \int\limits_0^\pi {\frac{{\left( {\pi - x} \right)\tan x + x\tan x}}{{\sec x + \tan x}}} dx\] \[2I = \pi \int\limits_0^\pi {\frac{{\tan x}}{{\sec x + \tan x}}dx} \] \[I = \frac{\pi }{2}\int\limits_0^\pi {\left( {\frac{{\tan x}}{{\sec x + \tan x}} \times \frac{{\sec x - \tan x}}{{\sec x - \tan x}}} \right)} dx\] \[I = \frac{\pi }{2}\int\limits_0^\pi {\frac{{\sec x\tan x - {{\tan }^2}x}}{{{{\sec }^2}x - {{\tan }^2}x}}} dx\] \[I = \frac{\pi }{2}\left[ {\int\limits_0^\pi {\sec x\tan xdx - \int\limits_0^\pi {{{\tan }^2}xdx} } } \right]\] \[I = \frac{\pi }{2}\left[ {\left[ {\sec x} \right]_0^\pi - \int\limits_0^\pi {\left( {{{\sec }^2}x - 1} \right)dx} } \right]\] \[I = \frac{\pi }{2}\left[ {\left( {\sec \pi - \sec 0} \right) - \left[ {\tan x} \right]_0^\pi + \left[ x \right]_0^\pi } \right]\] \[I = \frac{\pi }{2}\left[ {\left( { - 1 - 1} \right) - \left( {\tan \pi - \tan 0} \right) + \left( {\pi - 0} \right)} \right]\] \[I = \frac{\pi }{2}\left[ { - 2 - \left( {0 - 0} \right) + \pi } \right]\] \[I = \frac{\pi }{2}\left( {\pi - 2} \right)\]

Question (33)

\[\int\limits_1^4 {\left( {\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right|} \right)} dx\]

Solution

\[I = \int\limits_1^4 {\left( {\left| {x - 1} \right| + \left| {x - 2} \right| + \left| {x - 3} \right|} \right)} dx\] \[I = \int\limits_1^4 {\left| {x - 1} \right|dx} + \int\limits_1^4 {\left| {x - 2} \right|dx + \int\limits_1^4 {\left| {x - 3} \right|dx} } \] |x-1| = x-1; x > 1
|x-1| = -(x-1); x < 1
|x-2| = x - 2; x > 2
|x-2|= -(x-2); x < 2
|x-3| = x - 3; x > 3
|x-3|= -(x-3); x < 3
\[ = \int\limits_1^4 {\left( {x - 1} \right)dx + \int\limits_1^2 { - \left( {x - 2} \right)dx + \int\limits_2^4 {\left( {x - 2} \right)dx + \int\limits_1^3 { - \left( {x - 3dx} \right)} } + \int\limits_3^4 {\left( {x - 3} \right)dx} } } \]
\[\begin{array}{l} = \int\limits_1^4 {\left( {x - 1} \right)dx + \int\limits_1^2 { - \left( {x - 2} \right)dx + \int\limits_2^4 {\left( {x - 2} \right)dx} } } \\\quad + \int\limits_1^3 { - \left( {x - 3}dx \right)} + \int\limits_3^4 {\left( {x - 3} \right)dx} \end{array}\]
\[ = \frac{{\left[ {{x^2}} \right]_1^4}}{2} - \left[ x \right]_1^4 - \frac{{\left[ {{x^2}} \right]_1^2}}{2} + 2\left[ x \right]_1^2 + \frac{{\left[ {{x^2}} \right]_2^4}}{2} - 2\left[ x \right]_1^4 - \frac{{\left[ {{x^2}} \right]_1^3}}{2} + 3\left[ x \right]_1^3 + \frac{{\left[ {{x^2}} \right]_3^4}}{2} - 3\left[ x \right]_3^4\] \[\begin{array}{l} = \frac{1}{2}\left( {16 - 1} \right) - \left( {4 - 1} \right) - \frac{1}{2}\left( {4 - 1} \right) + 2\left( {2 - 1} \right) + \frac{1}{2}\left( {16 - 4} \right)\\ - 2\left( {4 - 2} \right) - \frac{1}{2}\left( {9 - 1} \right) + 3\left( {3 - 1} \right) + \frac{1}{2}\left( {16 - 9} \right) - 3\left( {4 - 3} \right)\end{array}\] \[I = \frac{{15}}{2} - 3 - \frac{3}{2} + 2 + \frac{{12}}{2} - 4 - \frac{8}{2} + 6 + \frac{7}{2} - 3\] \[I = \frac{{23}}{2} - 2\] \[I = \frac{{19}}{2}\] Prove the following (Exercises 34 to 39)

