How to integrate $\int {{e^{ – 3x}}} {\cos ^3}xdx$ ~~~~~[NCERT Exemp. Q. 42,Page 166 ]

$I = \mathop {\int {{e^{ – 3x}}} {{\cos }^3}xdx}\limits_{{\rm{I}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}} $


$ = {\cos ^3}x\int {{e^{ – 3x}}} dx – \int {\left( {\frac{d}{{dx}}{{\cos }^3}x\int {{e^{ – 3x}}} dx} \right)} dx$


$ = {\cos ^3}x \cdot \frac{{{e^{ – 3x}}}}{{ – 3}} – \int {\left( { – 3{{\cos }^2}x} \right)} \sin x \cdot \frac{{{e^{ – 3x}}}}{{ – 3}}dx$


$ = – \frac{1}{3}{\cos ^3}x{e^{ – 3x}} – \int {{{\cos }^2}} x\sin x{e^{ – 3x}}dx$


$ = – \frac{1}{3}{\cos ^3}x{e^{ – 3x}} – \int {\left( {1 – {{\sin }^2}x} \right)} \sin x{e^{ – 3x}}dx$


$ = – \frac{1}{3}{\cos ^3}x{e^{ – 3x}} – \int {\sin } x{e^{ – 3x}}dx + \mathop {\int {{{\sin }^3}} x{e^{ – 3x}}dx}\limits_{{\rm{I}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}} $


$ = – \frac{1}{3}{\cos ^3}x{e^{ – 3x}} – \int {\sin } x{e^{ – 3x}}dx + {\sin ^3}x \cdot \frac{{{e^{ – 3x}}}}{{ – 3}} – \int 3 {\sin ^2}x\cos x \cdot \frac{{{e^{ – 3x}}}}{{ – 3}}dx$


$ = – \frac{1}{3}{\cos ^3}x{e^{ – 3x}} – \int {\sin } x{e^{ – 3x}}dx – \frac{1}{3}{\sin ^3}x{e^{ – 3x}} + \int {\left( {1 – {{\cos }^2}x} \right)} \cos x{e^{ – 3x}}dx$


$I = – \frac{1}{3}{\cos ^3}x{e^{ – 3x}} – \mathop {\int {\sin } \,x\,{e^{ – 3x}}}\limits_{{\rm{I}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}} – \frac{1}{3}{\sin ^3}x{e^{ – 3x}} + \int {\cos } x{e^{ – 3x}}dx – \int {{{\cos }^3}} x{e^{ – 3x}}dx$

 

$2I = \frac{{{e^{ – 3x}}}}{3}\left[ {{{\cos }^3}x + {{\sin }^3}x} \right] – \left[ {\sin x \cdot \frac{{{e^{ – 3x}}}}{{ – 3}} – \int {\cos } x \cdot \frac{{{e^{ – 3x}}}}{{ – 3}}dx} \right] + \int {\cos } x{e^{ – 3x}}dx$

 

$2I = \frac{{{e^{ – 3x}}}}{{ – 3}}\left[ {{{\cos }^3}x + {{\sin }^3}x} \right] + \frac{1}{3}\sin x \cdot {e^{ – 3x}} – \frac{1}{3}\int {\cos } x \cdot {e^{ – 3x}}dx + \int {\cos } x{e^{ – 3x}}dx$


$2I = \frac{{{e^{ – 3x}}}}{{ – 3}}\left[ {{{\cos }^3}x + {{\sin }^3}x} \right] + \frac{1}{3}\sin x{e^{ – 3x}} + \frac{2}{3}\int {\cos } x{e^{ – 3x}}dx$


Now, let ${I_1} = \mathop {\int {\cos } x{e^{ – 3x}}dx}\limits_{{\rm{I}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{II}}} $

${I_1} = \cos x \cdot \frac{{{e^{ – 3x}}}}{{ – 3}} – \int {( – \sin x)} \cdot \frac{{{e^{ – 3x}}}}{{ – 3}}dx$

${I_1} = \frac{{ – 1}}{3}\cos x \cdot {e^{ – 3x}} – \frac{1}{3}\int {\sin } x \cdot {e^{ – 3x}}dx$


$ = – \frac{1}{3}\cos x \cdot {e^{ – 3x}} – \frac{1}{3}\left[ {\sin x \cdot \frac{{{e^{ – 3x}}}}{{ – 3}} – \int {\cos } x \cdot \frac{{{e^{ – 3x}}}}{{ – 3}}dx} \right]$


$ = – \frac{1}{3}\cos x \cdot {e^{ – 3x}} + \frac{1}{9}\sin x \cdot {e^{ – 3x}} – \frac{1}{9}\int {\cos } x \cdot {e^{ – 3x}}dx$


${I_1} + \frac{1}{9}{I_1} = – \frac{1}{3}{e^{ – 3x}} \cdot \cos x + \frac{1}{9}\sin x \cdot {e^{ – 3x}}$


$\left( {\frac{{10}}{9}} \right){I_1} = – \frac{1}{3}{e^{ – 3x}} \cdot \cos x + \frac{1}{9}\sin x \cdot {e^{ – 3x}}$


${I_1} = \frac{{ – 3}}{{10}}{e^{ – 3x}} \cdot \cos x + \frac{1}{{10}}{e^{ – 3x}}\sin x$


$2I = – \frac{1}{3}{e^{ – 3x}}\left[ {{{\sin }^3}x + {{\cos }^3}x} \right] + \frac{1}{3}\sin x \cdot {e^{ – 3x}} – \frac{3}{{10}}{e^{ – 3x}} \cdot \cos x + \frac{1}{{10}}{e^{ – 3x}} \cdot \sin x + C$


therefore,$I = – \frac{1}{6}{e^{ – 3x}}\left[ {{{\sin }^3}x + {{\cos }^3}x} \right] + \frac{{13}}{{30}}{e^{ – 3x}} \cdot \sin x – \frac{3}{{10}}{e^{ – 3x}} \cdot \cos x + C$


$ = \frac{{{e^{ – 3x}}}}{{24}}[\sin 3x – \cos 3x] + \frac{{3{e^{ – 3x}}}}{{40}}[\sin x – 3\cos x] + C$

 

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