12th NCERT INTEGRALS Miscellaneous Exercise Questions 1 to 24
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Integrate the functions in Exercise 1 to 24

Question (1)

\[\frac{1}{{x - {x^3}}}\]

Solution

\[I = \int {\frac{1}{{x - {x^2}}}} dx\] \[I = \int {\frac{1}{{x\left( {1 - {x^2}} \right)}}} dx\] \[I = \int {\frac{1}{{x\left( {1 - x} \right)\left( {1 + x} \right)}}} dx\] \[Let\quad \frac{A}{x} + \frac{B}{{1 - x}} + \frac{C}{{1 + x}} = \frac{1}{{x\left( {1 + x} \right)\left( {1 - x} \right)}}\] \[A\left( {1 - x} \right)\left( {1 + x} \right) + Bx\left( {1 + x} \right) + Cx\left( {1 - x} \right) = 1\] If x = 0 ⇒ A = 1
If X = 1 ⇒ 2B = 1 ⇒ B = 1/2
If x = -1 ⇒ C(-1)(2) = 1 ⇒ C = -1/2 \[I = \int {\frac{1}{{x\left( {1 - x} \right)\left( {1 + x} \right)}}} dx\] \[I = \int {\left( {\frac{1}{x} + \frac{{\frac{1}{2}}}{{1 - x}} + \frac{{\frac{{ - 1}}{2}}}{{1 + x}}} \right)} dx\] \[I = \int {\frac{1}{x}dx + \frac{1}{2}\int {\frac{1}{{1 - x}}dx - \frac{1}{2}\int {\frac{1}{{1 + x}}dx} } } \] \[I = \log \left| x \right| + \frac{1}{2}\frac{{\log \left| {1 - x} \right|}}{{ - 1}} - \frac{1}{2}\log \left| {1 + x} \right| + c\] \[I = \log \left| x \right| - \frac{1}{2}\log \left| {1 - x} \right| - \frac{1}{2}\log \left| {1 + x} \right| + c\] \[I = \frac{1}{2}\left[ {2\log \left| x \right| - \log \left| {1 - x} \right| - \log \left| {1 + x} \right|} \right] + c\] \[I = \frac{1}{2}\left[ {\log \left| {\frac{{{x^2}}}{{1 - {x^2}}}} \right|} \right] + c\] \[I = \frac{1}{2}\log \left| {\frac{{{x^2}}}{{1 - {x^2}}}} \right| + c\]

Question (2)

\[\frac{1}{{\sqrt {x + a} + \sqrt {x + b} }}dx\]

Solution

\[I = \int {\frac{1}{{\sqrt {x + a} + \sqrt {x + b} }}dx} \] \[I = \int {\frac{{\sqrt {x + a} - \sqrt {x - a} }}{{\left( {x + a} \right) - \left( {x - b} \right)}}} dx\] \[I = \frac{1}{{a - b}}\left[ {\int {{{\left( {x + a} \right)}^{\frac{1}{2}}}dx - \int {{{\left( {x + b} \right)}^{\frac{1}{2}}}dx} } } \right]\] \[I = \frac{1}{{a - b}}\left[ {\frac{{{{\left( {x + a} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}} - \frac{{{{\left( {x + b} \right)}^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right] + c\] \[I = \frac{2}{{3\left( {a - b} \right)}}\left[ {{{\left( {x + a} \right)}^{\frac{3}{2}}} - {{\left( {x + b} \right)}^{\frac{3}{2}}}} \right] + c\]

Question (3)

\[\int {\frac{1}{{x\sqrt {ax - {x^2}} }}dx} \]