Question (34)

Prove that \[\int\limits_1^3 {\frac{{dx}}{{{x^2}\left( {x + 1} \right)}}} \]

Solution

\[I = \int\limits_1^3 {\frac{{dx}}{{{x^2}\left( {x + 1} \right)}}} \] \[Let \quad \frac{A}{x} + \frac{B}{{x + 1}} + \frac{C}{{{x^2}}} = \frac{1}{{{x^2}\left( {x + 1} \right)}}\] \[Ax\left( {x + 1} \right) + B{x^2} + C\left( {x + 1} \right) = 1\] If x = 0 ⇒ C = 1
If x = -1 ⇒ B = 1
If x = 1 ⇒ 2A+B+2C =1
2A+1+2 = 1
A = -1 \[I = \int\limits_1^3 {\frac{{dx}}{{{x^2}\left( {x + 1} \right)}}} \] \[I = \int\limits_1^3 {\left[ {\frac{{ - 1}}{x} + \frac{1}{{x + 1}} + \frac{1}{{{x^2}}}} \right]dx} \] \[I = - \int\limits_1^3 {\frac{1}{x}dx + \int\limits_1^3 {\frac{1}{{x + 1}}dx + \int\limits_1^3 {\frac{1}{{{x^2}}}dx} } } \] \[I = - \left[ {\log x} \right]_1^3 + \left[ {\log \left| {x + 1} \right|} \right]_1^3 - \left[ {\frac{1}{x}} \right]_1^3\] \[I = - \left[ {\log 3 - \log 1} \right] + \left[ {\log 4 - \log 2} \right] - \left[ {\frac{1}{3} - 1} \right]\] \[I = - \left( {\log 3} \right) + \log \left( {\frac{\cancel{4}^2}{\cancel{2}}} \right) - \left( {\frac{{ - 2}}{3}} \right)\] \[I = \frac{2}{3} + \log 2 - \log 3\] \[I = \frac{2}{3} + \log \left( {\frac{2}{3}} \right)\]

Question (35)

\[\int\limits_0^1 {x{e^x}} dx = 1\]

Solution

\[I = \int\limits_0^1 {x{e^x}} dx\] By parts, u = x and v = ex \[I = \left[ {x\int {{e^x}dx} } \right]_0^1 - \int\limits_0^1 {\left( {\frac{d}{{dx}}x\int {{e^x}dx} } \right)} dx\] \[I = \left[ {x{e^x}} \right]_0^1 - \int\limits_0^1 {{e^x}} dx\] \[I = \left[ {1( e) - 0} \right] - \left[ {{e^x}} \right]_0^1\] \[I = e - e + 1\] \[I = 1\]

Question (36)

\[\int\limits_{ - 1}^{ + 1} {{x^{17}}{{\cos }^4}x \cdot dx = 0} \]

Solution

\[I = \int\limits_{ - 1}^{ + 1} {{x^{17}}{{\cos }^4}x \cdot dx} \] f(x) = x17 cos4
f(-x) = (-x)17 [cos(-x)]4
f(-x) = - x17 cos4x
f(-x) = -f(x) ⇒ f is odd function, I = 0

Question (37)

\[\int\limits_0^{\frac{\pi }{2}} {{{\sin }^3}xdx = \frac{2}{3}} \]

Solution

\[I = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^3}xdx} \] \[I = \int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}x \cdot \sin xdx} \] \[I = \int\limits_0^{\frac{\pi }{2}} {\left( {1 - {{\cos }^2}x} \right) \cdot \sin xdx} \] Let cosx = t; x = 0; t = cos0 = 1
-sinx dx = dt ; x = π/2; t = cosπ/2 = 0
sinx dx = -dt
\[I = \int\limits_1^0 {\left( {1 - {t^2}} \right)\left( { - dt} \right)} \]
\[\int\limits_a^b { - f\left( x \right)} dx = \int\limits_b^a {f\left( x \right)} dx\]
\[I = \int\limits_0^1 {\left( {1 - {t^2}} \right)dt} \] \[I = \int\limits_0^1 {dt - \int\limits_0^1 {{t^2}dt} } \] \[I = \left[ t \right]_0^1 - \frac{{\left[ {{t^3}} \right]_0^1}}{3}\] \[I = \left( {1 - 0} \right) - \frac{1}{3}\left( {1 - 0} \right)\] \[I = 1 - \frac{1}{3}\] \[I = \frac{2}{3}\]

