The correct option is A 14tan (7 4x)+C ^2( 7 4x)dx let #160; #160;7 #160; 4x #160; = #160; t then #160; #160; 4dx # ...

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Integration by Substitution

∫2 7 - 4x dx

Question

$\int {sec}^{2} (7 - 4 x) d x$

\frac{1}{4} {tan}^{2} (7 - 4 x) + C

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\frac{- 1}{4} tan (7 - 4 x) + C

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\frac{1}{16} tan (7 - 4 x) + C

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\frac{1}{8} {tan}^{2} (7 - 4 x) + C

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Solution

The correct option is A $\frac{- 1}{4} tan (7 - 4 x) + C$
${sec}^{2} (7 - 4 x) d x$
$l e t 7 - 4x = t$
$then - 4dx = dt$
$\int {sec}^{2} t \frac{d t}{- 4}$
$= - \frac{1}{4} \int {sec}^{2} t d t$

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

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

Suggest Corrections

Similar questions

\frac{1}{\sqrt{4 x + 1}} ⎡ ⎣ {(\frac{1 + \sqrt{4 x + 1}}{2})}^{7} - {(\frac{1 - \sqrt{4 x + 1}}{2})}^{7} ⎤ ⎦

Factorise: $1 - 16 (x - 2 y)^{2}$ .

\int (x + 3) \sqrt{3 - 4 x - x^{2}} d x = \frac{- 1}{3} {(3 - 4 x - x^{2})}^{3 / 2} + \frac{(x + 2) \sqrt{3 - 4 x - x^{2}}}{2} + \frac{7}{2} {sin}^{- 1} \frac{(x + 2)}{\sqrt{7}} + C

a, b, c

x^{3} + 4 x + 1 = 0

\frac{1}{a + b} + \frac{1}{b + c} + \frac{1}{c + a} =

\int \frac{2^{x}}{\sqrt{1 - 4^{x}}}

= k {sin}^{- 1} 2^{x} + c

k =

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