Solution

\[I = \int {\frac{1}{{x\sqrt {ax - {x^2}} }}dx} \] Let x = asin2θ
\[\frac{x}{a} = {\sin ^2}\theta \]\[\sin \theta = \sqrt {\frac{x}{a}} \] \[\sin \theta = \sqrt {\frac{x}{a}} \]
dx = a(2sinθcosθ)dθ
dx = 2asinθcosθdθ \[I = \int {\frac{{2\require{cancel}\cancel{a}\cancel{\sin \theta} \cos \theta }}{{\cancel{a}{{\sin }^\cancel{2}}\theta \sqrt {a\left( {a{{\sin }^2}\theta } \right) - \left( {{a^2}{{\sin }^4}\theta } \right)} }}d\theta } \] \[I = 2\int {\frac{{\cos \theta }}{{\sin \theta }}\frac{{d\theta }}{{a\sqrt {{{\sin }^2}\theta - {{\sin }^4}\theta } }}} \] \[I = \frac{2}{a}\int {\frac{{\cos \theta }}{{\sin \theta }}\frac{{d\theta }}{{\sqrt {{{\sin }^2}\theta \left( {1 - {{\sin }^2}\theta } \right)} }}} \] \[I = \frac{2}{a}\int {\frac{{\cancel{\cos \theta} }}{{\sin \theta \sin \theta }}\frac{{d\theta }}{{\cancel{\cos \theta} }}} \] \[I = \frac{2}{a}\int {\cos e{c^2}\theta d\theta } \] \[I = - \frac{2}{a}\cot \theta + c\] from figure 3 \[I = - \frac{2}{a}\frac{{\sqrt {a - x} }}{{\sqrt x }} + c\] \[I = - \frac{2}{a}\sqrt {\frac{{a - x}}{x}} + c\]

Question (4)

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

Solution

\[I = \int {\frac{1}{{{x^2}\frac{{{{\left( {{x^4} + 1} \right)}^{\frac{3}{4}}}}}{{{{\left( {{x^4}} \right)}^{\frac{3}{4}}}}} \times {x^3}}}dx} \] \[I = \int {\frac{1}{{{x^5}}}\frac{1}{{{{\left( {1 + \frac{1}{{{x^4}}}} \right)}^{\frac{3}{4}}}}}dx} \] \[Let\quad 1 + \frac{1}{{{x^4}}} = {t^4}\] \[ \Rightarrow \frac{{ - \cancel{4}}}{{{x^5}}}dx = \cancel{4}{t^3}dt\] \[\frac{{dx}}{{{x^5}}} = - {t^3}dt\] \[I = \int {\frac{{ - dt \cdot {t^3}}}{{{{\left( {{t^4}} \right)}^{\frac{3}{4}}}}}} \] \[I = - t + c\] \[I = - {\left( {1 + \frac{1}{{{x^4}}}} \right)^{\frac{1}{4}}} + c\]

Question (5)

\[\int {\frac{1}{{{x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}}} dx\]

Solution

\[I = \int {\frac{1}{{{x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}}} dx\] Let x = t6
∴ dx = 6t5 dt \[I = \int {\frac{{6{t^5}}}{{{t^3} + {t^2}}}} dt\] \[I = 6\int {\frac{{{t^5}}}{{{t^2}\left( {t + 1} \right)}}} dt\] \[I = 6\int {\frac{{{t^3}}}{{t + 1}}dt} \] $$ \require{enclose} \begin{array}{r1} t^2-t+1 \quad\; \\ t + 1 \enclose{longdiv}{\; {t^3}+{0} +\;0}\ \\ \underline {-{t^3} \pm \ t^2\mp \ 0}\ \\ {0 - \ t^2\mp \ 0}\ \\ \underline { 0 \mp t^2\mp \ t\phantom{00}} \\ { 0 + 0 \, + t\phantom{00}} \\ \qquad \underline {\qquad -t\pm \ 1}\\ -1 \end{array} $$
\[I = 6\int {\left( {{t^2} - t + 1 - \frac{1}{{t + 1}}} \right)dt} \] \[I = 6\int {{t^2}dt - 6\int {tdt + 6\int {dt - 6\int {\frac{1}{{t + 1}}dt} } } } \] \[I = \cancel{6}^2\frac{{{t^3}}}{\cancel{3}} - \cancel{6}^3\frac{{{t^2}}}{\cancel{2}} + 6t - 6\log \left| {t + 1} \right| + c\] \[I = 2{x^{\frac{1}{2}}} - 3{x^{\frac{1}{3}}} + 6{x^{\frac{1}{6}}} - 6\log \left| {{x^{\frac{1}{6}}} + 1} \right| + c\]

Question (6)

\[\int {\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}}} dx\]

