Question

$\frac{x. {sec}^{2}( {x}^{2} )}{ {tan}^{3}( {x}^{2} )}$
Integration class 12th...

good answer will be marked as brainliest
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Answer 1

ANSWER :–

$\\ \implies { \bold{ I = \int \dfrac{x \: {\sec}^{2} ( {x}^{2} )}{ {tan}^{3}( {x}^{2} ) }.dx }}$

• We should write this as –

$\\ \implies { \bold{ I = \int \dfrac{x \: {\sec}^{2} ( {x}^{2} )}{[ {tan}( {x}^{2} )] ^{3}} .dx }}$

▪︎ Now let's use substitution method –

• Put $\: \: { \bold{ tan ({x}^{2} ) = t \: \: - }}$

• Now Differentiate with respect to 't' –

$\\ \: \implies { \bold{(2x) \sec ^{2} ({x}^{2} ). \dfrac{dx}{dt} = 1 \: \: }}$

$\\ \: \implies { \bold{x.\sec ^{2} ({x}^{2} ).dx = \dfrac{1}{2} dt \: \: }}$

• So that –

$\\ \implies { \bold{ I = \int \dfrac{dt}{2(t) ^{3}} }}$

$\\ \implies { \bold{ I = \int \dfrac{ {t}^{ - 3}. dt}{2} }}$

$\\ \implies{ \bold{ I = \dfrac{1}{2} \int {{t}^{ - 3}. dt} }}$

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

$\\ \implies{ \bold{ I = \dfrac{1}{2} [\dfrac{ {t}^{ - 2} }{ - 2} ] + c}}$

$\\ \implies{ \bold{ I = - \dfrac{1}{4 {t}^{2} } + c}}$

• Now replace 't' –

$\\ \implies \large{ \red{ \boxed{ \bold{ I = - \dfrac{1}{4 { [tan( {x}^{2} )] }^{2} } + c}}}}$

$\rule{210}{2}$

USED FORMULA :–

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

$\\ { \pink { \bold{(2) \: \: \dfrac{d [ tan(x) ]}{dx} = { \sec}^{2}(x) }}}$

$\\ { \pink{\bold{(3) \: \: \int {x}^{n} .dx = \dfrac{ {x}^{n + 1} }{n + 1} }}}$

$\rule{210}{2}$

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star \: {\sf{\green{ \int \: \frac{x \: {sec}^{2} ( {x)}^{2} }{(tan {x}^{2} )^{3} } }}}$

${\sf{\underline{\blue{Now,}}}}$

$\: \star \: \boxed{\sf{\orange{ put \: tan {x}^{2} = t }}}$

${ \underline{\bf{ diff .\: w.r.t.x}}}$

${\implies{\tt{ 2x \: {sec}^{2} ({x}^{2}) \frac{dx}{dt} = 1}}} \\ \\ {\implies{\tt{ x \: {sec}^{2} dx = \frac{1}{2} dt}}}$

${\tt{ I = \int \frac{1}{2 {t}^{3} } dt }} \\ \\{\tt{ I = \int \frac{ {t}^{ - 3} }{2 }dt }}$

$\star{\sf{\underline{\blue{integrate \: w.r.to.t,}}}}$

${\tt{ I = \frac{1}{2} \bigg[ \frac{ {t}^{ - 3 + 1} }{ - 3 + 1} \bigg] }}$

${\tt{ I = - \frac{1}{ 4} {t}^{ - 2} + c}}$

${\tt{ I = - \frac{1}{ 4 {t}^{ 2} } + c}}$

$\boxed{\tt{ \: where \: t = tan {x}^{2} }}$

hence

${\tt{ I = - \frac{1}{ 4(tan {x}^{2}) ^{2} } + c}}$

$\star \: {\sf{\green {\underline{ selection \: for \: proper \: substitution }}}}$

1. If the integrand contains the t-ratios of f(x) or logarithmic of f(x) or an exponential of f(x) in which the index is f(x),put f(x) = t
2. If the integrand is rational function of e^x ,put e^x=t