As we know that ∫f(ax+b)dx=F(ax+b)a+C, where a,b and C are constants. We have here, f(x)=^2( 3x+4), where a= 3, b=4 So, nbsp;∫^2( 3x+4)dx= tan( 3x+4)3+C n ...

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Byju's Answer

Standard XIII

Physics

Antiderivative

Evaluate ∫2-3...

Question

Evaluate $\int {sec}^{2} (- 3 x + 4) d x$

\frac{- tan (- 3 x + 4)}{3} + C

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\frac{tan (- 3 x + 4)}{4} + C

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\frac{tan (- 3 x + 4)}{3} + C

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\frac{- cot (- 3 x + 4)}{4} + C

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Solution

The correct option is A $\frac{- tan (- 3 x + 4)}{3} + C$
As we know that $\int f (a x + b) d x = \frac{F (a x + b)}{a} + C$ , where $a, b$ and $C$ are constants.
We have here,
$f (x) = {sec}^{2} (- 3 x + 4)$ , where $a = - 3, b = 4$
So, $\int {sec}^{2} (- 3 x + 4) d x = \frac{- tan (- 3 x + 4)}{3} + C$

Suggest Corrections

Evaluate ∫sec2(−3x+4)dx

The correct option is A −tan(−3x+4)3+CAs we know that ∫f(ax+b)dx=F(ax+b)a+C, where a,b and C are constants. We have here, f(x)=sec2(−3x+4), where a=−3,b=4 So, ∫sec2(−3x+4)dx=−tan(−3x+4)3+C

Evaluate $\int {sec}^{2} (- 3 x + 4) d x$