Solution

\[I = \int {\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}}} dx\] \[Let \quad \frac{A}{{x + 1}} + \frac{{Bx + C}}{{{x^2} + 9}} = \frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}}\] \[A\left( {{x^2} + 9} \right) + \left( {Bx + C} \right)\left( {x + 1} \right) = 5x\] If x = -1 ⇒ 10A = - 5 ⇒ A = -1/2
f x = 0 ⇒ 9A + C = 0 ⇒ C = -9A = 9/2
if x = 1 ⇒ 10A+2B+2C = 5
-5 + 2B + 9 = 5
2B = 1 ⇒ B = 1/2 \[I = \int {\frac{{5x}}{{\left( {x + 1} \right)\left( {{x^2} + 9} \right)}}} dx\] \[I = \int {\left( {\frac{{\frac{{ - 1}}{2}}}{{x + 1}} + \frac{{\frac{1}{2}x + \frac{9}{2}}}{{{x^2} + 9}}} \right)} dx\] \[I = - \frac{1}{2}\int {\frac{1}{{x + 1}}dx + \frac{1}{2}\int {\frac{x}{{{x^2} + 9}}dx + \frac{9}{2}\int {\frac{1}{{{x^2} + 9}}dx} } } \] \[I = - \frac{1}{2}\log \left| {x + 1} \right| + \frac{1}{2}\int {\frac{x}{{{x^2} + 9}}dx + \frac{\cancel{9}^3}{2}\frac{1}{\cancel{3}}{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right) + c} \] x2+9 = t
2x dx = dt

xdx = dt/2
\[I = - \frac{1}{2}\log \left| {x + 1} \right| + \frac{1}{2}\int {\frac{{\frac{{dt}}{2}}}{t} + \frac{3}{2}{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right) + c} \] \[I = - \frac{1}{2}\log \left| {x + 1} \right| + \frac{1}{4}\log \left| t \right| + \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{x}{3}} \right) + c\] \[I = - \frac{1}{2}\log \left| {x + 1} \right| + \frac{1}{4}\log \left( {{x^2} + 9} \right) + \frac{3}{2}{\tan ^{ - 1}}\left( {\frac{x}{3}} \right) + c\]

Question (7)

\[\int {\frac{{\sin x}}{{\sin \left( {x - a} \right)}}} dx\]

Solution

\[I = \int {\frac{{\sin x}}{{\sin \left( {x - a} \right)}}} dx\] \[I = \int {\frac{{\sin \left[ {\left( {x - a} \right) + a} \right]}}{{\sin \left( {x - a} \right)}}} dx\] \[I = \int {\frac{{sin\left( {x - a} \right)\cos a + \cos \left( {x - a} \right)\sin a}}{{\sin \left( {x - a} \right)}}} dx\] \[I = \cos a\int {dx} + \sin a\int {\frac{{\cos \left( {x - a} \right)}}{{\sin \left( {x - a} \right)}}dx} \] \[I = \cos a\left[ x \right] + \sin a \cdot \log \left| {\sin \left( {x - a} \right)} \right| + c\]

Question (8)

\[\int {\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}} dx\]

Solution

\[I = \int {\frac{{{e^{5\log x}} - {e^{4\log x}}}}{{{e^{3\log x}} - {e^{2\log x}}}}} dx\]
\[{e^{a{{\log }_e}x}} = {x^{a{{\log }_e}e}} = {x^a}\]
\[I = \int {\frac{{{x^5} - {x^4}}}{{{x^3} - {x^2}}}dx} \] \[I = \int {\frac{{{x^4}\left( {x - 1} \right)}}{{{x^2}\left( {x - 1} \right)}}} dx\] \[I = \int {{x^2}dx} \] \[I = \frac{{{x^3}}}{3} + c\]

Question (9)

\[\int {\frac{{\cos x}}{{\sqrt {4 - {{\sin }^2}x} }}} dx\]

Solution

\[I = \int {\frac{{\cos x}}{{\sqrt {4 - {{\sin }^2}x} }}} dx\] Let sinx = t
cos x dx = dt\[I = \int {\frac{{dt}}{{\sqrt {4 - {t^2}} }}} \] \[I = \int {\frac{{dt}}{{\sqrt {4 - {t^2}} }}} \] \[I = {\sin ^{ - 1}}\left( {\frac{t}{2}} \right) + c\] \[I = {\sin ^{ - 1}}\left( {\frac{{sinx}}{2}} \right) + c\]

Question (10)

\[\int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}{{\cos }^2}x}}} dx\]

