Question

$\bold\pink{solve\:the\:above\:question}$

•useless answer/random letter answer not allowed.
•wrong answer also not allowed.​

Answer 1

$\sec ^{2} (7 - 4x)dx \\ let \: 7 - 4x = > t$

$then - 4dx = t$

$\sec^{2} t \frac{dt}{ - 4}$

$= - \frac{1}{4} \: \sec ^{2}t \: dt$

$= \: \frac{ - 1}{4} tant + c$

$= \: \frac{ - 1}{4} tan \: (7 - 4x) + c$

$\bold\purple{mark\: it\: brainliest}$

Answer 2

$\: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \boxed{{ \huge{ \bf{ S}}} \rm{OLUTION :-}}}$

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$: \longrightarrow \bold{ \blue{{ \: {sec}^{2}(7 - 4x) }}}$

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➨ Let 7 - 4x = m.

➨ Differentiate both sides by w, r, t, x.

$: \rightarrow \bold{ \: \dfrac{dm}{dx} = 0 - 4}$

$: \rightarrow \bold{ \dfrac{dm}{dx} = - 4}$

$: \rightarrow \bold{ dx = \dfrac{dm}{ - 4} }$

$: \rightarrow \bold{ dx = - \dfrac{dm}{4} }$

Now,

$: \rightarrow \bold{ { \huge{\int}} {sec}^{2} (7 - 4x) \: . \: dx}$

Since,

➨ 7 - 4x = t and,

➨ dx = - dm/4

On putting these values,

$: \rightarrow \bold{I = { \huge{ \int}} \: {sec}^{2}(m) \: . \: ( - \dfrac{dm}{4}) }$

Taking, - 1/4 as common,

$: \rightarrow \bold{I = - \dfrac{1}{4} \: { \huge{ \int}} \: {sec}^{2}(m) \: . \: dm}$

$: \rightarrow \bold {I = \dfrac{1}{4} \: tan(m) +C }$

since,

➨ m = 7 - 4x

$: \rightarrow \boxed{\bold{ \: \: I = \dfrac{1}{4} \: tan(7 - 4x) + C} \: \: }$

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@MissSolitary ✌️

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