Question (38)

\[\int\limits_0^{\frac{\pi }{4}} {2{{\tan }^3}xdx = 1 - \log 2} \]

Solution

\[I = \int\limits_0^{\frac{\pi }{4}} {2{{\tan }^3}xdx} \] \[I = 2\int\limits_0^{\frac{\pi }{4}} {\tan x \cdot {{\tan }^2}x \cdot dx} \] \[I = 2\int\limits_0^{\frac{\pi }{4}} {\tan x\left( {{{\sec }^2}x - 1} \right)dx} \] \[I = 2\int\limits_0^{\frac{\pi }{4}} {\tan x{{\sec }^2}xdx - 2\int\limits_0^{\frac{\pi }{4}} {\tan xdx} } \] \[I = 2{I_1} - 2\left[ {\log \left| {\sec x} \right|} \right]_0^{\frac{\pi }{4}}\] \[I = 2{I_1} - 2\left[ {\log \left| {\sec \frac{\pi }{4}} \right| - \log \left| {\sec 0} \right|} \right]\] \[I = 2{I_1} - 2\left[ {\log \sqrt 2 - \log 1} \right]\] \[I = 2{I_1} - 2\log \sqrt 2 + 0\] \[I = 2{I_1} - 2 \times \frac{1}{2}\log 2\] \[{I_1} = \int\limits_0^{\frac{\pi }{4}} {\tan x{{\sec }^2}xdx} \] tan x = t; sec2x dx = dt
x = 0, t = tan0 = 0
x = π/4, t = tanπ/4 = 1 \[{I_1} = \int\limits_0^1 {tdt} \] \[{I_1} = \left[ {\frac{{{t^2}}}{2}} \right]_0^1\] \[{I_1} = \frac{{1 - 0}}{2} = \frac{1}{2}\] \[I = 2{I_1} - \log 2\] \[I = 2 \times \frac{1}{2} - \log 2\] \[I = 1 - \log 2\]

Question (39)

\[\int\limits_0^1 {{{\sin }^{ - 1}}xdx = \frac{\pi }{2} - 1} \]

Solution

\[I = \int\limits_0^1 {{{\sin }^{ - 1}}xdx} \] \[I = \int\limits_0^1 {{{\sin }^{ - 1}}x \cdot 1dx} \] Let u = sin-1x and v = 1
\[I = \left[ {{{\sin }^{ - 1}}x\int {dx} } \right]_0^1 - \int\limits_0^1 {\left( {\frac{d}{{dx}}{{\sin }^{ - 1}}x\int {dx} } \right)} dx\] \[I = \left[ {x{{\sin }^{ - 1}}x} \right]_0^1 - \int\limits_0^1 {\frac{1}{{\sqrt {1 - {x^2}} }}x \cdot dx} \] \[I = \left[ {{{\sin }^{ - 1}}1 - 0} \right] - \int\limits_0^1 {\frac{x}{{\sqrt {1 - {x^2}} }} \cdot dx} \] \[I = \frac{\pi }{2} - {I_1}\] \[{I_1} = \int\limits_0^1 {\frac{x}{{\sqrt {1 - {x^2}} }}dx} \] Let 1-x2 = t2
-2xdx = 2tdt
xdx= -tdt
When x = 0; t2 = 1-0, t = 1
When x = 1; t2 = 1 -1 = 0; t = 0
\[{I_1} = \int\limits_1^0 {\frac{{ - \cancel{t}dt}}{\cancel{t}}} \] \[{I_1} = - \left[ t \right]_1^0\] \[I = - \left( {0 - 1} \right) = 1\] \[I = \frac{\pi }{2} - {I_1}\] \[I = \frac{\pi }{2} - 1\]

Question (40)