Solution

\[I = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}{{\cos }^2}x}}} dx\] \[I = \int {\frac{{{{\left( {{{\sin }^4}x} \right)}^2} - {{\left( {{{\cos }^4}x} \right)}^2}}}{{1 - 2{{\sin }^2}{{\cos }^2}x}}} dx\] \[I = \int {\frac{{\left( {{{\sin }^4}x + {{\cos }^4}x} \right)\left( {{{\sin }^4}x - {{\cos }^4}x} \right)}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}} dx\] \[= \int {\frac{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^2}x - {{\cos }^2}} \right)\left[ {\left( {{{\sin }^2}x + {{\cos }^2}x} \right) - 2{{\sin }^2}x{{\cos }^2}x} \right]}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}dx} \] \[ = \int {\frac{{\left( 1 \right)\left( { - \cos 2x} \right)\left( {1 - 2{{\sin }^2}xco{x^2}x} \right)}}{{\left( {1 - 2{{\sin }^2}x{{\cos }^2}x} \right)}}} dx\]\[I = - \int {\cos 2xdx} \] \[I = - \int {\cos 2xdx} \] \[I = - \frac{{sin2x}}{2} + c\]

Question (11)

\[\int {\frac{{dx}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} \]

Solution

\[I = \int {\frac{{dx}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}} \] \[I = \frac{1}{{\sin \left( {a - b} \right)}}\int {\frac{{\sin \left[ {\left( {x + a} \right) - \left( {x + b} \right)} \right]}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}dx} \] \[ = \frac{1}{{\sin \left( {a - b} \right)}}\int {\frac{{\sin \left( {x + a} \right)\cos \left( {x + b} \right) - \cos \left( {x + a} \right)\sin \left( {x + b} \right)}}{{\cos \left( {x + a} \right)\cos \left( {x + b} \right)}}dx} \] \[ = \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\int {\tan \left( {x + a} \right)dx - \int {\tan \left( {x + b} \right)dx} } } \right]\] \[ = \frac{1}{{\sin \left( {a - b} \right)}}\left[ { - \log \left| {\cos \left( {x + a} \right)} \right| - \left( { - \log \left| {\cos \left( {x + b} \right)} \right|} \right)} \right] + c\] \[ = \frac{1}{{\sin \left( {a - b} \right)}}\left[ {\log \left| {\cos \left( {x + b} \right)} \right| - \log \left| {\cos \left( {x + a} \right)} \right|} \right] + c\] \[I = \frac{1}{{\sin \left( {a - b} \right)}}\log \left| {\frac{{\cos \left( {x + b} \right)}}{{\cos \left( {x + a} \right)}}} \right| + c\]

Question (12)

\[\int {\frac{{{x^3}}}{{\sqrt {1 - {x^8}} }}dx} \]

Solution

\[I = \int {\frac{{{x^3}}}{{\sqrt {1 - {x^8}} }}dx} \] \[I = \int {\frac{{{x^3}}}{{\sqrt {1 - {{\left( {{x^4}} \right)}^2}} }}dx} \] Let x4 = t
4x3dx = dt
x3 dx = dt/4 \[I = \int {\frac{{\frac{{dt}}{4}}}{{\sqrt {1 - {t^2}} }}} dt\] \[I = \frac{1}{4}\int {\frac{1}{{\sqrt {1 - {t^2}} }}dt} \] \[I = \frac{1}{4}{\sin ^{ - 1}}\left( t \right) + c\] \[I = \frac{1}{4}{\sin ^{ - 1}}\left( {{x^4}} \right) + c\]

Question (13)

\[\int {\frac{{{e^x}}}{{\left( {{e^x} + 1} \right)\left( {{e^x} + 1} \right)}}dx} \]

Solution

\[I = \int {\frac{{{e^x}}}{{\left( {{e^x} + 1} \right)\left( {{e^x} + 1} \right)}}dx} \] Let ex = t
∴ ex dx = dt \[I = \int {\frac{{dt}}{{\left( {t + 1} \right)\left( {t + 2} \right)}}} \] \[I = \int {\frac{{\left( {t + 2} \right) - \left( {t + 1} \right)}}{{\left( {t + 1} \right)\left( {t + 2} \right)}}} dt\] \[I = \int {\frac{1}{{t + 1}}dt - \int {\frac{1}{{t + 2}}dt} } \] \[I = \log \left| {t + 1} \right| - \log \left| {t + 2} \right| + c\] \[I = \log \left| {\frac{{t + 1}}{{t + 2}}} \right| + c\] \[I = \log \left| {\frac{{{e^x} + 1}}{{{e^x} + 2}}} \right| + c\]

Question (14)

\[\int {\frac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}} dx\]

