The correct option is C 1/3tan(3x+5)+c ∫ sec^2(3x+5)dx let #160; #160; 3x+5=t 3 dx = dt dx= ^1/_3dt ∫ sec^2 t #160;·13· dt ...

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Integration by Substitution

∫ sec23x+5dx=

Question

$\int s e c^{2} (3 x + 5) d x =$

tan (3 x + 5) + c

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{tan}^{2} (3 x + 5) + c

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\frac{1}{3} tan (3 x + 5) + c

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\frac{1}{3} {tan}^{2} (3 x + 5) + c

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Solution

The correct option is C $\frac{1}{3} tan (3 x + 5) + c$
$\int s e c^{2} (3 x + 5) d x$
$l e t 3 x + 5 = t$
$3 d x = d t$
$d x =^{1} /_{3} d t$
$\int s e c^{2} t \cdot \frac{1}{3} \cdot d t$
$=dfrac13∫sec2t⋅dt$
$= \frac{1}{3} (t a n t) + c$
$= \frac{1}{3} (t a n (3 x + 5)) + c$
$\int s e c^{2} (3 x + 5) d x = \frac{1}{3} (t a n (3 x + 5)) + c$

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