$\int {cosec}^{2} (4 x + 9) d x$ is equal to
(where $C$ is the constant of integration)

\frac{- cot (4 x + 9)}{9} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{cot (4 x + 9)}{9} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{cot (4 x + 9)}{4} + C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{- cot (4 x + 9)}{4} + C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{- cot (4 x + 9)}{4} + C$
$\int {cosec}^{2} (4 x + 9) d x = - cot (4 x + 9) \times \frac{1}{4} + C$
$(∵ \int {cosec}^{2} (a x + b) d x = \frac{- cot (a x + b)}{a} + C, a \neq 0)$
$= \frac{- cot (4 x + 9)}{4} + C$

Suggest Corrections

Similar questions

\int \frac{8 x + 13}{\sqrt{4 x + 7}} d x

C

\int e^{{sin}^{2} x} sin x (cos x + {cos}^{3} x) d x

C

\int \frac{sin x}{sin x - cos x} d x

C

\int e^{x} sec x (1 + tan x) d x

c

\int \sqrt{sec x - 1} d x

C

Join BYJU'S Learning Program