Solution

\[I = \int {\frac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}} dx\] \[Let \quad \frac{A}{{{x^2} + 1}} + \frac{B}{{{x^2} + 4}} = \frac{1}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}\] A(x2 + 4 ) + B(x2 + 1) = 1
If x2 = -1 ⇒ 3A = 1 ⇒ A = 1/3
If x2 = -4 ⇒ -3B = 1 ⇒ B = -1/3 \[\therefore I = \int {\left( {\frac{{\frac{1}{3}}}{{{x^2} + 1}} + \frac{{\frac{{ - 1}}{3}}}{{{x^2} + 4}}} \right)} dx\] \[I = \frac{1}{3}\int {\frac{1}{{{x^2} + 1}}dx - \frac{1}{3}\int {\frac{1}{{{x^2} + 4}}dx} } \] \[I = \frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{{3 \times 2}}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + c\] \[I = \frac{1}{3}{\tan ^{ - 1}}x - \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{x}{2}} \right) + c\]

Question (15)

\[\int {{{\cos }^3}x \cdot {e^{\log \sin x}}dx} \]

Solution

\[I = \int {{{\cos }^3}x \cdot {e^{\log \sin x}}dx} \] \[I = \int {{{\cos }^3}x\sin {x^{{{\log }_e}e}}} dx\] \[I = \int {{{\cos }^3}x\sin xdx} \] Let cosx = t
-sinx dx = dt
sinx dx = -dt \[I = \int {{t^3}\left( { - dt} \right)} \] \[I = - \frac{{{{\cos }^4}x}}{4} + c\]

Question (16)

\[\int {{e^{3\log x}}{{\left( {{x^4} + 1} \right)}^{ - 1}}dx} \]

Solution

\[I = \int {{e^{3\log x}}{{\left( {{x^4} + 1} \right)}^{ - 1}}dx} \] \[I = \int {\frac{{{x^{3\log e}}}}{{\left( {{x^4} + 1} \right)}}dx} \] \[I = \int {\frac{{{x^3}}}{{{x^4} + 1}}} dx\] Let x4 + 1 = t
4x3 = dt
x3 = dt/4 \[I = \int {\frac{{\frac{{dt}}{4}}}{t}} \] \[I = \frac{1}{4}\int {\frac{{dt}}{t}} \] \[I = \frac{1}{4}\log \left| t \right| + c\] \[I = \frac{1}{4}\log \left| {{x^4} + 1} \right| + c\]

Question (17)

\[\int {f'\left( {ax + b} \right){{\left[ {f\left( {ax + b} \right)} \right]}^n}dx} \]

Solution

Let f(ax+b) = t
∴ f'(ax+b). a dx = dt
f'(ax+b)dx = dt/a \[I = \int {{t^n}\frac{{dt}}{a}} \] \[I = \frac{1}{a}\int {{t^n}} dt\] \[I = \frac{1}{a}\frac{{{t^{n + 1}}}}{{n + 1}} + c\] \[I = \frac{1}{a}\frac{{{{\left[ {f\left( {ax + b} \right)} \right]}^{n + 1}}}}{{n + 1}} + c\]

Question (18)

\[\int {\frac{1}{{\sqrt {{{\sin }^3}x.\sin \left( {x + \alpha } \right)} }}} dx\]

Solution

\[I = \int {\frac{1}{{\sqrt {{{\sin }^3}x.\sin \left( {x + \alpha } \right)} }}} dx\] \[I = \int {\frac{1}{{\sqrt {{{\sin }^3}x\left[ {\frac{{\sin \left( {x + \alpha } \right)}}{{sinx}}} \right]\sin x} }}} dx\] \[I = \int {\frac{1}{{\sqrt {{{\sin }^4}x\frac{{\sin \left( {x + \alpha } \right)}}{{\sin x}}} }}} dx\] \[I = \int {\frac{1}{{{{\sin }^2}x}} \cdot \frac{1}{{\sqrt {\frac{{\sin \left( {x + \alpha } \right)}}{{\sin x}}} }}} dx\] \[Let \quad \frac{{\sin \left( {x + \alpha } \right)}}{{\sin x}} = {t^2}\] \[\left[ {\frac{{\sin x\cos \left( {x + \alpha } \right) - \sin \left( {x + \alpha } \right)\cos x}}{{{{\sin }^2}x}}} \right]dx = 2tdt\] \[\frac{1}{{{{\sin }^2}x}}\left[ {\sin \left( {x - \left( {x + \alpha } \right)} \right)} \right]dx = 2tdt\] \[\frac{1}{{{{\sin }^2}x}}\sin \left( { - \alpha } \right)dx = 2tdt\] \[\frac{{dt}}{{{{\sin }^2}x}} = \frac{{2tdt}}{{ - \sin \alpha }}\] \[I = \int {\frac{{2tdt}}{{ - \sin \alpha }}} \cdot \frac{1}{{\sqrt {{t^2}} }}\] \[I = \frac{{ - 2}}{{\sin \alpha }}\int {\frac{{\cancel{t} \cdot dt}}{\cancel{t}}} \] \[I = \frac{{ - 2}}{{\sin \alpha }}t + c\] \[I = \frac{{ - 2}}{{\sin \alpha }}\sqrt {\frac{{\sin \left( {x + \alpha } \right)}}{{\sin x}}} + c\]

