Question

sinA=1.141cosA=tanA
​

Answer 1

answer:

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Answer 2

ANSWER :–

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

• We should write this as –

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

▪︎ Now let's use substitution method –

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

• Now Differentiate with respect to 't' –

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

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

• So that –

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

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

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

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

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

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

• Now replace 't' –

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