Evaluate \[\int\limits_0^1 {{e^{2 - 3x}}dx} \] as a limit of sum

Solution

a = 0; b= 1
f(x) = e2-3x
f(a+ih) = f(0+ih)
f(a+ih) = f(ih)
f(a+ih) = e2-3ih
f(a+ih) = e2 e-3ih
Divide [0, 1] in n sub intervals of length 'h' each
\[h = \frac{{1 - 0}}{n} = \frac{1}{n} \Rightarrow nh = 1\] By defination \[I = \mathop {\lim }\limits_{h \to 0} h\sum\limits_{i = 1}^n {f\left( {a + ih} \right)} \] \[I = \mathop {\lim }\limits_{h \to 0} h\sum\limits_{i = 1}^n {{e^2} \cdot {e^{ - 3ih}}} \] \[I = \mathop {\lim }\limits_{h \to 0} {e^2}h\sum\limits_{i = 1}^n {{e^{ - 3ih}}} \] \[I = {e^2}\mathop {\lim }\limits_{h \to 0} h\left[ {{e^{ - 3h}} + {e^{ - 6h}} + {e^{ - 9h}} + ... + {e^{ - 3nh}}} \right]\]
e-3h + e-6h + e-9h + ... + e-3nh
are G.P. a = e-3h-3h \[{S_n} = \frac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} = \frac{{{e^{ - 3h}}\left[ {{{\left( {{e^{ - 3h}}} \right)}^n} - 1} \right]}}{{{e^{ - 3h}} - 1}}\]
\[I = {e^2}\mathop {\lim }\limits_{h \to 0} h\left[ {\frac{{{e^{ - 3h}}\left( {{e^{ - 3nh}} - 1} \right)}}{{{e^{ - 3h}} - 1}}} \right]\] Replace nh = 1 and divide by 'h'
\[I = {e^2}\left[ {\frac{{\mathop {\lim }\limits_{h \to 0} \cancel{h}{e^{ - 3h}}\frac{{\left( {{e^{ - 3}} - 1} \right)}}{\cancel{h}}}}{{\mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - 3h}} - 1}}{h}}}} \right]\] \[I = \frac{{{e^2}\left[ {{e^0}\left( {{e^{ - 3}} - 1} \right)} \right]}}{{\left[ {\mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - 3h}} - 1}}{{ - 3h}}} \right] \times \left( { - 3} \right)}}\]
Use \[\mathop {\lim }\limits_{h \to 0} \frac{{{e^h} - 1}}{h} = 1\]
\[I = \frac{{{e^2}\left( 1 \right)\left( {{e^{ - 3}} - 1} \right)}}{{1 \times \left( { - 3} \right)}}\] \[I = \frac{{{e^2}}}{{ - 3}}\left( {\frac{1}{{{e^3}}} - 1} \right)\] \[I = \frac{{{\cancel{e^2}}\left( {1 - {e^3}} \right)}}{{ - 3{e^\cancel{3}}}}\] \[I = \frac{{1 - {e^3}}}{{ - 3e}}\] \[I = \frac{1}{3}\left( {{e^2} - \frac{1}{e}} \right)\]

Choose the correct answer in Exercises 41 to 44

Question (41)

\[\int {\frac{{dx}}{{{e^x} + {e^{ - x}}}}} \text{ is equal }\] Options (A) tan-1(ex) + c
(B) tan-1(e-x) + c
(C) log (ex - e-x) + c
(D) log (ex + e-x) + c

Solution

\[I = \int {\frac{{dx}}{{{e^x} + \frac{1}{{{e^x}}}}}} {\rm{ }}\] \[I = \int {\frac{{{e^x}dx}}{{{e^{2x}} + 1}}} {\rm{ }}\] Let ex = t
ex dx = dt \[I = \int {\frac{{dt}}{{{t^2} + 1}}} {\rm{ }}\] I = tan-1(t)
I = tan-1(ex) + c
So option (A) is correct option

Question (42)

\[\int {\frac{{\cos 2x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}dx} \] Options \[(A)\quad \frac{{ - 1}}{{\sin x + \cos x}} + c\] \[(B)\quad \log \left| {\sin x + \cos x} \right| + c\] \[(C)\quad \log \left| {\sin x - \cos x} \right| + c\] \[(D)\quad \frac{1}{{{{\left( {\sin x + \cos x} \right)}^2}}}\]