Question (19)

\[\int {\frac{{{{\sin }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{{{\sin }^{ - 1}}\sqrt x + {{\cos }^{ - 1}}\sqrt x }}} dx\]\[I = \int {\frac{{{{\sin }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{{{\sin }^{ - 1}}\sqrt x + {{\cos }^{ - 1}}\sqrt x }}} dx\]

Solution

\[I = \int {\frac{{{{\sin }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{{{\sin }^{ - 1}}\sqrt x + {{\cos }^{ - 1}}\sqrt x }}} dx\] \[{\sin ^{ - 1}}\sqrt x + {\cos ^{ - 1}}\sqrt x = \frac{\pi }{2}\] \[\therefore \quad {\sin ^{ - 1}}\sqrt x = \frac{\pi }{2} - {\cos ^{ - 1}}\sqrt x \] Replacing values we get \[I = \int {\frac{{\frac{\pi }{2} - {{\cos }^{ - 1}}\sqrt x - {{\cos }^{ - 1}}\sqrt x }}{{\frac{\pi }{2}}}} dx\] \[I = \int {\frac{{\frac{\pi }{2} - 2{{\cos }^{ - 1}}\sqrt x }}{{\frac{\pi }{2}}}} dx\] \[I = \int {dx} - \frac{4}{\pi }\int {{{\cos }^{ - 1}}\sqrt x } dx\] \[I = x - \frac{4}{\pi }{I_1} + c\]\[{I_1} = \int {{{\cos }^{ - 1}}\sqrt x } dx\] \[{I_1} = \int {{{\cos }^{ - 1}}\sqrt x } dx\] Let x = cos2θ
dx = 2cosθ (-sinθ)dθ
dx = -2sinθcosθ dθ \[I_1 = \int {\theta \left( { - \sin 2\theta } \right) \cdot d\theta } \]
θ = u and sin2θ = v
\[I_1 = - \left[ {\theta \int {\sin 2\theta dx - \int {\left( {\frac{d}{{d\theta }}\theta \int {\sin 2\theta d\theta } } \right)d\theta } } } \right]\] \[I_1 = - \theta \frac{{\left( { - \cos 2\theta } \right)}}{2} + \int {1 \cdot \left( {\frac{{ - \cos 2\theta }}{2}} \right)d\theta } \] \[I_1 = \frac{{\theta \cos 2\theta }}{2} - \frac{1}{2}\int {\cos 2\theta d\theta } \] \[I_1 = \frac{{\theta \left( {2{{\cos }^2}\theta - 1} \right)}}{2} - \frac{1}{2} \cdot \frac{{\sin 2\theta }}{2}\] \[I_1 = \frac{{{{\cos }^{ - 1}}\sqrt x \left( {2x - 1} \right)}}{2} - \frac{1}{\cancel{4}} \cdot \cancel{2}\sin \theta \cos \theta \] \[{I_1} = \frac{{\left( {2x - 1} \right){{\cos }^{ - 1}}\sqrt x }}{2} - \frac{{\sqrt {1 - x} \cdot \sqrt x }}{2}\] \[{I_1} = \frac{{\left( {2x - 1} \right){{\cos }^{ - 1}}\sqrt x }}{2} - \frac{{\sqrt {x - {x^2}} }}{2}\] \[I = x - \frac{4}{\pi }\left[ {\frac{{\left( {2x - 1} \right)co{s^{ - 1}}\sqrt x }}{2} - \frac{{\sqrt {x - {x^2}} }}{2}} \right]\] \[I = x - \frac{2}{\pi }\left( {2x - 1} \right){\cos ^{ - 1}}\sqrt x + \frac{2}{\pi }\sqrt {x - {x^2}} + c\]

Question (20)

