Question

Integration it(jee Advance 2013)​

Answer 1

Question :–

$\\ { \bold{Solve: \int \frac{ \cos(x).dx }{( { \sin(x) )}^{3} \times {(1 + ( \sin(x) ) {}^{6}) }^{ \frac{2}{3} } } }}$

ANSWER :–

$\\ \implies { \bold{I = \int \frac{ \cos(x).dx }{( { \sin(x) )}^{3} \times {(1 + ( \sin(x) ) {}^{6}) }^{ \frac{2}{3} } } }}$

• Let's use substitution method –

$\\ { \huge{.}} \: \: \: { \bold{put \: \: \sin(x) = t \: - }}$

• Differentiate with respect to 'x' –

$\\ \: \implies { \bold{ \cos(x) \frac{dx}{dt} = 1 \: }}$

$\\ \: \implies { \bold{ \cos(x). dx = dt \: }}$

• So that –

$\\ \implies { \bold{I = \int \frac{ dt }{({t )}^{3} \times { [1 + (t ) ^{6}] }^{ \frac{2}{3} } } }}$

• We should write this as –

$\\ \implies { \bold{I = \int \frac{ dt }{({t )}^{3} \times( {t}^{6} ) {}^{ \frac{2}{3} } \times { [1 + (t ) ^{ - 6}] }^{ \frac{2}{3} } } }}$

$\\ \implies { \bold{I = \int \frac{ dt }{({t )}^{3} \times( {t}^{4} ) \times { [1 + (t ) ^{ - 6}] }^{ \frac{2}{3} } } }}$

$\\ \implies { \bold{I = \int \frac{ dt }{({t )}^{7} \times { [1 + (t ) ^{ - 6}] }^{ \frac{2}{3} } } }}$

$\\ \: \: { \huge{. }}\: \: \: { \bold{ Again \: \: put \: \: (1 + {t}^{ - 6} ) = p \: - }}$

• Now Differentiate with respect to 't' –

$\\ \implies { \bold{ - 6 {t}^{ - 7}. \frac{dt}{dp} = 1 \: }}$

$\\ \implies { \bold{ \dfrac{dt}{ {t}^{7} } = \frac{dp}{ - 6} \: }}$

• So that –

$\\ \implies { \bold{I = \int \frac{ dp }{( - 6) \times { (p)}^{ \frac{2}{3} } } }}$

$\\ \implies { \bold{I = \int \frac{{ (p)}^{ - \frac{2}{3} } .dp }{( - 6) } }}$

$\\ \implies { \bold{I = - \frac{1}{6} \int { (p)}^{ - \frac{2}{3} } .dp } }$

$\\ \implies { \bold{I = - \frac{1}{6} [ { \dfrac{ {p}^{ - \frac{2}{3} + 1} }{ - \frac{2}{3} + 1 } ] + c}}}$

$\\ \implies { \bold{I = - \frac{1}{6} [ { \dfrac{ {p}^{ \frac{1}{3}} }{ \frac{1}{3} } ] + c}}}$

$\\ \implies { \bold{I = - \frac{3}{6} ({ { {p}^{ \frac{1}{3}} } ) + c}}}$

$\\ \implies { \bold{I = - \frac{1}{2} ({ { {p}^{ \frac{1}{3}} } ) + c}}}$

• Now replace 'p' –

$\\ \implies \large { \pink{ \boxed{ \bold{I = - \frac{1}{2} { { {[1 + {( \sin(x)) }^{ - 6} ]}^{ \frac{1}{3}} } + c}}}}}$

$\\ \rule{220}{2}$

USED FORMULA :–

$\\ \: { \blue { \bold{(1) \: \: \dfrac{d( \sin(x) )}{dx} = \cos(x) }}}$

$\\ \: { \blue { \bold{(2) \: \: \dfrac{d( x )}{dx} = 1 }}}$

$\\ \: { \blue{ \bold{(3) \: \: \dfrac{d( x {}^{n} )}{dx} = n {x}^{n - 1} }}}$

$\\ \: { \blue{ \bold{(4) \: \: \int {x}^{n} .dx = \dfrac{ {x}^{n + 1} }{n + 1} }}}$

$\\ \rule{220}{2}$