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Q.

$\int \sec (3 x + 5) \tan (3 x + 5) . d x =$

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a

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

b

$\frac{- 1}{3} \sec (3 x + 5) + c$

c

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

d

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

answer is B.

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Detailed Solution

$\int \sec (3 x + 5) \tan (3 x + 5) . d x$

$3 x + 5 = t$

$3. d x = d t$

$d x = \frac{d t}{3}$

$= \int \sec t \tan t . \frac{d t}{3}$

$= \frac{1}{3} \int \sec t \tan t . d t$

$= \frac{1}{3} \sec t + c$

$= \frac{1}{3} \sec (3 x + 5) + c$

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