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

Solution

\[I = \int {\sqrt {\frac{{1 - \sqrt x }}{{1 + \sqrt x }}} } dx\]
x = cos2
∴ dx = 2 cos2θ (-sin2θ).2
dx = -4 sin2θ cos2θ dθ \[I = \int {\sqrt {\frac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }}} } \times \left( { - 4\sin 2\theta \cos 2\theta d\theta } \right)\] \[I = \int {\sqrt {\frac{{\cancel{2}{{\sin }^2}\theta }}{{\cancel{2}{{\cos }^2}\theta }}} } \times \left( { - 4 \cdot 2\sin \theta \cos \theta \cdot \cos 2\theta } \right)d\theta \] \[I = - 8\int {\frac{{\sin \theta }}{{\cancel{\cos \theta }}}} \cdot \sin \theta \cancel{\cos \theta} \cos 2\theta d\theta \] \[I = - 8\int {{{\sin }^2}\theta \cos 2\theta d\theta } \] \[I = - 8\int {\left( {\frac{{1 - \cos 2\theta }}{2}} \right)} \cos 2\theta d\theta \] \[I = - 4\int {\left( {1 - \cos 2\theta } \right)} \cos 2\theta d\theta \] \[I = - 4\int {\left( {\cos 2\theta - {{\cos }^2}2\theta } \right)} d\theta \] \[I = - 4\int {\cos 2\theta d\theta + 4\int {{{\cos }^2}2\theta d\theta } } \] \[I = \frac{{ - 4\sin 2\theta }}{2} + 4\int {\frac{{1 + \cos 4\theta }}{2}d\theta } \] \[I = - 2sin2\theta + 2\int {d\theta + 2\int {\cos 4\theta d\theta } } \] \[I = - 2\sin 2\theta + 2\theta + 2\frac{{\sin 4\theta }}{4} + c\] \[I = - 2\sqrt {{{\sin }^2}2\theta } + {\cos ^{ - 1}}\sqrt x + \frac{{\cancel{2} \cdot \cancel{2}\sin 2\theta \cos 2\theta }}{\cancel{4}} + c\] \[I = - 2\sqrt {1 - x} + {\cos ^{ - 1}}\sqrt x + \sqrt {1 - x} \sqrt x + c\] \[I = - 2\sqrt {1 - x} + {\cos ^{ - 1}}\sqrt x + \sqrt {x\left( {1 - x} \right)} + c\] \[I = - 2\sqrt {1 - x} + {\cos ^{ - 1}}\sqrt x + \sqrt {x - {x^2}} + c\]

Question (21)

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

Solution

\[I = \int {\left( {\frac{{2 + \sin 2x}}{{1 + \cos 2x}}} \right)} {e^x}dx\] \[I = \int {{e^x}} \left[ {\frac{{2 + \sin 2x}}{{2{{\cos }^2}x}}} \right]dx\] \[I = \int {{e^x}} \left[ {\frac{\cancel{2}}{{\cancel{2}{{\cos }^2}x}} + \frac{{\sin 2x}}{{2{{\cos }^2}x}}} \right]dx\] \[I = \int {{e^x}} \left[ {\frac{1}{{{{\cos }^2}x}} + \frac{{2\sin x\cancel{\cos x}}}{{2{{\cos }^\cancel{2}}x}}} \right]dx\] \[I = \int {{e^x}\left( {{{\sec }^2}x + \tan x} \right)} dx\] Let f(x) = tanx
f'(x) = sec2x \[\therefore I = \int {{e^x}} \left[ {f'\left( x \right) + f\left( x \right)} \right]dx\] \[I = {e^x}f\left( x \right) + c\] \[I = \tan x \cdot {e^x} + c\]

Question (22)

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

Solution

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

Question (23)

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

Solution

\[I = \int {{{\tan }^{ - 1}}\sqrt {\frac{{1 - x}}{{1 + x}}} } dx\] Let x = cos2θ
dx = -sin2θ 2 dθ
\[I = \int {{{\tan }^{ - 1}}\left( {\sqrt {\frac{{1 - \cos 2\theta }}{{1 + \cos 2\theta }}} } \right)} \times \left( { - 2{{\sin }^2}\theta } \right)d\theta \] \[I = - 2\int {{{\tan }^{ - 1}}\left( {\sqrt {\frac{{\cancel{2}{{\sin }^2}\theta }}{{\cancel{2}{{\cos }^2}\theta }}} } \right)} \cdot \sin 2\theta d\theta \] \[I = - 2\int {{{\tan }^{ - 1}}\left( {\tan \theta } \right)} \cdot \sin 2\theta d\theta \]
u = θ and v = sinθ and use integration by parts
\[I = - 2\int {\theta \sin 2\theta d\theta } \] \[I = - 2\left[ {\theta \int {\sin 2\theta d\theta - \int {\left( {\frac{d}{{d\theta }}\theta \int {\sin 2\theta d\theta } } \right)d\theta } } } \right]\] \[I = - \cancel {2}\theta \frac{{\left( { - \cos 2\theta } \right)}}{\cancel{2}} + 2\int {1 \cdot \left( {\frac{{ - \cos 2\theta }}{2}} \right)} d\theta \] \[I = \theta \cos 2\theta - \int {\cos 2\theta d\theta } \] \[I = \theta \cos 2\theta - \frac{{\sin 2\theta }}{2} + c\] \[I = \frac{1}{2}{\cos ^{ - 1}}x \cdot x - \frac{{\sqrt {1 - {x^2}} }}{2} + c\] \[I = \frac{x}{2}co{s^{ - 1}}x - \frac{{\sqrt {1 - {x^2}} }}{2} + c\]