Solution

\[I = \int {\frac{{\cos 2x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}dx} \] \[I = \int {\frac{{{{\cos }^2}x - {{\sin }^2}x}}{{{{\left( {\sin x + \cos x} \right)}^2}}}dx} \] \[I = \int {\frac{{\left( {\cos x - \sin x} \right)\left( {\cancel{\cos x + \sin x}} \right)}}{{{{\left( {\sin x + \cos x} \right)}^\cancel{2}}}}dx} \] \[I = \int {\frac{{\left( {\cos x - \sin x} \right)}}{{\left( {\sin x + \cos x} \right)}}dx} \] sinx + cosx = t
(cos x - sinx) dx = dt \[I = \int {\frac{{dt}}{t}} \] \[I = \log \left| t \right| + c\] \[I = \log \left| {\sin x + \cos x} \right| + c\] Option (B) is correct

Question (43)

If f(a+b-x) = f(x) , then \[\int\limits_a^b {xf\left( x \right)} dx \text{ is equal to}\] Options \[(A)\quad \frac{{a + b}}{2}\int\limits_a^b {f\left( {b - x} \right)dx} \] \[(B)\quad \frac{{a + b}}{2}\int\limits_a^b {f\left( {b + x} \right)dx} \] \[(C)\quad \frac{{b - a}}{2}\int\limits_a^b {f\left( x \right)dx} \] \[(D)\quad \frac{{a + b}}{2}\int\limits_a^b {f\left( x \right)dx} \]

Solution

\[I = \int\limits_a^b {xf\left( x \right)} dx\quad - - - (1)\] By property \[I = \int\limits_a^b {\left( {a + b - x} \right)f\left( {a + b - x} \right)} dx\] \[I = \int\limits_a^b {\left( {a + b - x} \right)f\left( x \right)} dx\quad - - - (2)\] Add (1) and (2) \[2I = \int\limits_a^b {\left[ {\left( {a + b - x} \right)f\left( x \right) + xf\left( x \right)} \right]} dx\] \[2I = \int\limits_a^b {\left( {a + b \cancel{-x} + \cancel{x}} \right)f\left( x \right)} dx\] \[2I = \left( {a + b} \right)\int\limits_a^b {f\left( x \right)} dx\] \[I = \frac{{a + b}}{2}\int\limits_a^b {f\left( x \right)} dx\] So option (D) is correct

Question (44)

The value of \[\int\limits_0^1 {{{\tan }^{ - 1}}\left( {\frac{{2x - 1}}{{1 + x - {x^2}}}} \right)dx} \quad \text{is} \] Options (A) 1 (B) 0 (C) -1 (D) π/4

Solution

\[I = \int\limits_0^1 {{{\tan }^{ - 1}}\left( {\frac{{2x - 1}}{{1 + x - {x^2}}}} \right)dx} \] \[I = \int\limits_0^1 {{{\tan }^{ - 1}}\left[ {\frac{{x - \left( {1 - x} \right)}}{{1 + x\left( {1 - x} \right)}}} \right]} dx\]
use property \[{\tan ^{ - 1}}\left( {\frac{{a - b}}{{1 + ab}}} \right) = {\tan ^{ - 1}}a - {\tan ^{ - 1}}b\]
\[I = \int\limits_0^1 {\left[ {{{\tan }^1}x - {{\tan }^{ - 1}}\left( {1 - x} \right)} \right]} dx\quad - - - (1)\] By property \[I = \int\limits_0^1 {\left[ {{{\tan }^{ - 1}}\left( {1 - x} \right) - {{\tan }^{ - 1}}\left[ {1 - \left( {1 - x} \right)} \right]} \right]} dx\] \[I = \int\limits_0^1 {\left( {{{\tan }^{ - 1}}\left( {1 - x} \right) - {{\tan }^{ - 1}}x} \right)dx\quad - - - (2)} \] Add (1) and (2) \[2I = \int\limits_0^1 {\left[ {{{\tan }^{ - 1}}x - {{\tan }^{ - 1}}\left( {1 - x} \right) + {{\tan }^{ - 1}}\left( {1 - x} \right) - {{\tan }^{ - 1}}x} \right]} dx\] \[2I = \int {0dx} \] 2I = 0 ⇒ I = 0
So Option (B) is correct
Miscellaneous Exercise Question 1 to 24 /a>
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