Question (24)

\[\int {\sqrt {{x^2} + 1} \left[ {\frac{{\log \left( {{x^2} + 1} \right) - 2\log x}}{{{x^4}}}} \right]} dx\]

Solution

\[I = \int {\sqrt {{x^2} + 1} \left[ {\frac{{\log \left( {{x^2} + 1} \right) - 2\log x}}{{{x^4}}}} \right]} dx\] \[I = \int {\frac{{\sqrt {{x^2} + 1} }}{{{x^4}}}\log \left[ {\frac{{{x^2} + 1}}{{{x^2}}}} \right]dx} \] \[I = \int {\frac{{\sqrt {{x^2} + 1} }}{{{x^4}}}\log \left( {1 + \frac{1}{{{x^2}}}} \right)dx} \] \[I = \int {\frac{1}{{{x^3}}}\sqrt {\frac{{{x^2} + 1}}{{{x^2}}}} } \log \left( {1 + \frac{1}{{{x^2}}}} \right)dx\] \[I = \int {\frac{1}{{{x^3}}}\sqrt {1 + \frac{1}{{{x^2}}}} } \log \left( {1 + \frac{1}{{{x^2}}}} \right)dx\] \[Let \quad 1 + \frac{1}{{{x^2}}} = {t^2}\] \[ \therefore - \frac{\cancel{2}}{{{x^3}}}dx = \cancel{2}tdt\] \[\therefore \frac{1}{{{x^3}}}dx = - tdt\] \[I = \int {\left( { - tdt} \right)} \sqrt {{t^2}} \log {t^2}\] \[I = - 2\int {{t^2}\log tdt} \] \[I = - 2\left[ {\log t\int {{t^2}dt - \int {\left( {\frac{d}{{dt}}\log t\int {{t^2}dt} } \right)dt} } } \right]\] \[I = - 2\log t\cdot\frac{{{t^3}}}{3} + 2\int {\frac{1}{\cancel{t}} \cdot \frac{{{\cancel{t^3}t^2}}}{3}dt} \]\[I = - \frac{2}{3}{t^3}\log t + \frac{2}{3} \cdot \frac{{{t^3}}}{3} + c\] \[I = \frac{2}{3}{t^3}\left( {\log t - \frac{1}{3}} \right) + c\] \[I = \frac{2}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{3}{2}}}\left[ {\log {{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{1}{2}}} - \frac{1}{3}} \right] + c\] \[I = - \frac{2}{3}{t^3}\log t + \frac{2}{3} \cdot \frac{{{t^3}}}{3} + c\] \[I = -\frac{{2{t^3}}}{3}\left( {\log t - \frac{1}{3}} \right) + c\] \[I = -\frac{2}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{3}{2}}}\left[ {\log {{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{1}{2}}} - \frac{1}{3}} \right] + c\] \[I = -\frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{3}{2}}}\left[ {2\log {{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{1}{2}}} - \frac{2}{3}} \right] + c\] \[I = -\frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{3}{2}}}\left[ {\cancel{2} \cdot \frac{1}{\cancel{2}}\log \left( {1 + \frac{1}{{{x^2}}}} \right) - \frac{2}{3}} \right] + c\] \[I =- \frac{1}{3}{\left( {1 + \frac{1}{{{x^2}}}} \right)^{\frac{3}{2}}}\left[ {\log \left( {1 + \frac{1}{{{x^2}}}} \right) - \frac{2}{3}} \right] + c\]
Exercise 7.11 ⇐
⇒ Miscellaneous Exercise Question 25 to 44